50999882VAKX_Electromagnetic Field Theory_Solution Manual

1

Vector Analysis

Solutions of Examples for Practice Example 1.6.3 Solution : The origin O (0, 0, 0) while P (3, – 3, – 2) hence the distance vector OP is, OP = ( 3 - 0) a x + ( -3 - 0) a y + ( -2 - 0) a z = 3 a x - 3 a y - 2 a z \

OP

=

2 2 ( 3) 2 + ( - 3) + ( - 2) = 4.6904

Hence the unit vector along the direction OP is, 3ax - 3ay - 2az OP a OP = = 4.6904 | OP| = 0.6396 a x – 0.6396 a y – 0.4264 a z Example 1.6.4 Solution : The starting point is A and terminating point is B. Now \

A = 2 a x + 2 a y + a z and B = 3 a x - 4 a y + 2 a z AB = B - A = ( 3 - 2) a x + ( -4 - 2) a y + ( 2 - 1) a z

\ AB = a x - 6 a y + a z This is the vector directed from A to B. Now

AB = (1) 2 + ( - 6) + (1) 2 = 6.1644 2

Thus unit vector directed from A to B is, ax - 6ay +az AB a AB = = 6.1644 AB

= 0.1622 a x – 0.9733 a y + 0.1622 a z

It can be cross checked that magnitude of this unit vector is unity i.e.

( 0.1622) 2 + ( - 0.9733) 2 + ( 0.1622) 2 = 1. Example 1.7.3

Kept this unsolved example for student's practice.

Example 1.7.4 Solution : Consider the upper surface area, the normal to which is a z . So the differential surface area normal to z direction is r df dr. Consider the Fig. 1.7.8. (1 - 1) TM

TECHNICAL PUBLICATIONS - An up thrust for knowledge

Electromagnetic Field Theory

\

2p R

S1 =

ò ò

1-2

r df dr =

0 0

2p

ò 0

Vector Analysis

R

ér2 ù R2 2p 2 ê 2 ú df = 2 ´ [f] 0 = p R ë û0

The bottom surface area S 2 is same as S 1 i.e. p R 2 . For remaining surface area consider the differential surface area normal to r direction which is r df dz. L 2p

S3 =

ò ò 0

0

L 2p

=

r df dz but r = R is constant

ò ò 0

2p

R df dz = R [f] 0 [z] L0 = 2 p RL

0

Total surface area = S 1 + S 2 + S 3 = p R 2 + p R 2 + 2 p RL = 2 p R ( R + L). Example 1.8.3


Example 1.8.4


Example 1.8.5 Solution : Consider the spherical shell of radius a hence r = a is constant. Consider differential surface area normal to r direction which is radially outward. dS r = r 2 sin q dq df = a 2 sin q dq df ... as r = a But f is varying between 0 to a while for spherical shell q varies from 0 to p. \

a p

Sr = a 2 ò

ò

0 0

=

a2

p

a

sin q dq df = a 2 [ - cos q ] 0 [f] 0

[

]

× - cos p - ( - cos 0) a = 2 a 2 a

So area of the region is 2

a 2 a.

If a = 2 p, the area of the region becomes 4 p a 2 , as the shell becomes complete sphere of radius a when f varies from 0 to 2 p . Example 1.10.4


Example 1.10.5


Example 1.10.6 Solution : The dot product is, A · B = A x Bx + A y By + A z Bz = ( 2 ´ 3) + ( -5)(5) + ( -4)( 2) = -27 As A · B is negative, it is expected that the angle between the two is greater than 90°. |A| =

( 2) 2 + ( -5) 2 + ( -4) 2 = 45 and |B| = ( 3) 2 + (5) 2 + ( 2) 2 = 38

TM



1-3

Vector Analysis

ì A· B ü -1 ì -27 ü = q = cos -1 í ý 130.762º ý = cos í î|A| |B|þ î 45 38 þ

\

Example 1.10.7 Solution : A = 5 a x and B = 4 a x + B y a y , q AB = 45° Now A · B = A x B x + A y B y + A z B z = (5 ´ 4) + ( 0) + ( 0) = 20 But A · B = |A| |B|cos q AB

(5) 2 ´ ( 4) 2 + (B y ) ´ cos 45° i.e. 2

\

20 =

\

B 2y = 16

Now Still \ \

i.e.

16 + B 2y = 5.6568

By = ± 4

B = 4 a x + By a y + Bz a z A · B = 20 20 =

(5) 2 ´ ( 4) 2 + (B y ) + ( B z ) 2 ´ cos 45° 2

16 + B 2y + B 2z = 5.6568

i.e.

B 2y + B 2z = 16

This is the required relation between B y and B z . Example 1.11.4 Solution : Note that the unit vector normal to the plane containing the vectors A and B is the unit vector in the direction of cross product of A and B. ax ay az 4 -5 3 -5 3 4 Now 3 4 -5 = a x -ay +az A´B = 2 4 -6 4 -6 2 -6 2 4 = 26 a x + 18 a y + 30 a z \

aN =

A´B A´B

=

26 a x - 18 a y + 30 a z

( 26) 2 + (18) 2 + ( 30) 2

= 0.5964 a x + 0.4129 a y + 0.6882 a z

This is the unit vector normal to the plane containing A and B. Example 1.11.5 Solution : The perpendicular vector to the plane containing A and B is given by their cross product. ar A´B = Ar Br \

af Af Bf

az ar af Az = 2 p 3p Bz -1 2

az 7p 1 =a + 3a f + 4pa z 2 r -2

a n = Unit vector in the direction A ´ B

TM



1-4

Vector Analysis

7p a + 3a f + 4pa z -3.5 pa r + 3a f + 4pa z 2 r = - 0.648 a r + 0.1768 a f + 0.74 a z = 16.9651 2 æ 7 p ö + ( 3) 2 + ( 4p) 2 ç ÷ è 2 ø

-

=


Example 1.11.6 Example 1.12.3

Solution : The scalar triple product is, 2 0 -1 A · B ´ C = 2 -1 2 = 14 2 -3 1

(

)

The vector triple product is, A ´ BĆ = B A· C - C A· B

(

)

(

)

(

)

A · C = ( 2)( 2) + ( 0) ( - 3) + ( - 1) (1) = 3

\ A · B = ( 2)( 2) + ( 0) ( - 1) + ( - 1) ( 2) = 2 A ´ B ´ C = 3 B -2 C = 3 2 a x - a y + 2 a z - 2 2 a x - 3 a y + a z

(

[

)

] [

] = 2a x + 3a y + 4 a z

Example 1.13.9 Solution : A = y a x - xa y + za z \ A r = A a r = ya x a r - xa y a r + z a z a r = y cosf - x sinf (Refer Table 1.13.1)

× × × A = A ×a = y a ×a - x a ×a = - y sinf - x cosf A = A ×a = y a ×a - x a ×a Now x = r cosf, y = r sinf, z = z f

f

z

z

x

x

f

z

×

f

y

y

z

× + z a ×a

… (1)

+ za z a f

z

… (2) z

= z

… (3)

1

Using in equations (1), (2), (3) we get, A = [r sinf cosf - r sinf cosf] a r + [- r sin 2 f - r cos 2 f] a f + z a z = 0 - r (sin 2 f + cos 2 f) a f + z a z = - r a f + z a z Example 1.13.10 Solution : \

B = Br =

10 a + r cos q a q + a f r r 10 , r

B q = r cos q,

Bf = 1

é Bx ù é sin q cos f cos q cos f - sin fù ê B ú = ê sin q sin f cos q sin f cos f ú ê yú ê ú - sin q 0 úû êë B z úû êë cos q

TM

… In spherical é 10 ù ê r ú ê r cos q ú ê ú êë 1 úû



1-5

Vector Analysis

\

Bx =

10 sin q cos f + r cos 2 q cos f - sin f r

… (1)

\

By =

10 sin q sinf + r cos 2 q sin f + cos f r

… (2)

\

Bz =

10 cos q - r sin q cos q r

… (3)

But

\

r =

sinq =

x2 + y 2 + z2 , x2 + y 2 x2 + y 2 + z2

z

cosq =

,

x2 + y 2 + z2 y

sin f =

y x x

cos f =

,

x2 + y 2

tan f =

,

x2 + y 2

Using equations (1), (2) and (3), B in cartesian system is : where, B = Bx a x + By a y + Bz a z Bx =

By =

Bz =

10 x 2

2

x +y +z

2

10 y 2

2

x +y +z

2

10 z x2 + y 2 + z2

+

+

-

xz 2 (x 2

+

y2)

(x 2

+

y2

+

z2)

yz 2 (x 2 + y 2 ) (x 2 + y 2 + z 2 )

-

+

y x2

… (4)

+ y2 x

… (5)

x2 + y 2

z x2 + y 2

… (6)

x2 + y 2 + z2

At (– 3, 4, 0), x = – 3, y = 4, z = 0 \ B = - 2 ax + ay

… In cartesian

For transforming spherical to cylindrical use, é Br ù é sin q êB ú = ê 0 ê fú ê êë B z úû êë cos q \

Now And \

cos q 0ù é B r ù ê ú 0 1ú êBq ú ú - sin q 0úû êë B f úû

Br = sin q B r + cos q B q =

10 sin q + r cos 2 q, B f = B f = 1 r

B z = cos q B r - sin q B q =

10 cos q - r sin q cos q r

r = r sin q, z = r cosq, f = f, r = r 2 + z 2 , q = tan - 1 tanq =

r hence sin q = z

r r2 + z2

, cos q =

z r2 + z2

B = Br a r + B f a f + B z a z where, TM


r z


1-6

10 r

Br =

2

r +z

2

+

z2 r2 + z2

Vector Analysis

B f = 1, B z =

,

10 z 2

r +z

2

-

rz r2 + z2

p p and z = – 2 At given point æç5, , - 2ö÷ , r = 5, f = 2 è 2 ø \

10 ´ 5

Br =

5 2 + ( - 2) 2 10 ´ ( - 2)

Bz =

52

+ (-

2) 2

+

-

( - 2) 2

= 2.467 ,

5 2 + ( - 2) 2 5 ´ ( - 2)

Bf = 1

= 1.167

5 2 + ( - 2) 2

B = 2.467 a r + a f + 1.167 a z

\

… In cylindrical

Example 1.13.11 Solution : Q ( – 2, 6, 3) i.e. x = – 2, y = 6, z = 3 x2 + y 2 =

1) In cylindrical, r = f = tan –1

y = tan x

–1

4 + 36 = 6.3245,

6 = – 71 . 565° but as x is negative f must be in second quadrant –2

hence add 180°. \ f = – 71.565° + 180° = 108.435°, \ In spherical,

z=3

Q ( 6.3245, 108. 435°, 3) r =

x2 + y 2 + z2 =

… Cylindrical

4 + 36 + 9 = 7

z 3 q = cos –1 é ù = cos –1 é ù = 64.623°, f = 108.435° as above êë 7 úû êë r úû \

Q (7 , 64. 623° , 108. 435° )

… Spherical

2) B in spherical co-ordinates Br = B · a r = y (a x · a r ) + (x + z) (a y · a r ) = y sin q cos f + ( x + z) sin q sin f But

x = r sin q cos f, y = r sin q sin f, z = r cos q Br = 2r sin 2 q sin f cos f + r sin q cos q sin f B q = B · a q = y(a x · a q ) + ( x + z) (a y · a q ) = y cos q cos f + ( x + z) cos q sin f

\

B q = 2r sin q cos q sin f cos f + r cos 2 q sin f B f = B · a f = y(a x · a f ) + (x + z) (a y · a f ) = y (– sin f) + ( x + z) cos f TM



1-7

Vector Analysis

\

B f = – r sin q sin 2 f + r sin q cos 2 f + r cos q cos f

\

B f = + r sin q cos 2 q + r cos q cos f

\

… cos 2 f – sin 2 f = cos 2 q

B = Br a r + Bq a q + Bf a f

At Q, r = 7, q = 64.623°, f = 108.435° hence B at Q is, B = – 0.8571 a r – 0.4064 a q – 6 a f

… At point Q

Example 1.13.12 Solution : Refer example 1.13.11 for P in cylindrical system. \ P (6.3245, 108.43°, 3)

… Cylindrical

To convert A to cylindrical, Ar = A · a r = y(a x · a r ) + (x + z) (a y · a r ) = y cos f + ( x + z) sin f and x = r cos f , y = r sin f, z = z \

Ar = 2r sin f cos f + z sin f = r sin 2 f + z sin f A f = A · a f = y (a x · a f ) + ( x + z) (a y · a f ) = y (– sin f) + ( x + z) (cos f) … cos 2 f – sin 2 f = cos 2 f

= r cos 2 f + z cos f \

A = [r sin 2 f + z sin f] a r + [r cos 2 f + z cos f] a f + 0 a z

\ At P,

A = – 0.9485 a r – 6 a f

Example 1.15.3 Solution : From given A, A x = 2 xy, A y = z, A z = yz 2 ¶Ax ¶Ay ¶Az + + Ñ A = div A = ¶x ¶y ¶z

×

=

¶ ¶ ¶ [2 xy ]+ ¶ y [z] + ¶ z ¶x

[yz ] = 2y + 0 + 2zy = 2y + 2yz 2

At P (2, – 1, 3), x = 2, y = – 1, z = 3

×

\ Ñ A = (2) ( - 1) + ( 2)( -1)( 3) = - 8 Example 1.15.4 Solution : Given A in cylindrical system, 1 ¶ 1 ¶Af ¶Az div A = r A r )+ + \ ( r ¶r r ¶f ¶z where \

A r = r z sin f, div A =

[

A f = 3 r z 2 cos f,

]

1 ¶ 1 ¶ r 2 z sin f + r ¶r r ¶f

[3 r z TM

2

Az = 0

]

cos f + 0



= At point P,

1-8

1 1 × z sin f × 2r + × 3r z 2 [ - sin f] = 2z sin f - 3 z 2 sin f r r p , 2

f=

r = 5,

z=1

div A = 2 ´ 1 ´ sin

\

Vector Analysis

p p - 3 ´ 1 ´ sin = – 1 at P. 2 2

Example 1.16.3 Solution : The outward flux is given by, f =

×

Surface S1

ò F dS over a closed surface S S

z=1

r=4

The cylindrical surface is shown in the Fig. 1.1. The total surface is made up of, 1. Top surface S1 for which z = 1, r varies from 0 to 4 and f varies from 0 to 2p .

Surface S2

2. Lateral surface for which z varies from 0 to 1, f from 0 to 2p and r = 4.

z=0

3. Bottom surface S3 for which z = 0, r varies from 0 to 4 and f varies from 0 to 2p . For S1, dS = r dr df az

x

For S2,

dS = r dz df ar

For S3,

dS = r dr df (– az)

\

×dS = ò F× dS = ò F× d S = òF

S1

S3 S2

Surface S3

Fig. 1.1

× + z sin f a ) ×[r dr df ( -a + z sin f a ) ×( r dz df a )

ò (r

2

cos 2 fa r + z sin f a f ) ( r dr df a z ) = 0

ò (r

2

cos 2 fa r

ò (r

2

cos 2 fa r

S1

S3 S2

1

=

y

2p

ò òr

2

f

f

z )]

=0

r

×

cos 2 fr dz df

= (4)3

1

ò

2p

z = 0 f= 0

\

òF S

×dS =

1

2p

0

f= 0

2 ò dz cos f df = 64 ò dz

= 64 × [z] 10 ´

×

… a r a z = 1, a f a r = 0 r = 4

z = 0 f= 0

ò

1 + cos 2f df 2

1 ïì 2 p é sin 2f ù 2 p ïü = 64p ´ [f] + + 2 íï 0 êë 2 úû 0 ýï î þ

0 + 64p + 0 = 64p

TM



1-9

Vector Analysis

Let us verify divergence theorem which states that,

òF S

×dS = ò (Ñ× F) dv ×

Ñ F =

1 ¶ 1 ¶ ( r ´ r 2 cos 2 f) + ( z sin f) + 0 r ¶r r ¶f

=

z cos f cos 2 f z 2 ´ 3r 2 + ( + cos f) = 3 r cos f + r r r

×

v

1 ¶ 1 ¶ Ff ¶ Fz (r Fr) + + r ¶r r ¶f ¶z

=

ò (Ñ F) dv =

\

where dv = r dr df dz

v

1

2p

4

ò ò ò

z = 0 f= 0 r = 0

z cos f ö æ 2 ç 3r cos f + r ÷ r dr df dz è ø 4

2p

é3r3 ù = ò ò ê cos 2 f + z cos f r ú df dz 3 úû 0 z = 0 f = 0 êë 1

2p

1

=

ò

z= 0

= Thus

ì 3 é 1 + cos 2f ù ü ò íî4 ëê 2 ûú + 4z cos fýþ df dz = f= 0

1

1

z= 0

z= 0

ìï é sin 2f ù 2 p 2p + 4z [sin f] 0 32 f + ò íï êë 2 ûú 0 z= 0 î 1

üï ý dz þï

ò { 32 ´ [2p + 0] + 4z[0]} dz = ò 64 p dz = 64p

òF S

×dS = ò (Ñ× F) dv and divergence theorem is verified. v

Example 1.16.4 Solution : Using divergence theorem

ò

A · dS =

S

ò

( Ñ · A) dv

v

To evaulate A · d S it is necessary to consider all six faces of the cube. Let us find dS for each surface, for a cube shown in the Fig. 1.2. –ax Back

ax Front

(a) Cube

–ay

Left

az Top

Right

ay

–az Bottom

(b) Directions of dS

x = Constant planes (back and front)

y = Constant planes (sides)

Fig. 1.2 TM


z = Constant planes (top and bottom)


1 - 10

Vector Analysis

dS = dy dz a x

1. Front surface (x = 1), dS = dy dz, direction = a x ,

2. Back surface (x = 0), dS = dy dz, direction = - a x , dS = - dy dz a x 3. Right side (y = 1),

dS = dx dz, direction = a y ,

4. Left side (y = 0),

dS = dx dz, direction = - a y , dS = - dx dz a y

5. Top side (z = 1),

dS = dx dy, direction = a z ,

dS = dx dz a y dS = dx dy a z

6. Bottom side (z = 0), dS = dx dy, direction = - a z , dS = - dx dy a z A = xy 2 a x + y 3 a y + y 2 z a z For front, A · d S = xy 2 dy dz (x = 1) = y 2 dy dz For back, A · d S = - xy 2 dy dz (x = 0) = 0 For right, A · d S = y 3 dx dz (y = 1) = dx dz For left,

A · d S = - y 3 dx dz (y = 0) = 0

For top,

A · d S = y 2 z dx dy (z = 1) = y 2 dx dy

For bottom, A · d S = - y 2 z dx dy (z = 0) = 0

ò

\

1

1

ò

A · dS =

ò

z= 0 y= 0

S

1

ò

y 2 dy dz +

1

ò

z= 0 x= 0

1

1

ò

dx dz +

1

ò

y 2 dx dy

y= 0 x= 0 1

éy 3 ù éy 3 ù 1 5 1 1 1 1 1 = ê ú [z] 0 + [x] 0 [z] 0 + ê ú [x] 0 = + 1 + = 3 3 3 3 3 êë úû 0 êë úû 0 Ñ· A =

\

ò

¶A y ¶A z ¶A x + + = y 2 + 3y 2 + y 2 = 5y 2 ¶x ¶y ¶z 1

ò

( Ñ · A) dv =

1

1

ò

ò

1

5y 2 dx

z= 0 y= 0 x= 0

v

éy 3 ù 1 5 1 1 dy dz = 5ê ú [x] 0 [z] 0 = 5 ´ ´ 1 ´ 1 = 3 3 êë 3 úû 0

Thus divergence theorem is verified. Example 1.16.5 Solution : A = 2xy a x + y 2 a y + 4 yz a z Using Divergence theorem,

ò

S

A · dS =

ò

( Ñ · A) dv

v

¶A x ¶A y ¶A z + + = 2y + 2y + 4y = 8y Ñ· A = ¶x ¶y ¶z \

ò

S

A · dS =

ò

... dv = dx dy dz

(8y)dv

v

TM



1 - 11

1

1

ò

=

1

1

ò

Vector Analysis

ò

z= 0 y= 0 x= 0

éy2 ù 1 1 1 8y dx dy dz = 8 ê ú [x] 0 [z] 0 = 8 ´ = 4 2 2 êë úû 0

Example 1.16.6 Solution : The divergence theorem states that

×

A dS =

ò

é ù A dS = ê ò + ò + ò ú A dS ê ú ë side top bottom û

S

Now

×

ò (Ñ A) dv

ò

S

v

×

×

Consider dS normal to a r direction which is for the side surface. \

×

\

dS = r df dz a r

A dS =

( 30 e- r a r - 2z a z )× r df dz a r

×

= 30 r e - r ( a r a r ) df dz = 30 r e - r df dz

ò

\

×

2p

side

5

ò ò

A dS =

30 r e - r df dz

with

r=2

f = 0z= 0 2p

= 30 ´ 2 ´ e -2 ´ [f] 0 ´ [z] 0 = 255.1 5

The dS on top has direction a z hence for top surface, dS = r dr df a z \ \

× ( 30 e ò A ×dS = ò ò

-r

A dS =

2p

2

ar - 2 zaz

)× r dr df a z

- 2 z r dr df with z = 5

az

f = 0r= 0

top

×

... ( a z a z = 1)

= – 2 z r dr df

z=5

2

ér2 ù 2p = -2 ´ 5 ´ ê ú ´ [f]0 = - 40 p 2 ë û0

dS

While dS for bottom has direction - a z hence for bottom surface, dS = r dr df ( - a z )

×

\ A dS =

dS

( 30 e- r a r - 2 z a z )× r dr df ( - a z )

×

= 2 z r dr df ... ( a z a z = 1) But z = 0 for the bottom surface, as shown in the Fig. 1.3. \ ò A dS = 255.1 – 40 p + 0 S

×

= 129.4363

ar

z = 0 –az

Fig. 1.3 TM



1 - 12

Vector Analysis

This is the left hand side of divergence theorem. Now evaluate

×

ò (Ñ A) dv

v

×

Ñ A=

1 ¶ 1 ¶A f ¶ A z r A r )+ + ( r ¶r r ¶f ¶z

and A r = 30 e - r , A f = 0, A z = -2 z

×

\

Ñ A = =

\

ò(

×

)

1 ¶ r ¶r 1 r

{30 r (- e ) + 30 e -r

5

2p

2

ò ò ò

Ñ A dv =

v

( 30 r e- r ) + 0 + ¶¶z ( -2 z)

z = 0f = 0r = 0 5

p

-r

}

30

(1) + ( - 2) = - 30 e - r + r e - r - 2

æ -30 e - r + 30 e - r - 2ö r dr df dz ç ÷ r è ø

2

-r -r ò ò ò (- 30 r e + 30 e - 2r )

=

dr df dz

z = 0f = 0r = 0

é e - r ù é r 2 ùïü é e- r ù é e- r ù ïì 2p 5 = í - 30 r ê - ò ( - 30) ê dr + 30 ê ú ú ú - ê 2 2 úý [z]0 [f]0 1 1 1 ïî û ë û ë û ë ë ûïþ Obtained using integration by parts. =

[ 30 r e

-r

+ 30 e - r - 30 e - r - r 2

]

2 0

[5][ 2 p ] =

[60 e

-2

- 22

] [10 p] = 129.437

This is same as obtained from the left hand side. Example 1.16.7 in spherical Solution : The given D is co-ordinates. The volume enclosed is shown in the Fig. 1.4.

Dr

z ar

According to divergence theorem, ò D dS = ò Ñ D dv S

×

(

v

×

)

The given D has only radial component as given. 5 r2 while D q = D f = 0. Hence D r = 4 Hence D has a value only on the surface r = 4 m. Consider dS normal to the a r r 2 sin q dq df

45º

r=4m aq

y

Dq

direction i.e. x

Fig. 1.4 TM



1 - 13

Vector Analysis

dS = r 2 sin q dq df a r

\

×

\

D dS =

ò

\

S

×

æ

2p p / 4

ò ò

D dS =

f = 0q = 0

=

2

(r 2 sin q dq df) ççè 5r4

ö 5 4 ÷÷ = r sin q dq df ø 4

5 4 r sin q dq df 4

... r = 4m é - cos p - - cos 0 ù 2 p ( )ú [ ] êë 4 û

5 4 5 p/4 2p r [ - cos q ]0 [f]0 = ( 4) 4 4 4

= 588.896 C

×

... ( a r a r = 1)

×

To evaluate right hand side, find Ñ D.

×

Ñ D = =

¶ Df 1 ¶ 1 ¶ 1 r 2 Dr + sin q D q ) + ( r sin q ¶ q r sin q ¶ f r2 ¶ r

(

1 r

2

¶ ¶r

é 2 êë r

)

æ 5 r 2 öù + 0 + 0 = 5 ¶ ç ÷ è 4 ø úû 4 r2 ¶ r

(r 4 ) = 4 5r 2 (4 r 3 ) = 5 r

In spherical co-ordinates, dv = r 2 sin q dr dq df \

ò(

×

)

Ñ D dv =

v

2p p / 4 4

ò ò ò

f = 0 q = 0r = 0

(

)

(5 r ) r 2 sin q dr dq df

4

ér4 ù 44 é p p/4 2p = 5 ê ú [ - cos q ]0 [f]0 = 5 ´ ´ - cos - ( - cos 0) ù ´ 2 p êë úû 4 4 4 ë û0 = 588.896 C Example 1.16.8 Solution : The volume bounded by the given planes is a cube. To evaluate total charge use Gauss's law. Q = ò D dS

×

S

×

But to evaluate D d S, it is necessary to consider all six faces of the cube. Let us find dS for each surface. 1) Front surface (x = 2), dS = dy dz, direction = a x , dS = dy dz a x 2) Back surface ( x = 1), dS = dy dz, direction = - a x , dS = – dy dz a x 3) Right side (y = 3), dS = dx dz, direction = a y , dS = dx dz a y 4) Left side (y = 2), dS = dx dz, direction = - a y , dS = – dx dz a y 5) Top side (z = 4), dS = dx dy, direction = a z , dS = dx dy a z 6) Bottom side (z = 3), dS = dx dy, direction = - a z , dS = – dx dy a z TM



1 - 14

Vector Analysis

Key Point Remember that though the co-ordinates of x, y and z are positive, the directions of

unit vectors are with respect to region bounded by the planes, as shown in the Fig. 1.5 (b). –ax Back

az Top

–ay

ax Front

(a) Cube

y = constant planes (sides)

2

For left

ò

S

×

4

3

ò

D dS =

ò

4x dy dz +

z= 3 y= 2

4

x = 1

... a x

y = 3

... a y

y = 2

... a y

z = 4

... a z

z = 3

... a z

4

ò

2

ò

3y 2 dx dz +

z= 3 x= 1

Right, y = 3 +

3

2

ò ò

ò

- 4x dy dz

z= 3 y= 2

Front, x = 2 +

3

ò

2z 3 dx dz +

Back, x = 1 4

2

ò

ò

-3y 2 dx dz

z= 3 x= 1

Left, y = 2 3

2

ò

ò

- 2z 3 dx dz

y = 2x = 1

y= 2 x= 1

Top, z = 4

Bottom, z = 3

= ( 4)( 2)[ y ]2 [z]43 - ( 4)(1)[ y ]2 [z]43 + ( 3)( 3) 2 [x]21 [z]43 3

3

TM

× ×a ×a ×a ×a ×a

... a x a x = 1

3

For bottom

z = constant planes (top and bottom)

x = 2

3

For top

\

× D × d S = – 4x dy dz, D × d S = 3 y dx dz, D × d S = - 3 y dx dz, D × d S = 2 z dx dy, D × d S = - 2 z dx dy, 2

For right

ay

Fig. 1.5

D d S = 4x dy dz,

For back

Right


x = constant planes (back and front)

For front

Left

–az Bottom


x

=1

y

=1

y

=1

z

=1

z

=1


1 - 15

Vector Analysis

= - ( 3)( 2) 2 [x]21 [z]43 + ( 2)( 4) 3 [x]21 [ y ]2 - ( 2)( 3) 3 [x]21 [ y ]2 3

3

= 8 - 4 + 27 - 12 + 128 - 54 = 93 C This is the total charge enclosed. Let us verify by divergence theorem. ¶ Dx ¶ Dy ¶ Dz + + = 4 + 6 y + 6 z2 Ñ D = ¶x ¶y ¶z

×

ò(

×

)

Ñ D dv =

v

4

3

2

ò ò ò

z = 3y = 2 x = 1 4

=

3

ò ò

z = 3y = 2

(4 + 6 y + 6 z 2 ) dx dy dz

... Integrate w.r.t. x

(4 + 6y + 6z 2 )[x]21 dy dz

... Integrate w.r.t. y

3

é ù y2 = ò ê 4y + 6 + 6 z 2 y ú dz 2 úû 2 z = 3 êë 4

=

=

4

(

4

(4 + 15 + 6 )

)

é 4 3 - 2 + 6 3 2 - 2 2 + 6 z 2 3 - 2 ù dz ) 2 ( )ú êë ( û z= 3

ò ò

z= 3

z2

4

é z3 ù dz = ê19 z + 6 3 ú ë û3

(

= 19 ( 4 - 3) + 2 4 3 - 3 3

) = 93 C

Thus divergence theorem is verified. Example 1.17.6 Solution : t = x 2 y + ez Ñt = Gradient of t = At P, \

and

P(1, 5, – 2)

¶t ¶t ¶t a + a + a = 2xy a x + x 2 a y + e z a z ¶x x ¶y y ¶z z

x = 1, y = 5, z = – 2 Ñt = 10 a x + a y + e - 2 a z

Example 1.17.7 –z

Solution : 1) V = e sin 2x cosh y ¶V ¶V ¶V ax + ay + a \ ÑV = ¶x ¶y ¶z z ¶V ¶ –z –z = [e sin 2x cosh y] = 2 e cos 2x cosh y ¶x ¶x

TM



1 - 16

Vector Analysis

¶V ¶ = [e– z sin 2x cosh y] = e– z sin 2x sinh y ¶y ¶y ¶V ¶ –z –z = [e sin 2x cosh y] = – e sin 2x cosh y ¶z ¶z Ñ V = 2e - z cos 2x cosh y a x + e - x sin 2x sinh y a y - e - z sin 2x cosh y a z

\

U = r 2 z cos 2f

2) \

ÑU =

¶U ¶U 1 ¶U a + a + a ¶r r r ¶f f ¶z z

= 2 r z cos 2f a r +

1 ´ r 2 z( - sin 2f ´ 2) a f + r 2 cos 2f a z r

= 2 r z cos 2f a r - 2rz sin 2f a f + r 2 cos 2f a z 3)

W = 10 r sin2 q cos f ÑW =

¶W 1 ¶W 1 ¶W a + a + a ¶r r r ¶q q r sin q ¶f f

1 1 2 = 10 sin q cos f a r + ´ 10 r cos f ´ 2 sin q cos q a q + ´ 10r sin 2 q( - sin f) a f r r sin q 2

= 10 sin q cos f a r + 10 cos f sin 2q a q - 10 sin q sin f a f Example 1.17.8 Solution : Gradient of f in spherical system is, ¶f 1 ¶f 1 ¶f ar + aq + af Ñf = r ¶q r sin q ¶f ¶r Hence verify the answer as,

[

]

Ñf = 100 r 3 sinq cosf + 5 sinf a r + é 25r 3 cosq cosf – êë

5 cosf ù 2 sinq ù a q + é –25r 3 sinf + af êë sinq úû r úû

Example 1.18.7 Solution : A = x 2 a x + y 2 a y + y 2 a z Ñ· A =

¶A y ¶(x 2 ) ¶(y 2 ) ¶(y 2 ) ¶A x ¶A z + + = + + ¶x ¶y ¶z ¶x ¶y ¶z

= 2x + 2y + 0 = 2 (x + y)

Ñ Á =

ax ¶ ¶x x2

ay ¶ ¶y y2

... Divergence

az ¶ = a x [2y - 0] +a y [0 - 0] + a z [0 - 0] = 2y a x ¶z y2

TM


… Curl


1 - 17

Vector Analysis

Example 1.18.8 Solution : 1)

2 P = x yz a x + xz a z 2

Px = x yz, Py = 0, Pz = xz ¶Py ù é ¶P é ¶Py ¶Px ù é ¶Px ¶Pz ù ay + ê Ñ´P = ê z úax + ê úa ú ¶y ûú z ¶z ûú ¶x û ë ¶z êë ¶y êë ¶x

\

= [0 – 0] a x + [x2y – z] a y + [0 – x2z] a z = (x2y – z) a y – x2z a z Q = r sin f a r + r 2 z a f + z cos f a z

2)

Qr = r sin f, Qf = r 2 z, Qz = z cos f é ¶Q r ¶Q z ù é 1 ¶Q z ¶Q f ù é 1 ¶(rQ f ) 1 ¶Q r ar + ê af + ê Ñ´Q = ê ú ú r ¶f ¶z û ¶r û ë ¶z ë r ¶f ë r ¶r

\

ù úaz û

1 1 1 = é ´ ( - z sin f) - r 2 ù a r + [0 - 0] a f + é 3r 2 z - ´ r cos fù a z úû êë r úû êë r r æ - z sin f ö = ç - r 2 ÷ a r + (3rz - cos f)a z r è ø T =

3)

Tr = Ñ´T =

1 r sin q

1 r2 1 r2

cos q, Tq = r sin q cos f, Tf = cos q

é ¶Tf sin q ¶Tq ù 1 é ¶ ( rTq ) ¶Tr ù 1 é 1 ¶ Tr ¶( rTf ) ù ar + ê ú a q + r ê ¶r - ¶q ú a f ú ê r sin q r ¶q ¶f ¶ ¶ f û ë û û ë ë

=

¶ ( r cos q) ù ù 1 é ¶(sin q cos q) 1é 1 - r sin q( - sin f) ú a r + ê ( 0) úaq r sin q êë r sin q ¶r ¶q û û ë +

Þ

cos q a r + r sin q cos f a q + cos q a f

2 ù 1 é ¶( r sin q cos f) 1 ( - sin q) ú a f ê rê ¶r r2 úû ë

¶ (sin q cos q) 1 ¶( 2 sin q cos q) 1 ¶(sin 2q) 1 = = = ´ 2 cos 2q = cos 2q 2 2 2 ¶q ¶q ¶q

\Ñ ´ T =

sin q ù 1 1 1é [cos 2q + r sin q sin f] a r + [- cos q] a q + ê 2r sin q cos f + úaf r sin q r rë r2 û

cos 2q cos q sin q ù é =é + sin fù a r a q + ê 2 sin q cos f + úa f êë r sin q úû r ë r3 û

TM



1 - 18

Vector Analysis

Example 1.18.9 Solution :

E = yz a x + xz a y + xy a z ¶ Ey ¶ Ex ¶ Ez + + = 0+0+0 = 0 Ñ· E = ¶x ¶y ¶z

As

Ñ · E = 0,

ax ay ¶ ¶ ¶x ¶y yz xz

Ñ´ E = As Ñ ´ E = 0,

field E is solenoidal in nature. az ¶ ¶z xy

= a x ( x - x) - a y ( y - y) + a z ( z - z) = 0

field E is irrotational in nature.

Example 1.18.10 Solution : 1) A = yza x + 4xya y + ya z ax ay az ¶ ¶ ¶ = a x + ya y + [4y – z] a z \ Ñ Á = ¶x ¶y ¶z yz 4xy y 2) B = rz sin fa r + 3rz 2 cos fa f ar ¶ ¶r rz sinf

1 \ Ñ´B= r

az ra f ¶ ¶ = – 6rz cosfa r + r sinfa f + ( 6z 2 cosf – z cosf) a z ¶f ¶z 3r 2 z 2 cosf 0

Example 1.19.2

òA

Solution :

… Cylindrical

L

×dL = ò + ò + ò + ò ab

bc

cd

da

DL = dr a r + rdf a f + dz a z

… Cylindrical system

A r = r cos f, Af = sin f, Az = 0

×

For path ab, the direction is a f hence A d L = (sin f) ( r df) \

×

ò A dL =

ab

30 º

ò sin f r df

with r = 2

f = 60 º

30 º

= [- cos f] 60 º (2) = 2 [– 0.866 + 0.5] = – 0.732

×

For path bc, the direction is a r hence A d L = r cos f dr \

òA

bc

×dL =

5

ò r cos f d r

with f = 30º

r=2

TM


… From given A


1 - 19

Vector Analysis

5

ér 2 ù 25 - 4 ù = + 9.093 = (cos 30º) ê ú = 0.866 é ê 2 ë 2 úû êë úû 2

×

For path cd, the direction is a f hence A d L = (sin f) ( r df)

×

ò A dL =

\

cd

60 º

ò sin f

r df

with r = 5

f = 30 º

60 º

= 5 [- cos f] 30 º = 5 [– cos 60º + cos 30º] = 1.83

×

For path da, the direction is a r hence A d L = r cos f dr

òA

\

da

×dL =

2

ò r cos f d r

with f = 60º

r =5

2

ér 2 ù 4 - 25 ù = – 5.25 = (cos 60º) ê ú = 0.5 é ê ë 2 úû êë 2 úû 5

òA

\

×dL =

– 0.732 + 9.093 + 1.83 – 5.25 = 4.941

For Stoke's theorem, find Ñ ´ A in cylindrical system. é ¶A r ¶A z ù é 1 ¶(rA f ) 1 ¶Ar ù é 1 ¶A z ¶A f ù ar + ê af + ê a ÑÁ = ê ú ú r ¶f úû z r ¶f ¶z û ¶r û ë ¶z ë r ¶r ë 1 1 æ (1 + r) ö = [0 – 0] a r + [0 – 0] a f + é ´ sin f - ´ r( - sin f)]ù a z = sin f ç ÷a êë r úû r è r ø z dS = r d r d f a z

\

[Ñ ´ A]

× dS

ò ( Ñ ´ A)

…As surface is in x-y plane

1 +rö = sin f æç ÷ r d r df = sin f (1 + r) dr df è r ø 60 º

ò

=

5

ò

5

sin f (1 + r) dr df =

f = 30 º r = 2

S

60 º [- cos f] 30 º

é r2 ù êr + 2 ú úû 2 êë

25 4 = [– cos 60º + cos 30º] é5 + - 2 - ù = 4.941 2 2 ûú ëê

Thus

òA

L

×dL =

ò (Ñ ´ A) S

× dS hence Stoke's theorem is verified.

Example 1.19.3 Solution : A = r cos f a r + r 2 a z In cylindrical system, TM



1 - 20

Vector Analysis z

¶A f ù é ¶A r é 1 ¶A z ¶A z ù Ñ Á = ê ú a r + ê ¶z - ¶r ú a f r ¶f ¶ z ë û ë û

az

é 1 ¶ (rA f ) 1 ¶A r ù +ê ú a r ¶f ú z ¶r êë r û Given : A r = r cos f,

1 1

A z = r2

A f = 0,

y

r

x

Fig. 1.6

Ñ Á =

1 a r + [0 - 2r] a f + é 0 - ( - sin f) r ù a z = - 2 r a f + sin f a z úû êë r

[0 - 0]

As the surface is in x - y plane, dS = r dr d f a z \

ò (Ñ ´ A) · dS

=

S

òò ( sin f a z ) · [r dr df] a z p 2

=

1

… af · az = 0

p 2

ò ò

f= 0r = 0

sin fr dr df = [ - cos f] 0

1

ér 2 ù é1 ù 1 ê 2 ú = 0 + - 0 - ( - 1) ê 2 ú = 2 ë û êë úû 0

[

]

Example 1.19.4 z

Solution : The path L is shown in the Fig. 1.7.

ò F· dL

=

ò F· dL

=

ò

+

AB

AB

ò

ò[(x

2

ò

+

BC

+

CD

A

D

]

)

+ y 2 i – 2xy j · dx i

AB

ù éx3 = ò (x 2 + y 2 )dx = ê + y 2 xú 3 ûx = ë x = –a

x=–a y

y=0

y=b

a

a

=

ò F· dL

DA

B

x=+a

C

x

–a

Fig. 1.7

2a 3 3

… y = 0 for AB

ò F· dL =

BC

ò[(

]

)

x 2 + y 2 i – 2xy j · dy j =

BC

b

ò –2xy dy

y= 0

b

é 2xy 2 ù = ê– ú 2 ú êë ûy =

ò F· dL =

CD

= – ab 2

… x = + a for BC

0

2 2 ò[(x + y )i – 2xy j]· dx i =

CD

–a

ù éx3 = ê + y 2 xú 3 ûx = ë

+a

–a

ò (x

2

+ y 2 ) dx

x= a

ù é a3 a3 = ê– – ab 2 – – ab 2 ú 3 3 û ë TM



DA

Vector Analysis

2a 3 – 2ab 2 3

= –

ò F· dL =

1 - 21

… y = + b for CD

2 2 ò[(x + y )i – 2xy j]· dy j =

DA

0

ò –2xy dy

y= b

0

é 2xy 2 ù = ê– ú 2 ú êë ûy =

ò F· dL

\

=

S

j ¶ ¶y -2xy

… L.H.S.

k ¶ = (– 2y – 2y) k = – 4yk ¶z 0

ò – 4y k · ( dxdy) k = S

b

ò

+a

ò – 4y dxdy

y = 0 x = –a

b

b

=

… x = – a for DA

b

2a 3 2a 3 – ab 2 – – 2ab 2 – ab 2 = – 4ab 2 3 3

i ¶ Ñ´F = ¶x 2 x + y2

ò ( Ñ ´ F) · dS =

= – ab 2

4x]+a x = –a

ò [–

ydy =

–4[x]+a –a

y= 0

éy2 ù 2 ê ú = – 4ab 2 êë úû 0 … Stoke's theorem is verified

Example 1.19.5

z

Solution : According theorem, ò H · dL = ò L

to

Stoke's

(Ñ ´ H) · dS

y = –1

y

Let us evaluate left hand side. The integral to be evaluated on a perimeter of a closed path shown in b the Fig. 1.8. The direction is a-b-c-d-a x such that normal to it is positive a z according to right hand rule. ò H · dL = ò + ò + ò + ò H · dL ab

ò

ab

y=1

S

H · dL =

5

ò

x= 2

bc

cd

a

x=5 c

Fig. 1.8

da

(6xy a x - 3 y 2 a y )· dx a x TM

x=2

d



1 - 22 5

5

é x2 ù 6 xy dx = 6y ê ú 2 ë û2

ò

=

x= 2

=

Vector Analysis

6y [25 - 4] = 63 y 2

Now y = – 1 for path ab,

ò

H · dL = 63 ( -1) = -63

ab

Similarly ò H · dL = bc

ò

1

ò

- 3y 2 dy =

y= - 1

ò

x= 5

cd

[ ]

1

-1

= - [1 - ( -1)] = – 2

2

2

H · dL =

- 3y3 =- y3 3

é x2 ù 6y 6 xy dx = 6 ê ú ( y) = [4 - 25] = - 63y 2 2 ë û5

But y = 1 for path cd hence

ò

H · dL = – 63

cd

ò

[ ]

ò

H · dL = -63 - 2 - 63 + 2 = - 126 A

- 3y 2 dy = - y 3

y=1

Now evaluate right hand side. ax ay ¶ ¶ Ñ´H = ¶x ¶y 6 xy -3y 2 \ò

S

-1

ò

da

\

-1

H · dL =

1

[

]

= - ( -1) 3 - (1) 3 = - [-1 - 1] = + 2

az ¶ ¶z 0

= a x [0 - 0] + a y [0 - 0] + a z [0 - 6x] = - 6x a z Ñ ´ H · d S = ò ( -6 x a z ) · ( dx dy a z )

(

)

S

dS = dx dy a z normal to direction a z \ò

(Ñ ´ H) · dS

1

ò

=

y = -1

S

5

5

é x2 ù 1 6x dx dy = 6 ê 2 ú [y] - 1 ò ë û2 x= 2

6 [25 - 4][1 - ( -1)] = -3 ´ 21 ´ 2 = - 126 A 2 Thus both the sides are same, hence Stoke's theorem is verified. = -

Example 1.19.6 Solution : According to Stoke's theorem,

ò

L

H · dL =

ò (Ñ ´ H) · dS S

TM



1 - 23

Vector Analysis

In spherical system, dL = dr a r + r dq a q + r sin q df a f The closed path forming its perimeter is composed of three circular arcs. The first path 1 is r = 3, f = 0, 0 £ q £ 90° as shown in the Fig. 1.9. The second path 2 is r = 3, q = 90°, Path 1 f = 0º

Path 3 f = 90º dL= rdq aq

dL= rdq aq

r=3

O

r=3 y

r=3

x

dL= rsin qdf af

Path 2 f = 90º

Fig. 1.9

0 £ f £ 90° while the path 3 is r = 3, f = 90°, 0 £ q £ 90°. For all the three arcs r = 3 m. Let us evaluate

\

ò

H · dL =

ò

ò

H · dL over these three paths. H q r dq +

Path1

ò

H f r sin q df +

Path2

ò

H q r dq

Path3

Now, H r = 0, H q = 0, H f = 10 sin q Thus only second line integtal exists. \

ò

H · dL =

p/2

ò

... Given H

10 sin q r sin q df = 10 r sin 2 q [f]p0 / 2

... Path 2

f= 0

= 10 ´ 3 ´ [ sin 90° ] ´ 2

p = 47.1238 A 2

... r = 3 m, q = 90° for path 2

Now evaluate second side of Stoke's theorem. Ñ ´ H in spherical co-ordinates is, =

1 é ¶ (rH q ) ¶ H r ù 1 é ¶ H f sin q ¶ H q ù 1 é 1 ¶ H r ¶ (rH f ) ù ar + ê aq + ê a ú ú ê rë ¶r r sin q ë r ë sin q ¶ f ¶ q úû f ¶q ¶f û ¶r û

As H r = 0, H q = 0, H f = 10 sin q Ñ´H =

ù ¶ (r10 sin q) ù 1 é ¶ sin 2 q 1é 1 10 - 0ú a r + ê 0 ê ú a q + r [0 - 0] a f r sin q rë ¶ ¶ q r û û ë

TM



(Ñ ´ H) · dS (

=

1 1 10 ´ 2 sin q cos q ]a r + [ - 10 sin q ]a q r sin q [ r

=

10 10 sin 2 q a r sin q a q r sin q r

dS = r 2 sin q dq df a r

while \

1 - 24

)

=

\ ò Ñ ´ H · dS = S

Vector Analysis

... As given in a r direction

10 sin 2 q r 2 sin q dq df r sin q p/ 2 p/ 2

ò

ò

... a r · a r = 1

10 r sin 2q dq df

f= 0 q= 0

p/ 2 - cos p ( - cos 0) p é - cos 2 q ù = 10 r ê [ f ] p0 / 2 = 10 ´ r ´ éê 2 - 2 ùú éëê 2 ùûú ... r = 3 m ú 2 ë û0 ë û p 1 1 ... Thus Stoke's theorem is verified. = 10 ´ 3 ´ é + ù ´ = 47.1238 A êë 2 2 úû 2

Example 1.20.3


qqq

TM


2

Coulomb's Law and Electric Field Intensity Solutions of Selected Examples for Practice

Example 2.2.6 z

Solution : The force exerted on 0.3 mC charge is, Q 1Q 2 a 12 F2 = 4pe 0 R 212 a 12 =

Q2 (1, 2, 3) 0.3 mC Q1 (2, 0, 5)

R12

a12

– 0.1 mC

R12

R12 = (1 - 2) a x + ( 2 - 0) a y + ( 3 - 5) a z

x

= - ax + 2 ay - 2 az and \

Fig. 2.1

( -1) 2 + ( 2) 2 + ( -2) 2 = 3

R12 = F2 =

y

(0.3 ´ 10 -3 )(-0.1 ´ 10 -3 ) [a 4p ´ 8.854 ´ 10 -12 ´ ( 3) 2

12

]

é- ax + 2 ay - 2 az ù = – 29.9591 ê ú 3 û ë = – 9.986 a x + 19.9727 a y – 19.9727 a z N \

2

2

( -9.986) + (19.9727 ) + ( -19.9737 )

F2 =

z

2 C

Q

R1Q

= 24.4613 N

R2Q

Q1 A

Example 2.2.7

Q2

Solution : The charges are shown in the Fig. 2.2. The position vectors of the points A, B and C are, A = - 4ay + 3 az

B –4

–3

O

B = ay +az x

C = - 3 ay + 4 az

Fig. 2.2 (2 - 1) TM


y 1


2-2

\

R1Q = C - A = a y + a z

and

R2Q = C - B = - 4 a y + 3 a z

\

R1Q =

(1) 2 + (1) 2 = 2 And

R2Q =

( -4) 2 + ( 3) 2 = 5

\

F1 = Force on Q due to Q 1 =

and

F2 = Force on Q due to Q 2 =

\

Ft = F1 + F2 =

Coulomb's Law and Electric Field Intensity

Q Q1 4 p e 0 R 21Q

a 1Q

QQ 2

a 2Q

4 p e 0 R 22Q

é ù Q Q ê Q1 a 1Q + 2 a 2Q ú 4 p e0 ê R2 ú R 22Q ë 1Q û ù öú ÷ øú úû

=

é -9 Q ê 2 ´ 10 æç a y + a z ö÷ + Q 2 æç - 4 a y + 3 a z 2 è 4 p e0 ê 5 2 ø (5) 2 è 2 êë

=

Q Q é ù 7.071 ´ 10 -10 a y + a z + 2 - 4 a y + 3 a z ú 125 4 p e 0 êë û

( )

(

)

(

)

\ Total z component of Ft is, =

3 Q2 ù Q é 7.071 ´ 10 -10 + a ê 125 úû z 4 p e0 ë

To have this component zero, 3 Q2 = 0 as Q is test charge and cannot be zero. 7.071 ´ 10 -10 + 125 \

Q2 = -

7.071 ´ 10 -10 ´ 125 = – 29.462 nC 3 y

Example 2.2.8

C

Solution : Let the side of equilateral triangle is d and is placed in x-y plane as shown in the Fig. 2.3. l(AB) = l(BC) = l(AC) = d 2

l(CD) = \ A (0, 0, 0),

d d 2 - æç ö÷ = è2ø

d

d P

3d 2 A

B (d, 0, 0),

D

d 2

æ d 3d ö Cç , , 0÷ è2 2 ø

d

z

Fig. 2.3 TM


d 2

B

x


2-3


1 rd of height of 3 perpendicular drawn from any one corner to opposite side, from the side on which Key Point

For equilateral triangle, the centroid is at a distance of

perpendicular is drawn. \

1 l(CD) 3

l(DP) =

i.e.

l(DP) =

d 1 3d = = 0.2886d ´ 3 2 2 3

d \ Co-ordinates of centroid P æç , 0 . 2886 d, 0ö÷ è2 ø The charge at each corner is +Q. Let charge at P is QP. Then net force Ft on charge at A due to all other charges is, Ft = FB + FC + FP =

a BA

Ft

4p e 0 R 2BA

a BA +

QQ 4p e 0 R 2CA

-

a CA +

=

4p e 0 R 2PA

a PA

d a - 0 . 2886 da y 2 x 0 . 5773 d

é d 3d Q2 ê- a x - 2 a x - 2 a y = + 4pe 0 ê d 2 d2 ´ d ê ë =

Q QP

d 3d - ax ay RCA 2 2 = a CA = d |RCA |

- da x RBA = = = – ax , d |RBA |

a PA =

\

QQ

Q2 4pe 0

ù ú Q QP ú + 4pe 0 ú û

[- a x - 0.5 a x - 0.866 a y ] + 4pe d2

é - d a - 0 . 2886 d a ù yú ê 2 x ê (0.5773 d) 2 (0.5773 d) ú êë úû

Q QP

Q 4pe 0 d 2

0d

2

[- 2 .5987 a x - 1 .5 a y ]

{[ - 1.5 Q - 2.5987 Q P ] a x + [- 0.866 Q - 1 .5 Q p ] a y }

For keeping all charges in equilibrium, Ft = 0 \ \

– 1.5 Q – 2.5987 QP = 0 QP = – 0.5773 Q

Thus charge at centroid P must be negative and 0.5773 times the charge Q. Example 2.2.9 Solution : The square is kept in x-y plane with origin as one of its corners, as shown in the Fig. 2.4. The diagonals AC = BD = 8 m Let AD = DC = BC = AB = l m TM



2-4

Coulomb's Law and Electric Field Intensity z

\ l2 + l2 = 82 \ l = 5.656 mz

Q2 = 150 mC

l (AC) = 8 m l (BD) = 8 m

P

Hence the co-ordinates of various points are, A (0, 0, 0), B (0, 5.656, 0), C (5.656, 5.656, 0), D (5.656, 0, 0)

B

A

The point E is centroid hence E (2.828, 2.828, 0). The point P is 3 m above the centre E hence the co-ordinates at P are (2.828, 2.828, 3).

C x

To find force on charge at P which is Q 2 = 150 mC due to charges at A, B, C and D of Q 1 = 30 mC each. \ FP = FA + FB + FC + FD FA =

FC =

Q1 Q2 2 4 p e 0 RAP

Q1 Q2 4 p e 0 R 2CP

Q1 Q2

a AP =

2 4 p e 0 RAP

a CP =

Q1 Q2 4 p e 0 R 2CP

At A, B, C, D Q1 = 30 mC

E

D

y

Fig. 2.4

Q1 Q2 Q1 Q2 RAP RBP , FB = a BP = 2 2 | RAP| 4 p e 0 R BP 4 p e 0 R BP | RBP| Q1 Q2 Q1 Q2 RCP RDP , FD = a DP = 2 2 | RCP| 4 p e 0 R DP 4 p e 0 R DP | RDP|

RAP = (2.828 - 0) a x + (2.828 - 0) a y + (3 - 0) a z = 2.828 a x + 2.828 a y + 3 a z RBP = (2.828 - 0) a x + (2.828 - 5.656) a y + (3 - 0) a z = 2.828 a x - 2.828 a y + 3 a z RCP = (2.828 - 5.656) a x + (2.828 - 5.656) a y + (3 - 0) a z = - 2.828 a x - 2.828 a y + 3 a z RDP = (2.828 - 5.656) a x + (2.828 - 0) a y + (3 - 0) a z = - 2.828 a x + 2.828 a y + 3 a z | RAP| =| RBP| = | RCP| = | RDP| = \ FA + FB + FC + FD = \ FP =

Q1 Q2 4 p e 0 (5) 3

30 ´ 10 - 6 ´ 150 ´ 10 - 6 4 p ´ 8.854 ´ 10 - 12 ´ 5 3

(2.828) 2 + (2.828) 2 + 3 2 = 5

[RAP + RBP + RCP + RDP ] [12 a z ] = 3.8827 a z N z

Example 2.2.10 Solution. : The arrangement of charges is shown in the Fig. 2.5. The charge at P is test charge i.e. QP = 1C. Q 1Q P ... Force due to Q1 a R1 F1P = 4pe 0 R 21

P (–1, 1, 0) 0

R1 = [( -1) - (1)]a x + [1 - 2] a y + ( 0 - 0) a z = – 2 ax – ay

y R1

Q2(2, 0, 0)

\ R1 = R1 = 2 2 + 1 2 = 5

R2

Q1(1, 2, 0)

x

Fig. 2.5 TM



\ F1P =

=

Q1

( 5)

4pe 0

Q1 4pe 0 5 5 F2P

2

2-5


é -2 a x - a y ù ê ú 5 ë û

(-2a x - a y ) Q2 QP

=

4pe 0 R 22

... Force due to Q2

a R2

R2 = [(–1) – (2)] a x + [1 – 0] a y + [0 – 0] a z = –3 a x + a y R2 = R2 = ( 3) 2 + (1) 2 = 10 \

F2P =

\

Q2 4pe 0

(

10

)

2

é -3 a x + a y ù Q2 ê ú = 10 ë û 4pe 0 10 10

[-3a x + a y ]

FP = F1P + F2P =

3Q 2 ö Q2 ö ü æ - Q1 1 ì æ -2Q 1 + ÷ ax + ç ÷ayý íç 4pe 0 î è 5 5 è 5 5 10 10 ø þ 10 10 ø

i) If x component of FP must be zero then, -2Q 1 5 5

-

3Q 2 10 10

= 0

i.e. Q1 = – 0.5303 Q2

ii) If y component of FP must be zero then, - Q1 5 5

+

Q2 10 10

= 0

i.e. Q1 = 0.3535 Q2

Example 2.2.11 Solution : The charges are shown in the Fig. 2.6. Q 1Q 2 a AB F21 = 2 4pe 0 RAB RAB = (2 - 0) a x + (3 - 0) a y + (6 - 0) a z

z

RAB Q2 = 4.9 mC

= + 2a x + 3a y + 6a z |RAB| = a AB = \

F12 =

B (2, 3, 6) Q1 = 0.7 mC

4 + 9 + 36 = 7

A (0, 0, 0)

x

+ 2a x + 3a y + 6a z RAB = 7 |RAB| 0.7 ´ 10 - 3 ´ 4.9 ´ 10 - 6 é 2 a x + 3 a y + 6 a z ù ú ê 7 4 p ´ 8.854 ´ 10 - 12 ´ 7 2 ë û TM


Fig. 2.6

y


2-6


= 0.1797 a x + 0.2696 a y + 0.5392 a z N |F12| =

(0.1797) 2 + (0.2696) 2 + (0.5392) 2 = 0.6291 N

… Magnitude

Example 2.2.12

z P4(0, 0, 2 2 )

Solution : The charges are shown in the Fig. 2.7. Consider the P4 (0, 0, 2.828).

charge

at

R3

i.e.

P4 (0, 0, 2 2)

Let us find force on charge at P4 due to all the charges at P1 , P2 and P3 . 1. Force due to charge at P1 . Q1 Q4 a R1 F1 = 4pe 0 R 21 =

R2

P3(–1, – 3, 0)

P2(–1, 3, 0) y

R1 P1(2, 0, 0) x

Fig. 2.7

Q 1 Q 4 é R1 ù ê ú 4p e 0 R 21 ë|R1|û

… Q 1 = Q 4 = 1 mC

R1 = (0 - 2) a x + (0 - 0) a y + (2.828 - 0) a z = - 2a x + 2.828 a z \

F1

=

é - 2 a + 2.828 a x z ê 2 2 ´ [2 2 + 2.828 2 ] êë 2 + 2.828

1 ´ 10 - 3 ´ 1 ´ 10 -3 4p ´ 8.854 ´ 10 - 12

= - 432.423 a x + 611.446 a z 2. Force due to charge at P2 . Q2Q4 Q2Q4 a R2 = F2 = 2 4pe 0 R 2 4pe 0 R 22

ù ú úû

N é R2 ù ú ê ë|R2|û

R2 = [(0) - ( -1)] a x + [0 - 3]a y + [2 2 - 0] a z = a x - 1.732 a y + 2.828 a z \ \

|R2| = F2

=

1 2 + 1.732 2 + 2.828 2 =

12 = 3.4641

é a x - 1.732 a y + 2.828 a z ù ú ê 12 ´ 12 ë û

1 ´ 10 - 3 ´ 1 ´ 10 - 3 4p ´ 8.854 ´ 10 - 12

= 216.211 a x - 374.489 a y + 611.446 a z N 3. Force due to charge at P3 Q 3Q4 Q 3 Q 4 é R3 ù a R3 = F3 = ú ê 2 4p e 0 R 3 4p e 0 R 23 ë| R3|û R3 = [0 - ( - 1)]a x + [0 - ( - 3)] a y + [2 2 - 0] a z = a x + 3 a y + 2 2 a z \

|R3| =

1 2 + ( 3) 2 + (2 2) 2 =

12 = 3.4641

TM



\

F3 =

2-7


éa x + 3 a y + 2 2 a z ù ê ú 12 ´ 12 êë úû

1 ´ 10 -3 ´ 1 ´ 10 - 3 4p ´ 8.854 ´ 10 - 12

= 216.211 a x + 374.489 a y + 611.538 a z N \ The total force on charge at P4 is, F = F1 + F2 + F3 = 3 ´ 611.446 a z = 1834.338 a z N i.e. |F| = 1834.338 N The magnitude of force on each charge remains same as above due to symmetrical distribution of charges. y

Example 2.2.13

P 0.25 mC

Solution : The arrangement is shown in the Fig. 2.8.

2

2

10 – 5 = 8.66 cm

10 cm

The co-ordinates of the vertices of triangle are, Point O ® (0, 0, 0) Point Q ® (0.1, 0, 0)

Q

O 0.25 mC

Point P ® (0.05, 0.0866, 0).

5 cm

Let us find the force on P due to the charges at O and Q. Q 1Q 2 Q 1Q 2 ROP \ a OP = ´ F1 = 2 2 ROP| | 4 p e 0 R OP 4 p e 0 R OP \ ROP = 0.05 a x + 0.0866 a y , | ROP| = \

F1 =

F2 =

0.25 ´ 10 - 6 ´ 0.25 ´ 10 - 6 4 p ´ 8.854 ´ 10 Q 1Q 2 4pe 0 R 2QP

- 12

a QP =

´

0.1 2

´

Q 1Q 2 4 p e 0 R 2QP

5 cm R d 10 cm

0.25 mC

x

Fig. 2.8

(0.05) 2 + (0.0866) 2 = 0.1

[0.05 a x + 0.0866 a y ] 0.1

= 0.02808 a x + 0.0486 a y N

R ´ QP | RQP|

RQP = (0.05 - 0.1) a x + 0.0866 a y = - 0.05 a x + 0.0866 a y |RQP| =

(0.05) 2 + (0.0866) 2 = 0.1 0.25 ´ 10 - 6 ´ 0.25 ´ 10 - 6

\

F2 =

\

F = F1 + F2 = 0.09729 a y N

\

4p ´ 8.854 ´ 10 - 12 ´ 0.1 2

´

[- 0.05 a x + 0.0866 a y ] 0.1

= - 0.02808 a x + 0.0486 a y N … Direction a y

|F| = 0.09729 N

… Magnitude

TM



2-8


Example 2.3.6

a R1 = \

R1

( 2)

E2 =

Q2 4 p e 0 R 22 =

R2

2

=

´

ax + az 2

R2

ax + az

P (0, 0, 1) Q 1 R1 A (–1, 0, 0) y

2

Q2

1 = [a x + a z ] 2

B (1, 0, 0)

x

Fig. 2.9

Q R2

( 0 - 1) a x + (1 - 0) a z R2

4 p e0

( 2)

4 p e0

+ (1 - 0) a z

R1

8 p e0

R2

a R1

4 p e 0 R 21

[ 0 - ( -1)] a x

=

4 p e0

a R2 = \

R1

E1 =

E2 =

Q1

E1 =

Solution :

2

-ax + az

=

2

1 éa x + a z ù = [ -ax + az êë ú 2 2 û

]

E = E 1 + E 2 = 0.3535 a x + 1.0606 a z V m

\

Example 2.3.7

z

Solution : The various points and charges are shown in the Fig. 2.10. Q2

The position vectors of points A, B and P are, A = 2ax , B = - 2ax

aBP

P = ax + 2ay + 2az EA is field at P due to Q 1 , and will act along a AP . EB is field at P due to Q 2 and will act along a BP . Q1 Q1 P-A a AP = ´ \ EA = x 2 2 P-A 4 p e 0 RAP 4 p e 0 RAP \ EB =

Q2

a BP =

4 p e 0 R 2BP

Q2 4 p e 0 R 2BP

´

aAP

RBP

O

Q1 A(2, 0, 0)

RAP

é Q Q 1 P-A ê 1 + 2 2 4 p e0 ê R P - A R2 ë AP BP

é 1 ê = 4 p e0 ê êë

P-B P-B

[

P-B ù ú P-B ú û

1 -ax + 2ay + 2az

( 9)

2

]

(1) 2 + ( 2) 2 + ( 2) 2

+

[

(

17 TM

)

2

]

ù ú 2 2 2 ú ( 3) + ( 2) + ( 2) úû

Q2 3a x + 2a y + 2a z

P(1, 2, 2)

Fig. 2.10

\ E at P = EA + EB =

B(–2, 0, 0)


y


=

2-9

[


é-a + 2a + 2a Q2 3a x + 2a y + 2a z y z 1 ê x + 4 p e0 ê 27 70.0927 ë

]ùú ú û

The y component of E must be zero. 2 Q2 2 2 70.0927 = 0 i.e. + ´ = - 2.596 C Q2 = 27 70.0927 27 2 This is the required charge Q 2 to be placed at ( - 2, 0, 0) which will make y component of E zero at point P.

\

Example 2.3.8 Solution : The arrangement is shown in the Fig. 2.11. 3l 2

Let AB = BC = CA = l. So CP = æl A(0, 0, 0), B (l, 0, 0), C ç , è2

3l ö , 0÷ 2 ø

y

and R3

æl ö P ç , 0, 0÷ è2 ø

l

E1 =

Now

E2 =

3Q 4pe 0 ( 0 . 5l) 2

3Q

ax =

E2 =

And

E3 =

l 2

P

R2

B

l 2

1Q

l

z

Fig. 2.11

1.078 ´ 10 11 Q l2

ax

æl ö a R2, R2 = ç - l÷ a x + 0 a y + 0 a z = – 0.5 l a x è2 ø 4p e 0 R2 Q2

|R2| = 0.5 l, a R2 = \

R1

A

æl ö R1 = ç - 0÷ a x + 0 a y + 0 a z = è2 ø 0.5 l a x R1 = = = ax 0.5 l |R1|

\

l Ö3l 2

E at P is to be obtained. Q1 E1 = a R1 4p e 0 R 21

0.5 l ax, a R1

–2Q

C

1Q 4p e 0 ( 0 . 5 l) 2

R2 = – ax |R2| (- a x ) =

- 3 . 595 ´ 10 10 Q l2

ax

æ 3 ö æl lö a R3 , R3 = ç - ÷ a x + ç 0 l÷ a y + 0 a z 2 2 2 è ø 4p e 0 R3 è ø Q3

TM


x


2 - 10


\

R3 = – 0.866 l a y ,|R3| = 0.866l, a R3 =

\

E3 =

\

- 2Q 4p e 0 ( 0 . 866 l) 2

E at P = E1 + E2 + E3 =

(- a y ) = Q l2

R3 = – ay |R3|

2 . 3968 ´ 10 10 Q l2

ay

[7.185 ´ 10 10 a x + 2.3968 ´ 10 10 a y ] V/m

... Ans.

z

Example 2.3.9 Solution : Let the square is in x-y plane as shown in the Fig. 2.12. The co-ordinates of various points are,

A

A(0, 0, 0), B(0, 0.1, 0), C(0.1, 0.1, 0), D(0.1, 0, 0)

10 cm

10 cm 1 mC D

To find E at the vacant corner A.

1 mC B

y

10 cm 10 cm

x

C 1 mC

Fig. 2.12 3

E =

å

i=1

\

EB =

\

ED =

\

EC =

Qi 4pe 0 Ri 2 Q 4pe 0 R B 2 Q 4pe 0 R D 2 Q 4pe 0 R C 2

a Ri a RB where R B = – 0.1 a y , a RB = –a y a RD where R D = – 0.1 a x , a RD = – a x a RC where R C = – 0.1 a x – 0.1 a y , a RC =

EA = EB + ED + EC = =

1 ´ 10 – 6 4p ´ 8.854 ´ 10 –12

Q 4pe 0

– 0.1 a x – 0.1a y ( 0.1) 2 + ( 0.1) 2

[– 0.1 a x – 0.1 a y ] ù é ay ax + êú 2 (0.1) 2 [( 0.1) 2 + ( 0.1) 2 ] 3 / 2 úû êë (0.1)

[–100 a y – 100 a x – 35. 35 a x – 35. 35 a y ] = –1216.49 a x – 1216.49 a y kV m

Example 2.3.10 Solution : Consider the circle consisting of charges placed in xy plane and charge of - 20 mC is on z axis, 2 m from the plane of the circle. This is shown in the Fig. 2.13. The charges are placed equally i.e. at an interval of 360°/10 = 36° between each other. Five pairs of charges which are dimetrically opposite to each other, exists on the circumference of a circle. Consider a pair A and B. The field EA due to Q at A, at point P is shown in the Fig. 2.13. l (OQ) = 2 m, l (OP) = 2 m hence Ð PAO = 45° TM



2 - 11

Coulomb's Law and Electric Field Intensity EA

\ y component of EA i.e. EA y = EA cos 45° Similarly l (OB) = 2 m, l (OP) = 2 m hence Ð PBO = 45°

z

EB

45º

\ y component of EB i.e. EBy = EB cos 45° But EAy is in - a y direction while EBy is in a y direction. From symmetry of the arrangement EAy = EBy . Hence they cancel

45º

EBy P(0,0,2)

EAy R

R 2

each other.

B Q

Q

Q 45º

Q

Q 45º

O 2

2

While z components of EA and EB help Q each other as both are in a z direction. Q Q EAz = EBz = EA or EB sin 45° a z x Similarly there are 4 more pairs of charges Fig. 2.13 which will behave identically and their y components are going to cancel while z components are going to add.

(

A Q

y

Q

)

Thus total z component of E at P is, Etotal = (E due to any charge) ´ 10 ´ sin 45° a z = where

R =

\

Etotal =

Q 4 p e0 R2

´ 10 ´ sin 45° a z

( 2) 2 + ( 2) 2 = 8 500 ´ 10 -6 4 p e0 ´

( ) 8

2

´ 10 ´ sin 45° a z = 3.972 ´ 10 6 a z

V/m

FP = Q P Etotal = -20 ´ 10 -6 ´ 3.972 ´ 10 6 a z = – 79.44 (a z ) N

\

This is the force on the charge at P. In general, force acts normal to the plane in which circle is kept, i.e. – 79.44 a n where a n is unit vector normal to the plane containing the circle. Example 2.3.11 Solution : \

Now

\

a) A (2, –1, 3) and P (0, 0, 0) Q a AP E at P = 2 4 p e 0 RAP a AP =

E =

( 0 - 2) a x + [ 0 - ( -1)]a y + [0 - 3] a z ( -2) 2 + (1) 2 + ( 3) 2 5 ´ 10 -9 4 p ´ 8.854 ´ 10 -12 ´

(

14

)

2

é- 2ax +ay - 3az ù ú ê 14 û ë

= – 1.715 a x + 0.857 a y – 2.573 a z V/m TM



2 - 12


b) Let point P is now (x, 0, 0). \

a AP =

\

E =

( x - 2) a x + a y - 3 a z rAP = rAP ( x - 2) 2 + (1) 2 + ( -3) 2

4 p e0

=

é ( x - 2) a + a - 3 a x y z ×ê 2 ( x - 2) + 1 + 9 êë ( x - 2) 2 + 1 + 9 Q

[ [

]

5 ´ 10 - 9

]

2

4 p e 0 ( x - 2) + 10 =

| E| =

44.938

[(x - 2)

2

]

3/ 2

+ 10

44.938

[

]

3/ 2

( x - 2) 2 + 10

3/ 2

ù ú ú û

[( x - 2) a x + a y - 3 a z ]

[( x - 2) a x + a y - 3 a z ] 44.938 é x - 2 2 + 1 2 + -3 2 ù = ) ( ) ( ) ú êë ( û ( x - 2) 2 + 10

[

d | E| = 0 dx é ê - 2 ( x - 2) 44.938 ê ê ( x - 2) 2 + 10 êë

[

4.49

|E| max

ù ú = 0 2ú ú úû

]

\

V/m

|E| in V/m

To find x at which| E| is maximum,

\

]

x –10

0

2

10

Fig. 2.14

(x – 2) = 0

x = 2 \ The graph of| E| against x is shown in the Fig. 2.14.

where| E| is maximum.

c) Hence| E|max is at x = 2, \

| E|max =

44.938 = 4.4938 V/m 10

z


E= aR =

Q 4pe 0 R 2 RQP RQP

aR =

y

P- Q P

P- Q

[

]

P - Q = ( -0.2 - 0.2) a x + ( 0 - 0.1) a y + -2. 3 - ( -2 . 5) a z = -0.4 a x - 0.1 a y + 0.2 a z TM

x


5 nC Q

aR

Fig. 2.15


\

aR =

=

2 - 13


-0.4 a x - 0.1 a y + 0.2 a z

( -0.4) 2 + ( 0.1) 2 + ( 0.2) 2 -0.4 a x - 0.1 a y + 0.2 a z 0.45825

= -0.8728 a x - 0.2182 a y + 0.4364 a z \

R =

\

E =

P - Q = 0.45825 5 ´ 10 -9 4p ´ 8.854 ´ 10 -12 ´ ( 0.45825) 2

[a R ] = 214 a R

Substituting value of a R , E = -186.779 a x - 46.694 a y + 93.389 a z V/m

… E at P

Example 2.3.13 z

Solution : The arrangement is shown in the Fig. 2.16. Let the charges are placed at A, B and C while E is to be obtained at fourth corner O. \ E at 'O' = EA + E B + E C Q Q Q a RA + a RB + a RC = 2 2 4pe 0 RA 4pe 0 R B 4pe 0 R 2C

(0,0,0)

RA = – 0.05 a x , RA = 0.05, a RA = – a x

A

(0,0.05,0) RC

O RA

RB B

(0.05,0,0)

a RB = – 0.707 a x – 0.707 a y

=

(0.05,0.05,0)

Fig. 2.16

RC = – 0.05 a y , RC = 0.05, a RC = – a y E =

y

x

RB = – 0.05 a x – 0.05 a y , RB = 0.0707,

\

C

Q é ( – a x ) ( – 0.707 a x – 0.707 a y ) ( – a y ) ù + + ê ú 4pe 0 ê (0.05) 2 (0.05) 2 úû ( 0.0707) 2 ë 100 ´ 10 –9 4p ´ 8.854 ´ 10 –12

[–400 a x – 141.4 a x – 141.4 a y – 400 a y ]

= – 486.6 a x – 486.6 a y kV/m, |E| at 'O' = 688.156 kV/m Q2 = – 2 mC

Example 2.3.14 Solution : The arrangement is shown in the Fig. 2.17. F3 = F13 + F23 =

Q 1Q 3 4pe 0 R 213

a R13 +

Q2Q 3 4pe 0 R 223

(–1,–1,4)

z R23 Q3 = 10 nC (0,3,1) y

a R23

R R13 = – 3 a x + a y + 2 a z , R13 = 14, a R13 = 13 R13

R13 x

(3,2,–1) Q1 = 1 mC

Fig. 2.17 TM



2 - 14

R 26, a R23 = 23 R23

R23 = a x + 4 a y – 3 a z , R23 = \ F3 =


1 é 1 ´ 10 –3 ´ 10 ´ 10 –9 æ – 3 a x + a y + 2 a z ö –2 ´ 10 –3 ´ 10 ´ 10 –9 æ a x + 4 a y – 3 a z ç ÷+ ç ê 4pe 0 ê 14 26 è ø è ( 14 ) 2 ( 26 ) 2 ë

öù ÷ú ø úû

= – 6.503 a x – 3.707 a y + 7.4985 a z mN E3 =

F3 = – 650.3 a x – 370.7 a y + 749.85 a z KV/m Q3

Example 2.4.4 Given : r v = 10 z 2 e - 0.1 x sin p y C m 3 .

Solution :

Consider differential volume in cartesian system as, dv = dx dy dz dQ = r v dv = 10 z 2 e - 0.1 x sin py dx dy dz \ \

Q =

ò

r v dv

vol

But now it becomes triple integration \Q =

4

1

2

ò ò ò

10 z 2 e - 0.1 x sin py dx dy dz

z = 3 y = 0 x = -2 2

é e - 0.1 x ù 2 ò ò 10 z sin p y ê - 0.1 ú dy dz = û -2 ë z= 3 y = 0 4

=

1

4

é z 3 ù é - cos p - cos 0 ù 4.0267 = 10 = 10 ê p úû 3 ú êë p û3 ë

1 - 0.2 e + 0.2 ù 2 é cos p y ù é e dz 10 z ê ò ú ê p û 0 - 0.1 - 0.1 ú ë ë û z= 3 4

é43 - 33 ù é1 1ù ú ê p + p ú 4.0267 = 316.162 C ê 3 û û ë ë

Example 2.4.5 Solution : i) 0 < x < 5 m, r L = 12x 2 mC m Q =

ò r L dL =

ii) r S = rz 2 nC m 2 , Q =

5

ò

0

5

éx3 ù = 500 mC = 0.5 C 12 x 2 dx mC = 12 ê 3 ú û0 ë

r = 3, 0 < z < 4 m

ò r S dS = S

ò r S [r df dz] = r S

4

2p

ò ò

rz 2 ´ 10 –9 df dz

z = 0 f= 0

4

2

= ( 3) ´ 10 iii) r v =

10 C m 3, rsinq

–9

´ [f]

2p 0

éz3 ù ê 3 ú = 1.206 mC û0 ë

r=4m TM


... r = 3


ò

Q=

r v dv =

vol

ò

2 - 15 2p

r v r 2 sin q dr dq df =

p

Coulomb's Law and Electric Field Intensity 4

ò ò ò

f= 0 q= 0 r = 0

vol

10 ´ r 2 sin q dr dq df r sin q

4

ér2 ù = 10 ê ú [q] 0p [f] 20 p = 1579.136 C. 2 ë û0 Example 2.4.6

f 1000 electrons/m 3 cos r 4 1 electron = - 1.6 ´ 10 -19 C charge

Solution :

ne =

\

r v = n e ´ charge on 1 electron =

f - 1.6 ´ 10 - 16 cos C / m 3 r 4

The volume is defined as sphere of r = 2 m. dv = r 2 sin q dr dq df

\ \

Q =

ò

2p

r v dv =

vol

p

... spherical system 2

ò ò ò

f = 0q = 0r = 0

= - 1.6 ´ 10 - 16

f - 1.6 ´ 10 - 16 cos r 2 sin q dr dq df 4 r

2p é sin f ù 2 ér2 ù p ê 4ú ê 2 ú [- cos q]0 ê 1 ú ë û0 êë 4 úû 0

= - 1.6 ´ 10 - 16 ´ 2 ´ 2 ´ 4 ´ 1 = - 2.56 ´ 10 - 15 C Example 2.6.6 Solution : i) For origin let r = r1 E=

rL a 2pe 0 r1 r1

z

Parallel to x-axis

Point on the line is (x, 3, 5). Origin is (0, 0, 0) z=5

Do not consider x co-ordinate as the charge is parallel to x-axis. \ r1 = (0 – 3) a y + (0 – 5) a z = – 3 a y – 5 a z , |r 1 | = \

E=

30 ´ 10 -9 2p ´ 8.854 ´ 10 -12

O

34

y=3

é -3 a y - 5a z ù ê ú 34 ´ 34 êë úû

x

Fig. 2.18

= – 47.582 a y – 79.303 a z V/m ii) P(5, 6, 1) \

r2 = (6 – 3) a y + (1 – 5) a z = 3 a y – 4 a z , |r2 | = 5 TM


y


\

E=

2 - 16


é 3 ay - 4 az ù ê ú = 64.711 a y – 86.2823 a z V/m 5 ´5 ë û

30 ´ 10 -9 2p ´ 8.854 ´ 10 -12

z

Example 2.6.7 Solutino : Fig. 2.19.

–¥

a) The line charge is shown in the

2 P(6,–1,3)

It is parallel to the x axis as y = 1 constant and z = 2 constant. The line charge is infinite hence using the standard result, r L a E = 2 p e0 r r

r

¥

O 1

To find a r , consider a point on the line charge x (x, 1, 2) while P (6, –1, 3). As the line charge is parallel to x axis, do not consider x coordinate while finding a r . \

r = ( -2) 2 + (1) 2 = 5

r = ( -1 - 1) a y + ( 3 - 2) a z = - 2 a y + a z ,

\

ar =

\

E =

Fig. 2.19

- 2ay +az r = |r| 5 rL 2 p e0

(

)

-9 2ay +az é - 2 a y + a z ù 24 ´ 10 = ú ê 5 ë 5 2 p ´ 8.854 ´ 10 -12 ´ 5 û

= – 172.564 a y + 86.282 a z V/m b) Consider a point charge Q A at A (–3, 4, 1). The electric field due to Q A at P (6, –1, 3) is, EA =

QA 2 4 p e 0 RAP

a AP

RAP = [ 6 - ( -3)]a x + [-1 - 4] a y + [3 - 1] a z = 9 a x - 5 a y + 2 a z , RAP = 10.488 \

a AP =

\

EA =

RAP RAP

=

9ax -5ay + 2az 10.4888

QA 4 p e 0 ´ (10.4888) 2

é9ax -5ay + 2az ù ú ê 10.4888 û ë

The total field at P is now, Et = E + EA The y component of total Et is to be made zero. é ù 5 QA \ ê - 172.564 ú ay = 0 êë 4 p e 0 ´ (10.4888) 3 úû \

QA =

i.e.

5 QA 4 p e 0 ´ (10.4888) 3

= – 172.564

- 172.564 ´ 4 p ´ 8.854 ´ 10 -12 ´ (10.4888) 3 = – 4.4311 mC 5 TM


y


2 - 17


Example 2.6.8

z

Solution : The charge is shown in the Fig. 2.20.

z=3

Key Point Charge is not infinite hence basic method of

differential charge dQ must be used.

dQ

(0,0,z)

dQ = r L dL = r L dz r L dz dQ R dE = aR = 2 2 |R| 4p e 0 R 4p e 0 R

\ \

z=1 y

i) To find E at origin \

R = – z a z , |R| = z, a R = – a z

\

dE =

\

E =

r L dz( -a z ) 4p e o

z2

i.e.

-20 ´ 10 -9

E=

-rL 4p e o

x 3

dz

ò

2 z=1 z

az

Fig. 2.20

z= 3

4p ´ 8.854 ´ 10 -12 ii) To find E at P(4, 0, 0)

é- 1 ù a z = – 119.824 a z V/m ëê z ûú z = 1

\ R = (4 – 0) a x + (0 – z) a z = 4 a x – z a z , |R| = 16 + z 2 Example 2.6.9 P(2,3,15)

Solution : The line is shown in the Fig. 2.21. The line with x = – 3 constant and y = 4 constant is a line parallel to z axis as z can take any value. The E at P (2, 3, 15) is to be calculated. The charge is infinite line charge hence E can be obtained by standard result, rL a E= 2 p e0 r r

z

ar

4 –3 O

To find r, consider two points, one on the line which is (–3, 4, z) while P (2, 3, 15). But as line is parallel to z axis, E cannot have component in x a z direction hence z need not be considered while calculating r. r = [ 2 - ( -3)]a x + [3 - 4] a y = 5 a x - a y \ \

|r| =

\

ar =

(5) 2 + ( -1) 2 = 26 5ax -ay r = |r| 26

TM

¥

r


y

–¥

Fig. 2.21

... z not considered


2 - 18


[

]

25 ´ 10 -9 5 a x - a y rL 1 é5 a x - a y ù × E = ú = 2 p e 0 26 êë 26 û 2 p ´ 8.854 ´ 10 -12 ´ 26

\

= 86.42 a x - 17.284 a y V/m

z rL = 30 nC/m

Example 2.6.10 Solution : The charge is shown in the Fig. 2.22. 5

The charge is parallel to x-axis hence E cannot have any component in x direction hence do not consider x while calculating E.

(x, 3, 5) y 3

i) E at P(0, 0, 0) \ r = (0 - 3) a y + (0 - 5) a z = - 3 a y - 5 a z

Parallel to x-axis

x

\ r = | r |= 3 2 + 5 2 = 34 rL rL é r ù \ ar = E = 2p e 0 r 2p e 0 r êë| r|úû =

30 ´ 10 - 9 2p ´ 8.854 ´ 10 - 12

Fig. 2.22

é- 3ay - 5az ù ú = - 47.58 a y - 79.3 a z V m ê 34 ´ 34 ë û

ii) E at P(0, 6, 1) \ r = (6 - 3) a y + (1 - 5) a z = 3 a y - 4 a z , |r| = \

E =

30 ´ 10 - 9 2p ´ 8.854 ´ 10 - 12

32 + 42 = 5

é + 3a y - 4 a z ù ê ú = 64.71 a y - 86.28 a z V m 5 ´5ë û

iii) E at P(5, 6, 1) As E does not have any component in x direction and y, z, co-ordinates are same as in (ii) hence E also remains same as obtained in (ii). E = 64.71 a y - 86.28 a z V m \ Example 2.6.11

z

rL = 40 nC/m

Solution : The charge is shown in the Fig. 2.23.

· P(–2,2,8)

Key Point As charge is along z-axis, E can not have any

component in a z direction. Do not consider z co-ordinate while calculating r . r = ( -2 - 0) a x + (2 - 0) a y \ = -2 a x + 2 a y , r = \

E =

=

4+4 = 8

(0,0,z) x

rL rL r a = 2 p e0 r r 2 p e0 r r

(

40 ´ 10 -9 -2 a x + 2 a y 2 p ´ 8.854 ´ 10 -12 ´

Fig. 2.23

)

8´ 8

= - 179.754 a x + 179.754 a y V m TM



2 - 19


Example 2.6.12 Solution : Consider the charge along z-axis as shown in the Fig. 2.24. Consider the differential charge at a distance z. dQ = r L dl = r L dz \

dE =

r dz L

B (0, 0, z2) dl R P(0, –h,0)

R = 0 a x + ( - h - 0)a y + (0 - z)a z h 2 + z2 , a R =

dE =

\

E =

r dz

Fig. 2.24

R | R| - ha y - za z

L

2

2

4pe 0 (h + z )

h 2 + z2

z2 é z 2 - h dz a ù z a z dz ú y ê 4pe 0 ê ò (h 2 + z 2 ) 3/ 2 ò (h 2 + z 2 ) 3/ 2 ú z1 ë z1 û ß ß

r

L

I1

z2

I1 =

ò

z1 z2

I1 =

ò

z1

h dz (h 2

2

+ z 2 ) 3/ 2

h 2 sec 2 q dq h

sec 3

3

z2

ò

z1 z2

q

I2 =

ò

z1

z dz (h 2 + z 2 ) 3/ 2 u du u3

…(1)

I2

, z = h tan q, dz = h sec q dq

=

1 h

z2

ò

cos q dq =

z1

1 z [sin q] z 2 h 1

z é ù 2 z2 z1 1 1é z = = ê ê ú hê hê h 2 + z 22 h 2 + z 21 ë h 2 + z 2 úû z 1 ëê

I2 =

A (0, 0, z1)

x

= - ha y - za z

\

y

aR

4pe 0 R 2

| R| =

(0, 0, z) z

2

2

ù ú ú úû

Öh2+z2

2

, h + z = u , 2z dz = 2u du z

ù é ú = ê– ú ê úû êë

TM

1 h 2 + z 22

z

Fig. 2.24 (a)

é ù 2 1 z2 1 = é- ù = -ê ú êë u úû z ëê h 2 + z 2 úû z 1 1

é 1 1 = - ê ê 2 + 2 2 + 2 z2 h z1 êë h Using I1 and I2in equation (1),

q h

+

ù ú 2 2 ú h + z1 ú û


1


\

E =

2 - 20

é -rL ê z2 z1 4pe 0 ê 2 2 2 2 êë h h + z 2 h h + z 1


ù ú a + -rL ú y 4pe 0 úû

é 1 -1 ê + ê 2 2 2 h + z 21 êë h + z 2

ù úa V/ m ú z úû

Example 2.6.13 Solution : Q = 1 mC and placed A(0, 0, 1) and B(0, 0, 2) m. L = 2–1=1m \

z

between B

Q 1 = = 1 mC/m = \ r L L 1 Consider an elementary charge dQ at a distance z as shown in the Fig. 2.25. dQ = r dz \

(0, 0, 2)

dz

(0, 0, z) z

A

P2(0, 1, 1) (0, 0, 1) y

L

P1(0, 0, 0)

i) For point P1(0, 0, 0),

\

R = -z a z , a R = - a z | R| = z r L dz dQ (- a z ) aR = dE = 2 4pe 0 z 2 4pe 0 R z =2

\

E =

=

ò

r dz L

2 z = 1 4 pe 0 z

-r

4pe 0

= -

2

ò

L

Fig. 2.25

(- a z )

dz

z =1z

x

2

az

1 ´ 10 -6

2

4p ´ 8.854 ´ 10 -12

é - 1 ù a = 8987.7424 é 1 - 1ù a z êë 2 ûú z ëê z ûú 1

= – 4493.8712 a z V/m ii) For point P2(0, 1, 1) R = 0 a x + (1 - 0)a y + (1 - z)a z , | R| = 1 + (1 - z) 2 \

dE =

\

dE =

\

E =

dQ 4pe 0 R 2

aR =

r dz [a y + (1 - z)a z ] R L = 4pe 0 R 2 |R | 4pe 0 [1 + (1 - z) 2 ] 1+ (1 - z) 2 dQ

1 ´ 10 -6 4p ´ 8.854 ´ 10

-12

ò dE = 8987.7424

a y dz (1 - z) a z dz ïü ïì + í ý 2 3/ 2 ïî [1 + (1 - z) ] [1 + (1 - z) 2 ] 3/ 2 ïþ 2

ò

z =

dz a y ìï (1 - z) dz a z üï + í 2 3/ 2 [1 + (1 - z) 2 ] 3/ 2 ý ïþ î [1 + (1 - z) ] 1ï

TM



2 - 21

2

dz

ò

I1 =

2 3/ 2 1 [1 + (1 - z) ]

z =


2 Þ put 1 – z = tan q, – dz = sec q dq

For z = 1, q 1 = 0° and z = 2, q 2 = – 45° q2

\

ò

I1 =

q1

\

- sec 2 q dq [1 + tan 2 q] 3/ 2

= - [sin q]

I1

-45° 0°

2

z = 1

for z = 1, u2

\

ò

I2 =

- u du u3

q2

ò

cos q d q

q1

2

Þ put [1 + (1 – z) ] = u

2

(1 – z)dz = –u du

u1 = 1 and z = 2,

u1

\

i.e.

ò

1 dq = sec q

= - [sin ( - 45° )] = + 0.7071

[1 + (1 - z) 2 ] 3/ 2

\ 2(1 – z) (– dz) = 2u du

q2 q1

(1 - z) dz

ò

I2 =

=-

u2 =

2

1 2 é 1 ù = - é- ù =ê - 1ú = – 0.2928 êë u úû 1 ë 2 û

E = 8987.7424 [0.7071 a y – 0.2928 a z ] = 6355.2326 a y – 2631.6109 a z V/m z

Example 2.6.14 Solution : The Fig. 2.26.

charge

is

shown

as

in

rL

the

dQ

Key Point If rL is not distributed all along the

5

length then standard result can not be used. The

aR

basic procedure is to be used. R

As charge is not infinite, let us use basic procedure of considering differential charge. Consider the differential element dl in the z direction hence, dl = dz \ \

dQ = r dE =

L

dl = r dQ

4 p e0

R2

L

0

(2,0,0) P –5 x

dE

rL

dz

aR =

r L dz 4 p e0 R2

y

Fig. 2.26

aR

Any point on z axis is (0, 0, z) while point P at which E to be calculated is ( 2, 0, 0). R = ( 2 - 0) a x + ( 0 - z) a z = 2 a x - z a z |R| =

( 2) 2 + ( - z) 2 = 4 + z 2 , a R = TM

2ax - zaz R = |R| 4 + z2



\

dE =

2 - 22

r

L

r dz é2ax - zaz ù L ê ú= 2 êë úû 4 p e 4 + z 2 4+z 0

dz

4 p e 0 æç 4 + z 2 ö÷ è ø


(

2

)

3/ 2

(2 a x

- zaz )

Now there is no charge between – 5 to 5 hence to find E, dE to be integrated in two zones - ¥ to – 5 and 5 to ¥ in z direction. -5

\

E =

ò

-¥

¥

dE + ò dE 5

Looking at the symmetry it can be observed that z component of E produced by charge between 5 to ¥ will cancel the z component of E produced by charge between – 5 to - ¥. Hence for integration a z component from dE can be neglected. -5 ¥ r dz ( 2 a x ) r dz ( 2 a x ) L L +ò \ E = ò 3/ 2 3/ 2 2 5 4 p e0 4 + z2 -¥ 4 p e 0 4 + z

(

)

(

)

Solving, E = 13 a x V m To find cylindrical co-ordinates find the dot product of E with a r , a f and a z , at point P, referring table of dot products of unit vectors. \ E r = E a r = 13 a x a r = 13 cos f \

Ef =

\

Ez =

× E× a E× a

× = 13 a × a = 13 a × a

f z

x

f

= - 13 sin f

x

z

= 0

At point P, x = 2, y = 0, z = 0

x 2 + y 2 = 2 and

\

r =

\

cos f = 1

\

and

E r = 13,

f = tan -1

y = tan -1 0 = 0 ° x

sin f = 0

E f = 0,

Ez = 0

Hence the cylindrical co-ordinate systems E is, E = E r a r + E f a f + E z a z = 13 a r V/m Example 2.7.2

Kept this example for student's practice.


Case 1 : Point charge Q 1 = 6 mC at A (0, 0, 1) and P (1, 5, 2)

\

E1 =

Q1 2 4 p e 0 RAP

a AP =

Q1 2 4 p e 0 RAP

é RAP ù ú ê ë| RAP|û

RAP = (1 - 0) a x + (5 - 0) a y + ( 2 - 1) a z = a x + 5 a y + a z \

| RAP | =

(1) 2 + (5) 2 + (1) 2 = 27

TM



\

E1 =

2 - 23

6 ´ 10 -6 4p ´ 8.854 ´ 10 -12 ´

(

27

)

2


éa x +5a y + a z ù ú ê 27 û ë

\ E1 = 384.375 a x + 1921.879 a y + 384.375 a z V/m Case 2 : Line charge r L along x-axis. It is infinite hence using standard result, rL rL r ar = E2 = 2 p e0 r 2 p e0 r r Consider any point on line charge i.e. (x, 0, 0) while P (1, 5, 2). But as line is along x-axis, no component of E will be along a x direction. Hence while calculating r and a r , do not consider x co-ordinates of the points. r = (5 - 0) a y + ( 2 - 0) a z = 5 a y + 2 a z

\ \

| r| =

\

E2

(5) 2 + ( 2) 2 = 29

[

= 557.859 a y + 223.144 a z V/m Case 3 : Surface charge r S over the plane z = – 1. The plane is parallel to xy plane and normal direction to the plane is a n = a z , as point P is above the plane. At all the points above z = – 1 plane the E is constant along a z direction. \ E3 = =

]

180 ´ 10 - 9 5 a y + 2 a z é5 a y + 2 a z ù = ú = ê 29 2 p e 0 ´ 29 ë 2p ´ 8.854 ´ 10 -12 ´ 29 û rL

z

P(1, 5, 2)

rS a 2 e0 n 25 ´ 10 -9 2 ´ 8.854 ´ 10 -12

az

y

az

= 1411.7913 a z V/m

rS

x

Hence the net E at point P is, E = E1 + E2 + E3

Fig. 2.27

= 384.375 a x + 1921.879 a y + 384.375 a z + 557.859 a y + 223.144 a z + 1411.7913 a z = 384.375 a x + 2479.738 a y + 2019.3103 a z V/m Example 2.8.6 Solution :

The sheet of charge is shown in the Fig. 2.28.

Consider the differential area dS carrying the charge dQ. The normal direction to dS is a z hence dS z = r dr d f . TM



2 - 24

\

dQ = r S dS = r S r dr df =

\

dQ = 10 -4 dr df

\

dE =

10 -4 dr df


10 -4 × r dr df r

z

P(0, 0, 3)

aR

4 p e0 R2

R

Consider R as shown in the Fig. 2.29, which has two components in cylindrical system, 1. The component along - a r having radius r i.e. - r a r .

rS r=4

2. The component z = 3 along a z i.e. 3 a z . \ R = - r ar + 3az 2

\

aR =

dE =

x

Fig. 2.28

2

z

- rar + 3az R = |R| r2 + 9 10 -4

3

2

R az

It can be seen that due to symmetry about z-axis, all radial components will cancel each other. Hence there will not be any component of E along a r . So in integration a r need not be considered. \

2p

E =

10 -4 dr df

4

ò

ò

f= 0 r= 0

P(0, 0, 3)

é- rar + 3az ù ê ú êë r 2 + 9 úû

dr df

4 p e 0 æç r 2 + 9 ö÷ è ø

y

dS

( - r ) + ( 3) = r + 9

|R| = \

2

aR

O

(

)

4 p e0 r 2 + 9

3/ 2

O

y r –ar

x

Fig. 2.29

(3az )

As there is no r dr in the numerator, use dr = 3 sec 2 q dq

r = 3 tan q, For r = 0, For r = 4,

q2 = tan –1 4 / 3 2p

\

ò

E =

q2

ò

10 -4 3 sec 2 q dq df

f = 0 q1 = 0

\

2p

E =

ü ý ...Change of limits þ

q1 = 0

ò

q2

ò

f = 0 f = 0°

[

]

4 p e 0 9 tan 2 q + 9

3/ 2

299.5914 ´ 10 3 sec 2 q dq df

[1 + tan q] 2

3/ 2

TM

(3az )

az


Electromagnetic Field Theory 2p

ò

=

q2

ò

f = 0 q1 = 0°

2 - 25

299.5914 ´ 10 3 dq df a z = sec q 2p

= 299.5914 ´ 10 3 [f]0

[sin q ]qq12= 0°

= 1.8823 ´ 10 6 sin q 2 a z Now \

q 2 = tan -1

sin q 2

Coulomb's Law and Electric Field Intensity 2p

ò

q2

299.5914 ´ 10 3 dq df[ cos q ]a z

ò

f = 0 q 1 = 0°

... Separating variables

az ... sin 0º = 0

4 4 i.e. tan q 2 = 3 3

2

4 = 0.8 = 5

4

q2 3

E = 1.8823 ´ 10 6 ´ 0.8 a z

\

2

Ö4 +3 =5

Fig. 2.30

= 1.5059 ´ 10 6 a z V/m = 1.5059 a z MV/m Example 2.8.7

Sheet at x = 0

Solution : The sheet is shown in the Fig. 2.31.

z

The point P is on the back side of the plane.

P (–5, 0, 0) – ax

The normal to the plane in the direction of P is - a x .

\ EP =

y

O

\ an = -ax 10 -12

rS = 5 pC/m

x

rS 5 ´ a = (- a x ) 2 e0 n 2 ´ 8.854 ´ 10 -12

2

Fig. 2.31

\ EP = – 0.2823 a x V/m Example 2.8.8 Solution : Q = 100 mC, r = 10 cm = 0.1 m, area = pr 2 = 0.03141 m2 Q 100 ´ 10 – 6 = 3.1831 ´ 10 – 3 C m 2 = area 0.03141 z The disc is shown in the Fig. 2.32. rS =

Consider differential surface area dS. Using cylindrical system,

P

dS = r dr df, R = – r a r + za z aR =

z

P

R

– r a r + za z

zaz y

r 2 + z2

r

Key Point All radial components of E

at P will cancel each other due to

R

0 –rar

ds

x

symmetry.

(a)

(b)

Fig. 2.32 TM



E=

ò

S

dQ 4pe 0 R 2 2 p 0. 1

=

ò ò

f = 0r = 0

2 - 26


aR

r S [r dr df] 4pe 0

Use r 2 + z 2 = u 2

[r 2

zaz

+ z2]

i.e.

r 2 + z2

=

2p

rSz 4pe 0

0. 1

ò

ò

f= 0 r= 0

r dr df [r 2

+ z 2 ] 3/ 2

az

r dr = u du

Limits : r = 0, u 1 = z and r = 0.1, u 2 = 0.1 2 + z 2 \

rSz 4pe 0

E =

2p u 2

u du df

ò ò

u3

f= 0 u1

az =

u é– 1 ù 2 a z êë u úû u 1

rSz 1 ù é 1 ´ 2p ´ ê – ú az 4pe 0 u u 2û ë 1

=

0.05 = 0.2236

Using z = 20 cm = 0.2 m, u 1 = 0.2 and u 2 = \

rSz [f] 2 p 4pe 0 0

3.1831 ´ 10 – 3 ´ 0.2

E =

4p

´ 8.854 ´ 10 –12

1 1 ù a = 18.9723 a z MV/m ´ 2p ´ é – êë 0.2 0.2236 úû z

Example 2.8.9

z

Solution : The plane is shown in the Fig. 2.33 Consider the differential surface area dS carrying charge dQ. dQ = r S dS where dS = dxdy \ \ \

(

dQ = 2 dE =

R =

x2

+ y2

dQ 4p e o R 2

)

+9

3/2

z = –3 plane O

y R

dx dy nC

S

(–2, 2, –3)

dS

aR

x

P (2,–2,–3)

[0 – x] a x + [ 0 – y ] a y + [ 0 – ( –3)] a z

Q (2,2,–3)

Fig. 2.33

R = –x a x – y a y + 3 a z , | R| =

\

x 2 + y 2 + 9, a R =

dE =

(

)

2 x2 + y 2 + 9

(

R | R| 3/ 2

dx dy

)

4p e o x 2 + y 2 + 9

´

[–x a x – y a y + 3 a z ] ´ 10 –9 (x 2 + y 2 + 9)

Due to symmetrical distribution, x and y components of dE will cancel each other and only z component will exist.

TM



\

dE =

2 - 27

6 a z ´ 10 –9 dx dy 4p e o 2

\


2

ò

E =

ò

y = –2 x =–2

6 ´ 10 –9 6 ´ 10 –9 dx dy a z = [x] 2–2 [y] 2–2 a z = 862.82 a z V/m. 4p e o 4p e o z

Example 2.8.10

2

Solution : The sheets are shown in the Fig. 2.34. rS E = a 2 e0 N

rS3 = –8 nC/m

z=4

2

z=1

rS2 = 6 nC/m

i) PA = (2, 5, – 5)

2

rS1 = 3 nC/m

z = –4

y

It is below the plane z = – 4. Hence a N for this point due to all the sheets is –a z .

x

Fig. 2.34

\

Et

r S1 r r = ( –a z ) + 2 eS2 ( –a z ) + 2 eS3 ( –a z ) = –56.47 a z V/m 2 e0 0 0

ii) PB = (4, 2, –3) It is above z = – 4 and below other two plane. Hence a N = +a z for r S1 and –a z for r S2 and r S3 . \

Et = iii)

(

–8 ´ 10 –9 3 ´ 10 –9 6 ´ 10 –9 az )+ –a z ) + ( ( 2 e0 2 e0 2 e0

) (–a

z

) = 282.358 a z V/m

PC = (–1, –5, 2)

It is above z = 1 and below z = 4. Hence a N = +a z for r S1 and r S2 while –a z for r S3 . \

Et

(

–8 ´ 10 –9 3 ´ 10 –9 6 ´ 10 –9 = az + az + 2 e0 2 e0 2 e0

iv) PD = (–2, 4, 5) It is above all the planes hence a N = +a z for all. \

Et

(

–8 ´ 10 –9 3 ´ 10 –9 6 ´ 10 –9 = az + az + 2 e0 2 e0 2 e0

) (–a

z

) (a

) = + 56.47 a z V/m

z

) = 960.018 a z V/m

Example 2.8.11 Solution : i) For Q = 12 nC at P(2, 0, 6) RPO = – 2 a x – 6 a z , |RPO | = \

E1 =

Q 4p e 0 R 2PO

40

a RPO

TM



=

2 - 28

12 ´ 10 -9 4p ´ 8.854 ´ 10 -12


é- 2ax - 6az ù ê ú 40 û ´ 40 ë

12 nC (2,0,6) P

= – 0.8526 a x – 2.558 a z V/m

x = –2,y = 3 rL = 3 nC/m

ii) r L is parallel to z-axis. Any point on line charge is (– 2, 3, z) \

y O

r = [0 – (– 2) a x + [0 – 3] a y = 2 a x – 3 a y , |r| =

–ax Origin on back side

13

The z co-ordinate is not considered as line charge is parallel to z-axis.

x

x=2 rs = 0.2 nC/m

2

Fig. 2.35

\

E2 =

rL 3 ´ 10 -9 ar = 2p e 0 r 2p ´ 8.854 ´ 10 -12 ´ 13

é2ax - 3ay ù ú ê 13 û ë

= 8.2963 a x – 12.445 a y V/m iii) r S is at x = 2 and origin is on the back side of the sheet. Hence a n = – a x . \ \

E3 =

rS 0.2 ´ 10 -9 an = ( - a x ) = – 11.2943 a x V/m 2 e0 2 ´ 8.854 ´ 10 -12

E at origin = E1 + E2 + E3 = – 3.8506 a x – 12.445 a y – 2.558 a z V/m

qqq

TM


3

Electric Flux Density and Gauss’s Law Solutions of Examples for Practice

Example 3.5.3

z

Solution : a) A point charge of 40 mC at the origin. O(0,0,0) Q

P (6, 8, – 10) and O (0, 0, 0) \

r = ( 6 - 0) a x + ( 8 - 0) a y + ( -10 - 0) a z

r

r =

\

ar =

\

D =

P(6,8,–10)

x

= 6 a x + 8 a y - 10 a z \

y ar

Fig. 3.1 (a)

( 6) 2 + ( 8) 2 + ( -10) 2 = 200 6 a x + 8 a y - 10 a z 200 Q 4pr 2

ar =

ì 6 a x + 8 a y - 10 a z ü ý í 200 4p ´ ( 200 ) 2 î þ 40 ´ 10 -3

= 6.752 ´ 10 -6 a x + 9.003 ´ 10 -6 a y - 11.254 ´ 10 -6 a z C m 2 b) r L = 40 mC m along z-axis The charge is infinite hence, rL a E = 2pe 0 r r As the charge is along z-axis there can not be any component of E along z-direction. Consider a point on the line charge (0, 0, z) and P (6, 8, – 10). But while obtaining r do not consider z co-ordinate, as E and D have no a z component. \ r = ( 6 - 0) a x + ( 8 - 0) a y = 6 a x + 8 a y ( 6) 2 + ( 8) 2 = 10

\

r

\

rL é6 ax + 8 ay ù E = ú 2pe 0 (10) êë 10 û

=

ar =

hence

(3 - 1) TM

6 ax + 8ay 10



3-2

Electric Flux Density and Gauss's Law z

y Plane x=12

6 –ax

8 –10 P Back side of plane

x

Fig. 3.1 (b)

\

D = e0 E =

rL é6ax + 8ay ù -7 -7 2 ú = 3.819 ´ 10 a x + 5.092 ´ 10 a y C m 2p ´ 10 êë 10 û

c) r S = 57. 2 m C m 2 on the plane x = 12. The sheet of charge is infinite over the plane x = 12 which is parallel to yz plane. The unit vector normal to this plane is a n = a x . rS \ a E = 2e 0 n The point P is on the backside of the plane hence a n = - a x , as shown in the Fig. 3.6.2. rS \ E = (- a x ) 2e 0 But

D = e0 E

\

D =

rS ( - a x ) = - 28.6 ´ 10 -6 a x C m 2 2

Example 3.5.4 Solution : i) Case 1 : Point charge Q = 6 mC at P (0, 0, 0). While D to be obtained at A (0, 0, 4). \ \

r = ( 4 - 0) a z = 4 a z , r = ( 4) 2 = 4 , a r = D1 =

Q 4pr

2

ar =

6 ´ 10 -6 4 p ´ ( 4) 2

r = az r

a z = 2.984 ´ 10 - 8 a z C m 2

Case 2 : Line Charge r L = 180 nC/m along x-axis. So any point P on the charge is (x, 0, 0), while A (0, 0, 4). As charge is along x-axis, no component of D is along x-axis. So do not consider x co-ordinate while obtaining r. r r = ( 4 - 0) a z = 4 a z , r = 4 , a r = = az \ r TM



3-3

Electric Flux Density and Gauss's Law

As charge is infinite, \

D2 =

rL 180 ´ 10 - 9 ar = a z = 7.161 ´ 10 - 9 a z C m 2 2pr 2p´ 4

Case 3 : Uniform sheet of charge lies in z = 0 plane. So the direction normal to it is z direction as plane is xy plane. Hence a n = a z and r S = 25 nC/m 2 . As sheet is infinite, D3 = \

r

S

2

an =

25 ´ 10 - 9 a z = 12.5 ´ 10 - 9 a z C m 2 2

D = D 1 + D 2 + D3 = 49.501 ´ 10 - 9 a z C m 2

ii) The point at which D is to be obtained is now B (1, 2, 4). Case 1 : Point charge Q = 6 mC at P (0, 0, 0). \

r = (1 - 0) a x + ( 2 - 0) a y + ( 4 - 0) a z = a x + 2 a y + 4 a z =

(1) 2 + ( 2) 2 + ( 4) 2 =

\

ar =

ax + 2ay + 4az r = r 21

\

D1 =

\

r

Q 4pr 2

ar =

21

6 ´ 10 - 6 4p ´ (

21 ) 2

éa x + 2a y + 4a z ê 21 êë

ù ú úû

= 4.961 ´ 10 - 9 a x + 9.923 ´ 10 - 9 a y + 1.9845 ´ 10 - 8 a z C m 2 Case 2 : Line charge : The point on the charge is (x, 0, 0). As charge is along x-axis, do not consider x co-ordinate. \ \

\

r = ( 2 - 0) a y + ( 4 - 0) a z = 2 a y + 4 a z r

=

D2 =

( 2) 2 + ( 4) 2 =

20

hence

ar =

... as B ( 1, 2, 4) 2ay + 4az r = r 20

rL 180 ´ 10 - 9 é 2 a y + 4 a z ù ar = ê ú 2pr 2p ´ 20 êë 20 úû

= 2.8647 ´ 10 - 9 a y + 5.7295 ´ 10 - 9 a z C m 2 Case 3 : Infinite sheet of charge in z = 0 plane. The point B ( 1, 2, 4) is above z = 0 plane hence a n = a z and D3 remains same as before. r 25 ´ 10 - 9 a z = 12.5 ´ 10 - 9 a z C m 2 D3 = S a n = 2 2 \

D = D 1 + D 2 + D3 = 4.961 ´ 10 - 9 a x + 1.2786 ´ 10 - 8 a y + 3.807 ´ 10 - 8 a z C m 2 TM



3-4


iii) Let us find the total charge enclosed by a sphere of radius 4 m. Charge 1 : Q 1 = 6 mC at the origin. Charge 2 : The charge on that part of the line which is enclosed by the sphere. The line charge intersects sphere at x = ± 4. Hence charge on the length of 8 m is enclosed by the sphere. This is shown in the Fig. 3.2. z

–4 4m

Intersection of z = 0 plane with sphere y

+4

rL rS z = 0 plane

x

Fig. 3.2

\

Q 2 = r L ´ length enclosed = 180 ´ 10 -9 ´ 8 = 1.44 mC

Charge 3 : The intersection of z = 0 plane with a sphere is a circle with radius 4 m, in xy plane. The surface area of this circle is p r 2 . \

S = p ´ ( 4) 2 = 50.2654 m 2

Hence the total charge enclosed is, \

Q 3 = r S ´ S = 25 ´ 10 -9 ´ 50.2654 = 1.2566 mC

Hence the total charge enclosed by the sphere is, Q total = Q 1 + Q 2 + Q 3 = 8.6966 mC But

y = Q total = Total electric flux leaving the surface of sphere = 8.6966 mC

Example 3.5.5 Solution : Due to point charge at (3, 0, 0), Q ar D = 4p r 2 r = ( 2 - 3) a x + 0 a y + 3 a z TM



\ | r| =

1+9 =

3-5

Electric Flux Density and Gauss's Law z

10

- 4p ´ 10 -3 é -a x + 3a z ù úû 2 êë 10 4p ´ 10

\ D1 =

(

)

= 3.162 ´ 10

-5

a x - 9.486 ´ 10

P(2,0,3) -5

az C m

y r

The line charge is along y axis so there can not be any component at E along y direction.

– 4pmC x

Let point on line charge is (0, y, 0) and P (2, 0, 3).

E =

\

(3,0,0)

Fig. 3.3

r = ( 2 - 0) a x + ( 3 - 0) a z = 2 a x + 3 a z , \

rL = 2p mC/m

2

|r| = 13

rL 2p ´ 10 -3 é 2a x + 3a z ù ar = úû 2p e 0 r 13 2p ´ e 0 ´ 13 êë

D2 = e 0 E =

10 -3 2a + 3a z ] = 5.547 ´ 10 - 4 a x + 8.3205 ´ 10 - 4 a z C / m 2 13 [ x

2 D = D1 + D2 = 0.5863 a x + 0.7372 a z mC/m

\

Example 3.6.2 z

Solution : The cube is shown in the Fig. 3.4. As the origin is at the centre, x varies from – 1 to 1, y varies from – 1 to 1 and z varies from – 1 to 1, as each side of the cube is 2 m. p r v = 50 x 2 cos æç y ö÷ ´ 10 -6 C / m 3 è2 ø \Q =

ò

2m O

r v dv where dv = dx dy dz

2m

v 1

=

ò

1

ò

1

ò

z = -1 y = -1 x = -1

=

50 ´ 10 -6

p 50 x 2 cos æç y ö÷ ´ 10 -6 dx dy dz è2 ø

p ù1 1 é é x 3 ù ê sin 2 y ú 1 ê 3 ú ê p ú [z]-1 û -1 ê ë ë 2 úû -1

2m

x

Fig. 3.4

é 1 3 ( -1) 3 ù æ 2 ö é p p = 50 ´ 10 -6 ê ú ç ÷ ê sin - sin - ùú [1 - ( -1)] è ø p 2 2û 3 3 ë êë úû =

50 ´ 10 -6 2 (1 + 1) æçè p ö÷ø [1 - ( -1)]( 2) = 84.882 mC 3

TM


y


3-6


Example 3.6.3 Note that the r v is dependent on the variable r. Hence though the charge 4 distribution is sphere of radius 'a' we can not obtain Q just by multiplying r v by æç p a 3 ö÷ è3 ø Solution :

as r v is not constant. As it depends on r, it is necessary to consider differential volume dv and integrating from r = 0 to a, total Q must be obtained. Thus if r v depends on r, do not use standard results. ... Spherical system a) dv = r 2 sin q dr dq df \

Q =

ò

r v dv =

v

2p

p

a

æ r2 ö 2 r 0 ç1 ÷ r sin q dr dq df 2ø è a f= 0 q = 0 r = 0

ò ò ò

p

2p

= r 0 [ - cos q ]0 [f]0

a

ì 2 r4 ü ò ír - a 2 ý dr þ r= 0 î a

ér 3 r5 ù = r0 ´ 2 ´ 2 p ´ = r 0 [ - ( -1) - ( -1)][ 2 p ] ê 3 5 a2 ú ë û0

éa 3 a 3 ù ê 3 - 5 ú ë û

8p 2a3 r a3 C = 15 0 15 Outside sphere, r v = 0 so Q = 0 for r > a. b) The total charge enclosed by the sphere can be assumed to be point charge placed at the centre of the sphere as per Gauss's law. Q \ a r at r > a D = 4 p r2 = r0 ´ 4 p ´

\ Outside the charge distribution i.e. r > a, E

=

Q 4 p e0 r 2

8p 3 r0 a 3 2 r0 a 1 = 15 = 15 e 0 r 2 4 p e0 r 2 z

\

3 1 2 r0 a a r V/m E= 15 e 0 r2

Thus E varies with r, outside the charge distribution.

Gaussian surface

r=a r

y

c) For r < a, consider a Gaussian surface as a sphere r having r < a as shown in the Fig. 3.5. Consider dS at point P normal to a r direction, as D and E are in a r direction. dS = r 2 sin q dq df a r

P ar D x

dS

Fig. 3.5 TM



\

×

3-7


D = Dr a r

D d S = D r r 2 sin q dq df

\

Q1 =

ò

×

2p

D dS =

S

p

ò ò

D r r 2 sin q dq df

f= 0 q = 0 p

2p

= D r r 2 [ - cos q ]0 [f]0

= 4 p r 2 Dr

where

Q 1 = Charge enclosed by Gaussian surface

\

Dr =

Q1

i.e.

4 p r2

D=

Q1 4 p r2

and

ar

E=

Q1 D = ar e0 4 p e0 r 2

Let us find Q 1 , charge enclosed by Gaussian surface of radius r. 2p p r æ r2 ö 2 Q 1 = ò ò ò r 0 ç1 ÷ r sin q dr dq df 2ø è a f= 0 q = 0 r = 0 p

2p

= r 0 [ - cos q ]0 [f]0

r

ær 3 ìr 3 r5 ü r5 ö = 4 pr0 ç ÷C ý í 3 è 3 5 a2 ø 5 a 2 þ0 î

Using in the equation of E, field intensity for r < a is,

\

E=

ær 3 r5 ö 4 pr0 ç ÷ è 3 5 a2 ø 4 p e0 r 2

d) To find E to be maximum, inside d E d ìï r 0 = 0 i.e. dr íï e 0 dr î \

1 3 r2 = 0 3 5 a2 r2 =

\ \

E

max

=

5 a2 9

ar =

r0 é r r3 ù a V/m ê e0 3 5 a 2 ú r ë û

the sphere i.e. r < a obtain, ær r 3 öüï ç ÷ý = 0 è 3 5 a 2 øïþ

as r v ¹ 0,

e0 ¹ 0

i.e. r = 0.745 a

... Proved

r 0 é 0.745 a ( 0.745 a ) 3 ù 0.1656 a r 0 V/m ê ú= e0 ê 3 e0 5 a2 úû ë

Example 3.6.4 Solution : The charges are shown in the Fig. 3.6. Consider line charge along x-axis. Any point Q on this charge is (x, 0, 0). As the charge is infinite along x axis, E and hence D has no component in a x direction. \ Q (x, 0, 0) and P (3, 3, 3) TM



3-8


z

P(3,3,3) y rL

rL

x

Fig. 3.6

\ r = ( 3 - 0) a y + ( 3 - 0) a z \ \

... x co-ordinate need not be considered.

r = 3 a y + 3 a z and |r|= D1 =

9 + 9 = 18

rL 25 ´ 10 -6 é 3 a y + 3 a z ù ar = ú ê 2pr 2 p ´ 18 ë 18 û

= 6.6314 ´ 10 -7 a y + 6.6314 ´ 10 -7 a z C / m 2 Consider any point Q on charge along y axis. Hence Q (0, y, 0) and P (3, 3, 3). There is no component of E hence D along a y direction as charge is along y axis. So do not consider y co-ordinate. \ r = ( 3) a x + ( 3) a z and r = 9 + 9 = 18 \

D2 =

rL 25 ´ 10 -6 é 3 a x + 3 a z ù ar = úû 2pr 2 p ´ 18 êë 18

= 6.6314 ´ 10 -7 a x + 6.6314 ´ 10 -7 a z C / m 2 Hence total D at point P due to both the charges is, D = D1 + D2 = 0.6631 a x + 0.6631 a y + 1.3262 a z mC / m 2 Example 3.6.5 Solution :

a) The flux leaving is charge enclosed. y = Q=

ò

S

r S dS =

2p

5

ò ò

f = 0r = 0

5r r2

+1

r dr df

The dS = r dr df as the r S is in plane z = 2, to which the normal direction is a z , as shown in the Fig. 3.7. TM



3-9


z az r<5 rS

z=2 plane dS

y

x

Fig. 3.7

\

y =

2p

5

5 r2

ò ò

f= 0 r = 0

Now

\

ò

x 2 dx 2

ax +c

=

dr df

r2 +1

æ a öù x c é 1 tan -1 ç x ÷ c ø úû a a êë ac è

(

)

5

é r - tan -1 r ù [] ú êë 1 û0

2p

y = 5 [f]0

[

... a = c = 1

]

= 5 ´ 2 p ´ 5 - tan -1 5 = 113.932 nC

... use radian mode

b) Half of the flux leaves in a z direction while other half leaves in - a z direction. \

y leaving in - a z direction =

113.932 = 56.966 nC 2

Example 3.7.6 Solution : The arrangement is shown in the Fig. 3.8. z

z

rL1 r = 3m

r = 3m

rL1 A

rS

rS

B

(a) Charge distribution

(b) Gaussian surfaces

Fig. 3.8 TM



3 - 10


The spherical surface A shown in the Fig. 3.8 (b) is the Gaussian surface for the line charge. Let the differential surface area is dS = r df dz to which a r is normal. The D is directed radially outwards. The length of the Gaussian surface is L. and d S = r df dz a r \ D = Dr a r The radius r of Gaussian surface A is 0 < r < 3. \

Q =

ò

×

D dS =

S L

×

... ( a r a r = 1)

D r r df dz

S

2p

ò ò

=

ò

D r r df dz = D r r [f] 20 p [z] L0 = D r r 2pL

z= 0 f= 0

But charge on the line of length L is Q = r L1 ´ L \ r L1 L = D r r 2pL \

Dr =

r L1 2pr

and

D=

r L1 a 2pr r

0.3978 2.5 ´ 10 -6 for 0 < r < 3 m ar = ar mC m2 r 2pr The spherical surface B is the Gaussian surface enclosing both the charge distributions. r Due to the line charge, D 1 = L1 a r remains same. 2pr And due to cylinder of radius 3 m, let it be D 2 . The direction of D 2 is radially outwards. Consider differential surface area normal to a r which is r df dz. The length of Gaussian surface is L.

\

\ \

D =

D 2 = D 2r a r Q =

ò

S

D2

and

× dS =

L

d S = r df dz a r 2p

ò ò

D 2r r df dz = D 2r r 2 p L

z= 0 f= 0

... ( a r

×a

r

= 1)

Now charge on the surface of length L and radius r is, Q = r S ´ Surface area = r S ´ 2p rL where r = 3 m = Radius of charge distribution

(

)

= 2p ´ - 0.12 ´ 10 -6 ´ (3) ´ L = - 2.2619 ´ 10 -6 L C \ \ \

- 2.2619 ´ 10 -6 L = D 2r ´ r ´ 2p L

i.e.

D 2r =

- 0.36 a r mC m 2 r 0.0378 a r mC m 2 D = D1 + D2 = r

D2 =

- 2.2619 ´ 10 -6 - 0.36 = ´ 10 -6 2p r r for r > 3 for r > 3

Example 3.7.7 Solution : a) At r = 2 cm, it is inner side of inner sphere. It is seen that inside a spherical shell with surface charge E and D = 0. Now r = 2 cm is inside of all three spheres hence E = D = 0. TM



3 - 11


At r = 4 cm which is exterior to innermost sphere but inside of spheres having radii 5 and 7 cm. Hence at r = 4 cm, D and E exist due to sphere of r = 3 cm with r S = 200 mC/m 2 . E =

=

rS a 2 e0 r 2

... (Refer section 3.8.5)

ar

( ) a = 12.706 ´ 10 6 a V/m r 2 r 8.853 ´ 10 -12 ´ (4 ´ 10 -2 ) 200 ´ 10 -6 ´ 3 ´ 10 -2

2

Here

a = Radius of sphere = 3 cm and r = 4 cm is distance.

and

D = e 0 E = 112.5 a r m C m 2

At r = 6 cm, the E and D will be due to the two spherical shells having radii 3 and 5 cm. While due to sphere of r = 7 cm, D and E are zero at r = 6 cm. \ a 1 = 3 cm , r S1 = 200 mC m 2 r S1 (a 1 ) 2

( ) ar = 2 8.854 ´ 10 -12 ´ (6 ´ 10 -2 ) 200 ´ 10 -6 ´ 3 ´ 10 -2

\

E1 =

\

D 1 = e 0 E = 50 a r mC m 2

And a 2 = 5 cm,

e 0 (r) 2

2

a r = 5.6471 ´ 10 6 a r V/m

r S2 = - 50 mC m 2 r S2 (a 2 ) 2

( ) 2 8.854 ´ 10 -12 ´ (6 ´ 10 -2 ) - 50 ´ 10 -6 ´ 5 ´ 10 -2

\

E2 =

\

D 2 = e 0 E = - 34.722 a r mC m 2

e0

(r) 2

ar =

\

E = E 1 + E 2 = 1.7255 ´ 10 6 a r V/m

and

D = D 1 + D 2 = 15.278 a r mC m 2

2

a r = - 3.9216 ´ 10 6 a r V/m

Note that radial distance r is measured from the centre i.e. origin of the spheres. b) The spheres are shown in the Fig. 3.9. At r = 7.32 cm, all three shells produce D. \

D1 =

r S1 (a 1 ) 2 (r) 2

ar ,

D2 =

r S2 (a 2 ) 2 (r) 2

a r , D3 =

r x (a 3 ) 2 (r) 2

ar

But D = 0 at r = 7.32 cm as given. \

D = D 1 + D 2 + D3 = 0

i.e.

r S1 (a 1 ) 2 + r S2 (a 2 ) 2 + r x (a 3 ) 2 (r) 2 TM


ar = 0


3 - 12


a1 = 3 cm

a2 = 5 cm

a3 = 7 cm

r = 7.32 cm

Fig. 3.9

But r ¹ 0 and a r ¹ 0 \ r S1 (a 1 ) 2 + r S2 (a 2 ) 2 + r x (a 3 ) 2 = 0 \

rx

(

) ( ( )

) (

é -6 -2 2 + - 50 ´ 10 -6 ´ 5 ´ 10 -2 ê 200 ´ 10 ´ 3 ´ 10 = -ê 2 7 ´ 10 -2 êë = – 11.2244 mC m 2

Solution : Consider a sphere of radius 'a' as shown in the Fig. 3.10 (a).

+

\ Dr =

Q 4 p r2

i.e.

0 +

Gaussian surface

2

S

+

a

Fig. 3.10 (a)

= Dr r sin q dq df

ò D· d S =

+

+

2 \ dy = D · dS = Dr a r · r sin q dq df a r

Q = y=

ù ú ú ûú

dS +

Case [1] Consider point P outside sphere such that r > a. The Gaussian surface passes through point P. Now D is directed along a r direction hence D = D r a r . dS = r2 sin q dq df … Spherical system

\

2

r>a

Charged sphere

Example 3.7.8

)

2p

p

ò ò

2

2

Dr r sin q dq df = 4p r Dr

f= 0 q= 0

D=

Q 4 p r2

ar

TM


P +

D


3 - 13


Now total charge enclosed is, Q = ò r v dv v

2p

\

p

a

ò ò ò

Q =

3 2

k r r sin q dr dq df =

f= 0 q= 0 r= 0

\

D = Dr a r =

k a6 2

a r C/m

4pk a6 6

2

6r Case [2] Let point P is on the surface of sphere i.e. 4pk a6 Q and Q = \ ar D = 6 4pa2

… for r > a r=a

k a4 2 a r C/m 6 Case [3] Let point P is inside sphere i.e. Charged sphere r < a. The Gaussian surface passes through point P as shown in the Fig. 3.10 (b).

\

y = Q=

r

+

+

+

Again dS and D are directed radially outwards. 2 \ D · dS = Dr r sin q dq df \

… for r = a

D =

0

+

dS P

D = Dr a r

+

+

ò D· d S

a

S 2p

=

p

ò ò

2

Dr r sin q dq df

Gaussian surface

f= 0 q= 0

Fig. 3.10 (b)

= 4p r 2 D r \

Dr =

Q 4 p r2

i.e.

D=

Q 4 p r2

ar

Now charge enclosed by sphere of radius r only is to be considered and not the entire sphere. 2p p r kr6 Q = ò r v dv = ò ò ò k r3 r2 sin q dr dq df = \ 4p 6 f= 0 q= 0 r= 0

v

\

D =

kr6 4 p 6 4 p r2

ar =

k r4 2 a r C/m 6

The sketch of D against r is, 1 ¶ 2 Ñ· D = ( r D r ) as D is only in a r direction r2 ¶ r

TM


… for 0 < r < a


= Ñ· D =

3 - 14


1 ¶ é 2 ka 6 ù ú=0 êr ´ r 2 ¶ r êë 6 r 2 úû

… for r > a

1 k 5 1 ¶ é 2 kr4 ù 3 ú = 2 ´ 6r = k r êr ´ 2 ¶r 6 6 r êë ûú r

… for r £ a

3

Key Point As Ñ · D = r v = k r as given, the results are correct. ka 6

D in C/m2

4

4

ka

kr 6

6r

6

2

r in m

0

r=a

Fig. 3.10 (c)

Example 3.7.9 Solution : Given D is in cartesian co-ordinates so convert given point P(r = 20 m, f = 55º, z = 5 m) to cartesian. x = r cos f = 11.471, y = r sin f = 16.383, z = 5 At point P, D = 4x a x + 2 (1 – y) a y + 4z a z |P(x, y, z) \

2 D = 45.884 a x – 30.766 a y + 20 a z C/m

Given area 1 mm ´ 1 mm = 10

– 6

z

2

m is very

small i.e. differential dS hence dS = dS a n where a n is normal unit vector to dS.

r

According to gauss's law, dy = D · d S No need to integrate as area is differential.

A(0,0,5)

r

To find a n , consider the cylinder as shown in

dS P(11.471, 16.383, 5)

the Fig. 3.11.

5 an

The normal a n is a r at P. But to find a n in r

cartesian co-ordinates, extend point P radially to meet axis of the cylinder at A (0, 0, 5).

f=55º

The vector AP is now in radial direction at P and x

represents the direction of a n to dS at P.

Fig. 3.11 TM


y


\ an =

3 - 15

11.471 a x + 16.383 a y + 0 a z 11.471 2 + 16.383 2


= 0.5735 a x + 0.8191 a y

–6 \ dS = dS a n = 10 [0.5735 a x + 0.8191 a y ]

\

dy = D · d S at P = (45.884 a x – 30.766 a y + 20 a z ) · d S = 45.884 ´ 10– 6 ´ 0.5735 – 30.766 ´ 10– 6 ´ 0.8191 = 1.114 µC

This is the required flux. Example 3.7.10 Solution : i) The spherical surface at r = 5 encloses all the shells with r1 = 1, r2 = 2 and r3 = 3. Q = Charge enclosed by surface at r = 5 \ = Charge enclosed by surfaces (r1 + r2 + r3) = Q1 + Q2 + Q3 = r S 1 ´ 4 p r12 + r S 2 ´ 4 pr22 + r S 3 ´ 4 pr 32 = 4p [20 ´ 10 – 9 ´ 12 – 9 ´ 10– 9 ´ 22 + 2 ´ 10– 9 ´ 32] = 25.1327 nC \ Flux leaving the surface at (r = 5) = Q = 25.1327 nC ii) P(1, – 1, 2) is in cartesian form \

r =

(1) 2 + ( -1) 2 + ( 2) 2 = 2.4494 m

Hence shells r1 = 1 and r2 = 2 are enclosed. Refer section 2.8.5 where D =

rSa 2 r2

a r for a

spherical shell with radius a at r > a. \ D at P = D1 due to (r1 = 1) + D2 due to (r2 = 2) while D at P = 0 due to r3 = 3 as r = 2.4494 is inside the shell. r (a ) 2 r (a ) 2 \ D at P = S 1 1 a r + S 2 2 a r ( r) 2 ( r) 2 é 20 ´ 10 - 9 ´ (1) 2 ( -9 ´ 10 - 9 ) ´ (2) 2 ù 2 = ê + ú a r = – 2.667 a r nC/m 2 2 (2.4494) êë (2.4494) úû Example 3.7.11 Solution :

The region is shown in the Fig. 3.12 (a).

a) Region r < 2 m For this region, there is no charge enclosed hence D = 0. b) Region 2 < r < 4 m Consider Gaussian surface in cylindrical form of height L and radius r such that 2 < r < 4 as shown in the Fig. 3.12 (b) TM



3 - 16


3

r C/m

z ¥

Dotted shown is Gaussian surface

2m 4m

L

–¥

(a)

(b)

Fig. 3.12

D = Dr a r

... D is in radial direction.

d S = r df dz a r Q =

ò

×

D dS=

S

2p

L

ò ò

×

r dr df D r = r D r 2 p L

... a r a r = 1

z = 0f = 0

But Q enclosed by Gaussian surface at 2 < r < 4 is, Q = r ´ [Volume of Gaussian surface at 2 < r < 4] = r

L

ò

2p

r

ò ò

r dr df dz

z= 0 f= 0 r= 2

[

]

( ) r (r 2 - 4) =

= r ´ p r 2 - p ´ ( 2) 2 ´ L = p r r 2 - 4 ´ L

(

)

\p r r 2 - 4 ´ L = r D r 2 p L \

D =

(

r r2 - 4 2r

)

i.e.

Dr

2r

a r C/ m 2

... 2 < r < 4

c) Region r > 4 m Again Q = r Dr 2 p L But Q enclosed by Gaussian surface is Q enclosed by the entire cylindrical region of length L as r > 4 m. TM



\

3 - 17

Q = r ´ Volume enclosed = r

L

ò

Electric Flux Density and Gauss's Law 2p

4

ò ò

r dr df dz

z= 0 f= 0 r= 2

= r´ \

{p[4 ]- p [2] } ´ L = r p (16 - 4) L = 12 r p L 2

2

12r p L = r D r 2 p L

\

Dr =

6r r

6r a r C/ m 2 r

i.e. D =

... r > 4


a) To find Q tot use standard result as r v is constant. Q tot =

ò

4 p (r) 3 r v 3

r v dv =

v

=

...

r = 10 cm

ò

dv =

v

4 p (r) 3 3

4 p ( 0.1) 3 ´ 4 = 0.016755 mC 3 p

2p 0. 1 m

ò ò òrv r

Alternatively, Q tot =

2

sin q dr dq df = 0.016755 mC

q= 0 f= 0 r = 0

b) To find D r , consider a Gaussian surface as a sphere of radius r as shown in the Fig. 3.13. Consider dS at point P. r = 10 cm The D is in a r direction hence D = D r a r and dS normal to a r is r 2 sin q dq df. \ dS = r 2 sin q dq df a r \

Q =

×

ò

D dS =

S

\

Q = Dr

\

Dr =

2p

p

ò

ò

r2

p 0

[- cos q ]

Q

D r r sin q dq df

4 p r 3 rv 3

D

Fig. 3.13

×

... ( a r a r = 1)

[f]20 p

D =

and

4 p r2

ar P

Gaussian surface

2

f= 0 q= 0

dS r

Q 4 p r2

ar

But

Q =

\

4 p r 3 4 ´ 10 -6 = 1.333 r mC / m 2 D = 3 2 4pr

for a sphere of r

c) Let charge between 10 cm < r < a is Q 1 . \ Q1 =

ò

v

r v dv =

2p

ò

p

ò

a

ò

p

2p

r v r 2 sin q dr dq df = [ - cos q ]0 [f]0

f = 0 q = 0 r = 0.1

TM


a

ò

r = 0.1 r

-3 r 2 3

+ 0.001

dr


Put r 3 + 0.001 = u \

3 - 18


3 r 2 dr = du

i.e.

a

Q 1 = 2p ´ 2 ´

ò

r = 0.1

-

du a = 4 p [ - ln u ]r = 0.1 u

Resubstitute u = r 3 + 0.001, \

[

é a 3 + 0.001 ù = - 4 p ê ln ú nC 0.1 2 ´ 10 -3 û ë

]

Q 1 = -4 p ln r 3 + 0.001

a

Hence the total charge for 0 < r < a is, Q tot + Q 1 i.e. resultant charge Q R is é a 3 + 0.001 ù Q R = 0.016755 ´ 10 -6 - 4 p ln ê ´ 10 -9 C -3 ú ë 2 ´ 10 û But required Q R = 0 é a 3 + 0.001 ù \ 4 p ln ê ´ 10 -9 = 0.016755 ´ 10 -6 -3 ú ë 2 ´ 10 û a 3 + 0.001

\ \

2 ´ 10 -3

i.e.

= e 1.3333 = 3.7936

i.e.

é a 3 + 0.001 ù ln ê = 1.3333 -3 ú ë 2 ´ 10 û a 3 = 6.5872 ´ 10 -3

a = 0.1874 m = 18.74 cm

Example 3.7.13 Solution : The charge enclosed by the cylinder is given by, Q = Charge density ´ Area Let length of each cylinder is 'L' Area = 2p R ´ L \ For cylindrical sheet 1, Q 1 = 5 ´ 2p ´ 2 ´ L = 20pL C

... R = 2 m

For cylindrical sheet 2, Q 2 = - 2 ´ 2p ´ 4 ´ L = - 16pL C

... R = 4 m

For cylindrical sheet 3, Q 3 = - 3 ´ 2p ´ 5 ´ L = - 30 pL C Charge enclosed a D = Area of cylindrical shell considered r

... R = 5 m

The shells are shown in the Fig. 3.14. For R 1 = 1 m,

Q = Charge enclosed = 0 C

For R 2 = 3 m,

Q = Q 1 = 20p L C, D =

\ D = 0 C m2

Q 20pL 10 = = a C m2 2pR 2 L 2p ´ 3L 3 r

For R 3 = 4.5 m, Q = Q 1 + Q 2 = 20p L - 16pL = 4pL C

TM



3 - 19


R=5 R4 = 6 m R=4 R3 = 4.5 m

R=2

R2 = 3 m

R1 = 1 m

Fig. 3.14

Q 4pL = a = 0.444 a r C m 2 2pR 3 L 2p ´ 4.5 ´ L r

\

D =

For R 4 = 6 m,

Q = Q 1 + Q 2 + Q 3 = 20p L - 16pL - 30 p L = - 26 pL C

\

D =

- 26 pL Q = = - 4.333 a r C m 2 2pR 4 L 2p ´ 6 ´ L

Example 3.7.14

Solution :

1)

ær ö -9 ç a r ÷ ´ 10 è4 ø D = E = e0 8.854 ´ 10 - 12

\

E =

2)

Q =

0.25 ´ 10 - 9 4 ´ 8.854 ´ 10 - 12

ò

... r = 0.25 m

a r = 7.0589 a r V/m

D· d S

S

dS = r 2 sin q dq df \

Q =

ò

S

æ r a ö ´ 10 - 9 r 2 sin q dq df (a ) ç ÷ r è4 r ø

2p

=

ò

f= 0

=

... Normal to a r in spherical

p

ær 3 ö r3 -9 sin d d 10 q q f ´ 10 - 9 [- cos q] 0p [f] 02 p ç ÷ ´ = ò è4 4 ø q= 0

r3 ´ 10 - 9 ´ [1 + 1] [2 p] = r 3 ´ 10 - 9 ´ p C 4

For a sphere of radius r = 0.25 m, Q = (0.25) 3 ´ 10 - 9 ´ p = 49.087 pC

TM


... (1)


3 - 20


3) According to Gauss's law, the total flux leaving is same as the charge enclosed. \

y = Q = r 3 ´ 10 - 9 ´ p

\

y = (0.35) 3 ´ 10 - 9 ´ p = 134.695 pC

at

r = 0.35 m


y 2 z 3 a x + 2 xyz 3 a y + 3 xy 2 z 2 a z

D =

i) As x = 3 is constant, the differential surface area is dS = dydz and the direction is + a x . \ dS = dy dz a x

× dS = y z dy dz According to Gauss’s law, ò D× dS = Q = Flux passing

\

2

D

3

... a x

×a

x

= 1

S

1

\

ò ò

Q =

2

2

2

y z

z= 0 y= 0

ii) At P (3, 2, 1),

3

x = 3, y = 2, z = 1

\

E =

\

E at P =

1

é y 3 ù é z4 ù 1 dy dz = ê = 0.6666 pC ú ê ú = 2.667 ´ 3 4 4 úû 0 ë û 0 êë D = 4 a x + 12 a y + 36 a z

i.e.

D 1 = [4 a x + 12 a y + 36 a z ] ´ 10 - 12 e0 e0 1 ´ e0

4 2 + 12 2 + 36 2 ´ 10 - 12 =

1456 8.854 ´ 10 - 12

´ 10 - 12 = 4.3096 V/m

Example 3.9.7 Solution : i) Q =

ò

r v dv =

10 e - 2r r 2 sin q dr dq df

v 2p

p

ò

=

ò

r

ò ò

10 e - 2r r 2 sin q dr dq df

f = 0 q = 0 0 r

ò

e - 2r r 2 dr = r 2

0

ò

e - 2r dr - ò 2r ò e -2r dr dr

=

r 2 e - 2r 2r e - 2r r 2 e - 2r -ò + r ò e -2r dr - ò 1ò e -2r dr dr dr = -2 -2 -2

=

æ e - 2r ö r 2 e -2r e - 2r + r ç ÷-ò dr -2 -2 è -2 ø r

é r 2 e -2r r e -2r 1 -2r ù r 2 e -2r r e -2r 1 -2r 1 = ê- e + - e ú =2 2 4 2 2 4 4 ë û0 \

... By parts

é r 2 e -2r r e -2r 1 -2r 1 ù 2p Q = 10 ê - e + ú [- cos q] p0 [f] 0 2 2 4 4 û ë TM




é r 2 e -2r r e -2r 1 -2r 1 ù = 40 p ê - e + ú C 2 2 4 4 ë û 2r 2r é e e e - 2r Q 1 ù 2 a r = 10 ê D = + ú a r C/ m 2 2 2 2 2 r 4r 4r û 4p r ë

\ ii)

3 - 21

¶ 2 ... D only in r direction r Dr ¶r 1 ¶ = - 5 r 2 e - 2r - 5 r e - 2r - 2.5 e - 2r + 2.5 r 2 ¶r 1 = - 10 r e - 2r + 10 r 2 e - 2r - 5 e - 2r + 10 r e - 2r + 5 e - 2r + 0 2 r

Ñ· D =

1

r2

{

}

{

= -

}

10 e - 2r 5 e - 2r 10 e - 2r 5 e -2r = 10 e -2r = r v + 10 e - 2r + + r r r2 r2

Hence the result obtained is correct. Example 3.9.8 Solution : According to Gauss’s Law, Q =

ò

×

( Ñ D) dv

v

dv = 10 - 9 m 3

Here,

×

Ñ D =

at origin

i.e.

×

at origin

´ dv = [0 + 0 + 2] ´ 10 - 9 = 2 ´ 10 - 9 C = 2 nC


×

Use Gauss's law in point form, Ñ D = r v

Given D in spherical coordinates hence,

×

Ñ D = Now \

x=y=z=0

¶D x ¶D y ¶D z + + = - e - x sin y - e - x ( - sin y) + 2 ¶x ¶y ¶z

dQ = ( Ñ D)

\

×

dQ = ( Ñ D) dv

and

1 r2

D r = 10 sin q,

×

Ñ D = = =

1 r2 1 r2

¶ ¶r

(r 2 Dr ) + r sin1 q

D q = 2 cos q, ¶ ¶r

r2

Df = 0

(r 2 10 sin q) + r sin1 q

10 sin q

10 sin q

¶ Df ¶ 1 sin q D q ) + ( r sin q ¶ f ¶q

¶ ¶r

(r 2 ) + r sin1 q 1

¶ ( 2 sin q cos q) + 0 ¶q ¶ ( sin 2 q) ¶q

( 2r ) + r sin q [ 2 cos 2 q ] =

TM

20 sin q 2 cos 2 q + r r sin q


…Spherical


\

rv =

3 - 22


20 sin q 2 cos 2q C m3 + r r sin q

cos 2 q = cos 2 q - sin 2 q

Now \

rv

[

2 cos 2 q - sin 2 q 20 sin q = + r r sin q

]

=

20 sin 2 q + 2 cos 2 q - 2 sin 2 q 18 sin 2 q + 2 cos 2 q = r sin q r sin q

=

sin q é 2 cos 2 q ù sin q ú = ê18 + r r ê sin 2 q úû ë

[18 + 2 cot 2q ] C / m 3


i) The surface x = 2, 0 £ y £ 2 and 0 £ z £ 2 is shown in the Fig. 3.15. z

Key Point The plane is parallel to yz plane

and unit vector normal to the plane is a x . z=2

The direction away from the origin is a x and dS = dy dz. \ dS = dy dz a x \ as a x \

×

×

2 2

x=y=z=0 y=2

D d S = 2y z dy dz a x = 1 and a y a x = a z a x = 0 Q =

ò

×

×

D dS

S

x=2 x

2

=

×

ò

2

ò

x=2

ax

2y 2 z 2 dy dz

Fig. 3.15

z= 0 y = 0 2

2

éy 3 ù éz3 ù 2´ 8 8 = 2 ê ú ê ú = 3 ´ 3 = 14.22 pC 3 3 úû 0 ë êë û0 ii) r = 1 mm for the sphere. So incremental volume of the sphere is, dv =

(

4 4 p r 3 = ´ p ´ 1 ´ 10 - 6 3 3

)

3

= 4.1887 ´ 10 -18 m 3

\

dQ = charge contained in volume dv = r v dv

Now

rv =

[Ñ× D]at P (2, 2, 2)

¶D y ¶D z ù é ¶D = ê x + + ú ¶y ¶z û ë ¶x ( 2, 2, 2)

TM


y

Plane bounded by 0£y£2 0£z£2 x=2


=

3 - 23

[0 + 6xyz

2

]

+ 2xy

(2, 2, 2)


= 96 + 8 = 104 pC/m 3

dQ = 104 ´ 10 -12 ´ 4.1887 ´ 10 -18 = 4.3562 ´ 10 -16 pC

\

Example 3.9.11

×

Solution : i) r v = Ñ D =

¶ Dx ¶ Dy ¶ Dz = y 2 + x 2 + 1 C/ m3 + + ¶x ¶y ¶z

ii) Assume given D in spherical co-ordinate system.

×

rv = Ñ D = =

r2

¶ ¶r

(r 2 Dr ) + r sin1 q

¶ ¶q

1

( sin q D q ) + r sin q

¶ Df ¶f

[ ]

¶ 1 2 r2 + 0 + 0 = ´ 2r = C / m 3 2 r ¶r r

1 r

1

2

Example 3.9.12 Solution : D = 2r ( z + 1) cos f a r – r ( z + 1) sin f a f + r cos f a z mC m 2 rv = Ñ· D =

i)

=

1 ¶ rDr r ¶r

[

1 ¶ 1 ¶ ¶ r ´ 2r ( z + 1) cos f] + –r( z + 1) sin f] + [r cos f] [ [ r ¶r r ¶f ¶z

= 4 [z + 1] cos f + ii)

Q =

1 ¶ D f ¶D z + ¶f ¶z

] +r

ò

r v dv

1 ´ –r[z + 1] cos f + 0 = 3[z + 1] cos f C m 3 r

4 p 2 2

ò ò ò

3 ( z + 1) cos f r dr df dz

z= 0 f = 0 r = 0

v 2

4

2 ér 2 ù ù p 2 éz = 3ê ú [ sin f]0 ê + zú = 3 ´ 2 ´ 1 ´ 12 = 72 mC 2 2 ë û0 ë û0

iii) Q =

ò D · dS = ò D · dS + ò D · dS + ò D · dS + ò D · dS + ò D · dS + ò D · dS r=0

S

ò D · dS

=

r=0

ò D · dS =

r=2

r=2

f= 0

f= p 2

z= 0

z= 4

ò [2r ( z + 1) cos f][r df dz] = 0 4 p 2

ò ò [2r ( z + 1) cos f][(r d f dz)]

z = 0 f= 0

4

ù ê 2 + zú = 96 mC. û0 ë

p 2 éz

= 2 ´ ( 2) 2 [ sin f]0

2

TM


...r = 2


ò D · dS =

f= 0

ò D · dS =

f= p 2

3 - 24 4

4

2

2

4

ù p ö ér 2 ù é z 2 æ ò ò – r( z + 1) sin f dr dz = – çè sin 2 ÷ø ê 2 ú ê 2 + zú = –24 mC. û0 ë û0 ë z= 0 r = 0 p 2 2

ò (r cos f) ( –r dr df) = ò ò z= 0

ù p = –ê sin f]0 3 ú [ ë û0 p 2 2

ò ò

=

– r 2 cos f dr df

f= 0 r = 0

2

ér 3

2

=–

8 mC 3 2

r 2 cos f

f= 0 r = 0

Q = 0 + 96 + 0 – 24 –

\

+ r( z + 1) sin f dr dz = 0

z= 0 r = 0

z= 0

z= 4

2

ò[–r( z + 1)sin f][– dr dz] = ò ò

ò D · dS =

ò D · dS


ér 3 ù p sin f]0 dr df = ê 3 ú [ û0 ë

2

8 = + mC 3

8 8 + = 72 mC. 3 3

...Gauss's law is verified.

Example 3.9.13 3 Solution : D = 4 x a x – 2z a y – 2y a z

i) Ñ · D =

¶D x ¶D y ¶D z 2 3 = 12 x C/m + + ¶x ¶y ¶z

2 3 ii) r v = Ñ · D = 12 x C/m

iii) Q =

ò r v dv =

vol

1

ò

1

1

ò

2

ò

12 x dx dy dz

z = -1 y = - 1 x = -1 1

éx3 ù 2 1 1 = 12 ê [- y] -1 [z] -1 = 12 ´ ´ 2 ´ 2 = 32 C 3 3 ú ë û -1 iv) To find Q without finding r v . Q = ò D · dS \ S

Consider all faces of the region – 1 < x, y, z < 1. z

–ax back

y

x

–ay left

+az top

+ay right

+ax front

(a) –1 < x,y,z<1


Fig. 3.16 TM


bottom –az


3 - 25


1) For front surface, x = + 1, dS = dy dz a x 2) For back surface, x = – 1, dS = – dy dz a x 3) For right surface, y = + 1, dS = dx dz a y 4) For left surface, y = – 1, dS = – dx dz a y 5) For top surface, z = + 1, dS = dx dy a z 6) For bottom surface, z = – 1, dS = – dx dy a z 3 For front, D · d S = 4x dy dz = 4 dy dz

…x=1

For back, D · d S = – 4x3 dy dz = 4 dy dz For right D · d S = – 2z dx dz

…x=–1

For left D · d S = 2z dx dz For top D · d S = – 2y dx dy For bottom D · d S = 2y dx dy

ò D · dS = S

1

1

ò

1

ò

z = -1 y = - 1

1

ò

4 dy dz +

ò

z = -1 y = - 1

1

ò

1

ò

1

2z dx dz +

z = -1 x = - 1

ò

ò

(right)

1

ò

ò

– 2 z dx dz

z = -1 x = - 1

1

– 2y dx dy +

y = -1 x = - 1

(left)

1

ò

ò

2 y dx dy

y = -1 x = - 1

(top)

(bottom) 1

\

1

(back)

(front) +

1

4 dy dz +

1

é z2 ù é z2 ù 1 Q = 4 [y] -1 1 [z] -1 1 + 4[y] -1 1 [z] -1 1 - 2 ê ú [x] -1 + 2 ê ú [x] 1-1 2 2 ë û -1 ë û -1 1

1

éy2 ù éy2 ù 1 1 – 2ê + [x] 2 ú [x] -1 ê ú -1 2 2 úû -1 êë úû -1 êë = 4 ´ 2 ´ 2 + 4 ´ 2 ´ 2 + 0 + 0 + 0 + 0 = 32 C

qqq

TM


4

Electric Work, Energy and Potential Solutions of Selected Examples for Practice Kept this unsolved example for student's practice.

Example 4.3.4 Example 4.3.5

Solution : The work done is given by, W = -Q

A

ò

E · dL

B

Let us differential length dL in cartesian co-ordinate system is, dL = dx a x + dy a y + dz a z \

E · dL =

(- 8 xy a x - 4x 2 a y + a z )· (dx a x + dy a y + dz a z )

= - 8 xy dx - 4x 2 dy + dz As a x · a x = a y · a y = a z · a z = 1, other dot products are zero. A A A éA ù W = - Q ò - 8 xy dx - 4x 2 dy + dz = - Q ê ò - 8 xy dx - ò 4 x 2 dy + ò dzú \ êë B úû B B B Case 1 : The path is y = 3x 2 + z, z = x + 4 y = 3x 2 + x + 4 differentiate i.e. dy = (6x + 1) dx A

For

ò

- 8 xy dx ® The limits are x = 1 to x = 2.

B A

For

ò

- 4x 2 y ® The limits are y = 8 to y = 18

B A

For

ò

dz ® The limits are z = 5 to z = 6.

B

\

18 6 é 2 ù W = - Q ê ò - 8 xy dx - ò 4 x 2 dy + ò dzú ê x= 1 ú y= 8 z= 5 û ë

Using y = 3x 2 + x + 4 and dy = (6x + 1) dx and changing limits of y from 8 to 18 interms of x from 1 to 2 we get (4 - 1) TM



\

4-2

Electric Work, Energy and Potential

2 6 é 2 ù W = - Q ê ò - 8x 3x 2 + x+ 4 dx - ò 4 x 2 [6x+ 1] dx + ò dzú ê ú x= 1 z= 5 û ëx = 1

[

]

2 6 é 2 ù = - Q ê ò -24x 3 - 8x 2 - 32x dx - ò 24 x 3 + 4x 2 dx + ò dzú ê ú x= 1 z= 5 û ë x= 1

[

]

(

)

2 é ù 8 4 = - Q ê æç -6x 4 - x 3 - 16x 2 - 6x 4 - x 3 ö÷ + ( z)56 ú øx = 1 3 3 êë è úû

= - Q {- 256 + 1} = - 6 ´ -255 = 1530 J Case 2 : Straight line path from B to A. To obtain the equations of the straight line, any two of the following three equations of planes passing through the line are sufficient, B (1, 8, 5)

and A (2, 18, 6)

y - yB (y - y B ) = A ( x - x B ), xA - x B

(z - z B ) =

Using the co-ordinates of A and B, 18 - 8 y–8 = i.e. (x - 1) 2 -1 \ \ And \ Now

zA - z B x - xB y - y B ), (x - x B ) = A ( (z - z B ) yA - y B zA - z B

y – 8 = 10 ( x – 1)

y = 10x – 2

... (1)

dy = 10 dx z–5 =

6 -5 (y - 8) 18 - 8

i.e.

z–5=

1 (y - 8) 10

10 z = y + 42

... (2)

18 6 é 2 ù W = - Q ê ò - 8 xy dx - ò 4 x 2 dy + ò dzú ê x= 1 ú y= 8 z= 5 û ë 2 6 é 2 ù = - Q ê ò - 8 x(10x - 2) dx - ò 4 x 2 (10dx) + ò dzú ú ê x= 1 x= 1 z= 5 û ë 2 ìé ü ï -80 3 16x 2 40x 3 ù ï = - Q íê x + + [z]56 ý ú 2 3 3 ûx= 1 ïî ë ïþ

= - Q {- 213.33 + 32 - 106.667 + 26.667 - 8 + 13.33 + 1} = - Q [ - 255] = - 6 ´ -255 = 1530 J This shows that irrespective of path selected, the work done in moving a charge from B to A remains same.

TM



4-3


z

Example 4.3.6 Solution : The line charge along the z-axis Circular path and the circular path along which charge is having radius r1 moving is shown in the Fig. 4.1.

Movement of charge Q in z = 0 plane (xy plane) r1

The circular path is in xy plane such that its radius is r 1 and centered at the line charge. Consider cylindrical co-ordinate where line charge is along z-axis.

y r1

system

dL = rdf af

x

Infinite

line charge The charge is moving in a f direction. Fig. 4.1 \ dL = r df a f The field E due to infinite line charge along z axis is given in cylindrical co-ordinates as, rL a E = 2pe 0 r r

The circular path indicates that dL has no component in a r and a z direction. \ dL = r df a f \

W = -Q

final

ò

E · dL = - Q

initial

2p rL rL a r · r df a f = - Q ò df 2pe 0 r 2pe 0

2p

ò

0

(ar ·af ) = 0

0

As a r · a f = 0 as q = 90° between a r and a f . This shows that the work done is zero while moving a charge such that path is always perpendicular to the E direction. Example 4.3.7

z

Solution : The path along which the charge is moved is shown in the Fig. 4.2. As path is straight line from B to A and the line is in xy plane (z = 0 plane), the equation of line can be easily obtained as, y = mx ... Passing through origin Now, B (4, 2, 0) and A (0, 0, 0) x

y - yB 0-2 1 \ y = A x= x= x xA - x B 0-4 2 \

x = 2y

i.e.

A(0,0,0)

2

B(4,2,0)

Fig. 4.2

dx = 2 dy

dL = dx a x + dy a y + dy a z \

W = -Q

A

ò

B

×

y

4

E dL = - Q

A

ò

B

... Cartesian éæ x ù + 2y ö÷ a x + 2x a y êë çè 2 úû ø

TM

×[dx a


x

+ dy a y + dz a z

]


= -Q

4-4 A

æç x + 2y ö÷ dx + (2x) dy è2 ø

ò

B


... a x

×a

x

= ay

×a

y

=1

0 ì 0 ü x ï ï = - Q í ò æç + 2y ö÷ dx+ ò 2x dyý è2 ø ïî x= 4 ïþ y= 2

... (1)

Using x = 2y and dx = 2dy and changing limits interms of y, in equation (1), 0 ì 0 2y ü ï ï æ ö W = - Q í ò ç + 2y ÷ (2dy) + ò 2 ( 2y) dyý è 2 ø ïþ y= 2 îï y = 2 0 0 ü ì 0 é 6y 2 4y 2 ù ï ï = - Q í ò 6y dy + ò 4y dyý = - Q ê + 2 2 úú ê ï ïî y = 2 ë û2 y=2 þ

= - 20 ´ 10 -6 ´ [- 12 - 8] = + 400 mJ Example 4.3.8 Solution : Q = 2 C, B(2, 0, 0), A(0, 2, 0), E = 12 x a x - 4 y a y

×

A

W = –Q

ò

E dL

B

where dL = dx a x + dy a y + dz a z

×

(0, 2 , 0)

= -2

ò

(12 x a x - 4 y a y ) ( dx a x + dy a y + dz a z )

(2, 0, 0)

×

(0, 2 , 0)

= -2

ò

×

y= 2ü x= 0 ì é4 y2 ù ï ï é 12 x 2 ù - 2 íê ê 2 ú ý = -2 ´ - 32 = + 64 J 2 ú úû y = 0 ï û x = 2 êë ïî ë þ

=


×

… (a x a y = a x a z = a y a z = 0)

12 x dx - 4 y dy

(2, 0, 0)

×

×

W = - Q ò E d L = - ( - 2) ò [y a x + x a y ] [dx a x + dy a y + dz a z ] = 2

[ò y dx + ò x dy]

= 2

[ò y ´ 4y dy + ò 2y

…x = 2y 2 2

]

é = 2 4 ò y 2 dy + 2 ò y 2 dy = 2 ê 6 êë

[

…dx = 4y dy

dy

]

TM

2

ò

1

ù y 2 dy ú úû



4-5 2

éy 3 ù 3 3 = 12 ê ú = 4 (2) - (1) 3 êë úû 1

[


] = 4 (8 - 1) = 28 J

Example 4.3.10

z

Solution : The work done is given by,

×

A

P(2, p , p ) 4 2

W = - Q ò E dL B

dL = dr a r + r dq a

Now

+ rsin q df a f

q

III y

I

Consider the path as shown in the Fig. 4.3. Along path I, dq = df = 0

×

\ E dL = 5e - r

4

´ dr

Along path II, dr = dq = 0

×

\ E dL =

II

... (1) x

Fig. 4.3

10 ´ rsinq df rsinq

= 10 df

... (2)

Along path III, dr = df = 0

×

\ E dL = 0

... (3) p 2 p 4

\

2

ò ò ò

W = -Q

[5e - r 4 dr + 10 df + 0]

f = 0q = 0r = 0

ì ü - r 4 ù2 ï p 2 = - 5 ´ 10 í ê + 10[f] 0 + 0ý ú -1 4 û0 ïþ ïî ë ìé - r 4 ù 2 ü 10p ï ï 5e = - 5 ´ 10 - 6 í ê + -1 4 ú 2 ý û0 ïî ë ïþ 10p ü = - 5 ´ 10 - 6 ìí - 20 e - 0.5 + 20 e 0 + = - 117.88 mJ 2 ýþ î - 6 ï é 5e


E = VAB = -

6y x

2

A

ò

ax +

E · dL

6 a +5 az x y where

dL = dx a x + dy a y + dz a z

B

Now a x · a x = a y · a y = a z · a z = 1 and all other dot products are zero. TM



\

E · dL = -

\

VAB = -

4-6

6y x2 A

ò

B

-

dx + 6y x

2


6 dy + 5 dz x dx +

6 dy + 5 dz x

To obtain the integral as it does not depend on the path from B (4, 1, 2) to A ( - 7, 2, 1) we can divide the path as, Path 1,

B (4, 1, 2) to ( - 7 , 1, 2) ® only x varies, y = 1, z = 2.

Path 2,

( - 7 , 1, 2) ( - 7 , 2, 2)

Path 3, \

to to

( - 7 , 2, 2) ® only y varies, x = – 7, z = 2. A ( - 7 , 2, 1) ® only z varies, x = – 7, y = 2.

ìx = - 7 - 6 y ï dx + VAB = - í ò 2 x îï x = 4 y=1

y= 2

6 dy + x

ò

y= 1

z= 1

ò

z= 2

ü ï 5 dzý ïþ

x=–7

-7 1 2 ì ü 1 6 ï ï = - í- 6 ò dx dy + 5 dz ý ò ò 2 7 ïî ïþ x= 4 x y= 1 z= 2 üï ìï 1 -7 6 2 = - í- 6 é - ù - [ y ] 1 + 5 [z]12 ý êë x úû 4 7 ïî þï 1 1 6 ü ì = - í - 6 é + + ù - [2 - 1] + 5 [1 - 2]ý ê ú 7 4 7 ë û þ î = - {- 2.3571 - 0.85714 - 5} = + 8.2142 V Alternatively find the equations for straight line path from B to A by using, y - yB z -z y - yB = A ( x - x B ) and z - z B = y A - y B ( y - y B ) xA - x B A B

and using the relations between x, y and z solve the integrals. From the above equations we get, x = – 11 y + 15 and z = – y + 3 so use y interms of x for first integral and x interms of y for the second integral and integrate. Example 4.5.9 Solution :

\ \

Potential due to the point charge, Q V = 4 p e0 r Q Q and VB = VA = 4 p e 0 rA 4 p e 0 rB VAB = VA - VB = = -

… r = Distance of point from Q

Q é 1 1 ù 4 p e 0 êë rA rB úû

20 ´ 10 - 10 4 p ´ 8.854 ´ 10 - 12

é 1 - 1 ù = 143.8038 V êë 0.5 0.1 úû TM


... Q is negative


4-7


Example 4.5.10

+

Solution : The arrangement is shown in the Fig. 4.4. The potential difference spheres is given by, Q V = 4p e 0

+

between the two concentric

r2

+

+

+

+ +

Inner + sphere

… Refer equation (4.5.2)

+ + +

Outer sphere

+

r1 +

+ +

1ö æ1 çr - r ÷ è 1 2ø

+

+

+ +

+

Fig. 4.4

while E on the surface of inner sphere is, E =

Q 4p e 0 r12

ar

While E due to outer sphere at r = r1 is zero as E inside the spherical shell is zero. r But r1 = 2 … Given 2 Using in V,

V =

Q æ1 1 ö Q æ 1 ö = 4p e 0 çè r1 2r1 ÷ø 4p e 0 çè 2r1 ÷ø

Multiply both sides by

V Q 1 1 1é Q ù 1 = = ê × × ú = | E| r1 4p e 0 2 r 2 2 ê 4p e r 2 ú 2 0 1 û 1 ë

\

\

1 , r1

|E| =

r 2V for r1 = 2 on surface of inner sphere. 2 r1

Example 4.5.11 Solution : Q = 5 nC, V = 2 V at (0, 6, – 8), Q is at origin (0, 0, 0). i) A (– 3, 2 , 6) rA = – 3 a x + 2 a y + 6 a z , \

VA =

9 + 4 + 36 = 7

Q +C 4p e 0 rA

VR = 2 V at (0, 6, – 8)

\

rA =

hence rR = 6 a y – 8 a z , rR = 5 ´ 10 -9 +C 4 p e 0 ´ 10

VR =

Q +C 4 p e 0 rR

VA =

5 ´ 10 - 9 – 2.4938 = 3.926 V 4p e 0 ´ 7

i.e.

2=

i.e. C = – 2.4938

ii) B (1, 5, 7) \

rB = a x + 5 a y + 7 a z ,

1 + 25 + 49 =

rB = TM

6 2 + 8 2 = 10

75



\

VB =

iii)

4-8


Q 5 ´ 10 -9 +C = – 2.4938 = 2.6952 V 4 p e 0 rB 4 p e 0 ´ 75

VAB = VA – VB = 3.926 – 2.6952 = + 1.23 V

Example 4.5.12 Solution : The charges are shown in the Fig. 4.5. The distances between the charges and point P are, R1 =

( 0 - 1) 2 + 0 2 + ( z - 0) 2 = 1 + z 2

R 2 = ( 0) 2 + ( 0 - 1) 2 + ( z - 0) 2 =

z

1 + z2

P(0,0,z) R3

R 3 = ( 0 + 1) 2 + ( 0) 2 + ( z - 0) 2 = 1 + z 2 R4

R 4 = ( 0) 2 + ( 0 + 1) 2 + ( z - 0) 2 = 1 + z 2

Q3 Q4 D (0,–1,0)

The potential at P due to all charges is,

R1 O

a) VP = V1 + V2 + V 3 + V4 =

Q1

1 Q + Q2 + Q 3 + Q4 ] 4 p e0 R [ 1

A(1,0,0)

VP =

4 ´ 6 ´ 10 -9 4p ´ 8.854 ´ 10 -12

´

1 + z2

=

215.7058 1 +z2

V

b) To find VP ( max) , find dV/dz = 0 - 0.5 ù d é 2 215.7058 1 + z 2 \ ê úû = 215.7058 (– 0.5) 1 + z dz ë

(

Thus z = 0

)

(

)

- 1.5

(2z) = 0

or 1 + z 2 = 0 i.e. z = ± j but z coordinate can not be imaginary.

\ At z = 0, VP is maximum i.e. VP ( max ) = 215.7058 V - 1.5 215.7058 z dV = 215.7058 ( - 0.5) 1 + z 2 c) Now ( 2 z) = 1.5 dz 1 +z2

(

d) To find its maximum value, \

d dz

Q2 B(0,1,0)

Fig. 4.5

where R = R 1 = R 2 = R 3 = R 4 = 1 + z 2 \

C (–1,0,0)

R2

(

ì 2 í 215.7058 z 1 + z î

(

)

)

(

)

d ì dV ü =0 dz íî dz ýþ

- 1.5 ü

ý=0 þ

ì \ 215.7058 í z ( - 1.5) 1 + z 2 î

)

- 2.5

(

( 2 z) + 1 + z 2 TM

)

- 1.5 ü

ý=0 þ



\ 215.7058

(1 + z 2 )

- 1.5

4-9

ì 3 z2 ü - 3 z2 + 1ý = 0 i.e. +1 = 0 í2 1 + z2 î 1+z þ 1 = 0.5 2

\ -3 z 2 + 1 + z 2 = 0

i.e. z 2 =

\

0.5 = ± 0.7071

\

z = ± dV ( max) = dz


215.7058 ( 0.7071)

[1 + (0.7071) ]

1.5

2

dV ( max) dz

for

= 83.024 V/m

Example 4.5.13 Solution : The charges are placed as shown in the Fig. 4.6. y – ax P

Q

Q

Q

Q

1

2

4

8

x

Fig. 4.6

The potential at a point due to a point charge is, Q V = 4 p e0 R For charge at x = 1, R 1 = 1 = ( 2) 0 ,

For charge at x = 2, R 2 = ( 2) = ( 2) 1

For charge at x = 4, R 3 = 4 = ( 2) 2 ,

For charge at x = 8, R 4 = 8 = ( 2) 3

\ VP = V1 + V2 + V 3 + V4 + ... =

ù Q ´ 10 -9 é 1 1 1 1 + + + + ....ú ê 4 p e 0 ê ( 2) 0 ( 2) 1 ( 2) 2 ( 2) 3 úû ë

From the series, a + a r + a r 2 + ...+ a r ¥ the sum given by, S = Comparing the series, a = 1,

\ Similarly

VP = E =

r=

1 hence S = 2

Q ´ 10 -9 4 p ´ 8.854 ´ 10 -12 Q 4 p e0 R2

aR

1 1-

1 2

a 1-r

=2

´ 2 = 17.9754 Q V aR = -ax

where TM



\

EP

4 - 10

é Q ´ 10 -9 ê 1 1 = 1+ + 4 p e 0 ê ( 2) 2 22 êë

EP =

1

( ) (2 2 )

Comparing the series, a = 1,

\

+


r=

1

hence

22

Q ´ 10 -9 ´ ( 4 / 3) 4p ´ 8.854 ´ 10 -12

2

S=

3

ù ú + ...ú ( - a x ) úû

a = 1-r

1 1-

( - a x ) = – 11.983 Q a x

1

=

4 3

22

V/m

Example 4.5.14 Solution : \

rA = 5 m, rB = 15 m, Q = 500 pC at origin VA = Absolute potential at A =

Q = 0.8987 V 4pe 0 rA

VB = Absolute potential at B =

Q = 0.3 V. 4pe 0 rB

\

VAB = VA - VB = 0.8987 - 0.3 = 0.5987 V.

Also

VAB =

Q é 1 1 ù 500 ´ 10 -12 é 1 - 1 ù = 0.599 V = 4pe 0 êë rA rB úû 4p ´ 8.854 ´ 10 -12 ëê 5 15 ûú

Example 4.5.15 Solution : Q = 15 nC at origin i)

Find V1 at P(2, –3, –1)

\

V1 =

\

V1 =

Q 2 and r1 = ( 2 – 0) 2 + ( –3 – 0) + ( –1 – 0) 2 = 4 p e 0 r1 15 ´ 10 –9 4 p ´ 8.854 ´ 10 –12 ´

14

= 36.0311 V

14

ii) V = 0 at (6, 5, 4) \

V1 =

Q +C 4 p e 0 r1

\

0 =

Q +C 4 p e 0 r2

\

0 =

\

V1 =

and

V1 = 0 at (6, 5, 4)

where r2 = ( 6 – 0) 2 + (5 – 0) 2 + ( 4 – 0) 2 = 77

15 ´ 10 –9 4 p ´ 8.854 ´ 10 –12 ´

77

+C

i.e.

C = –15.3637

Q – 15.3637 = 36.0311 – 15.3637 = 20.6673 V 4 p e 0 r1 TM



4 - 11

Electric Work, Energy and Potential y

Example 4.6.3 Solution : r L = 1 nC/m along x-axis and find V at A (0, 1000) mm.

A(0,1)m VA = ?

Consider elementary charge dQ on length dx at distance x from origin. \ dQ = r L dx Distance of point A from charge dQ is, R = \ d VA

d

x2

Ö x2+1

R=

1m L

C

B

+1

0

(–0.5,0) in m

(+0.5,0) in m

x

dQ = 4 p e0 R =

r L dx

\

ò

- 0.5

= =

r L dx 4 p e0

2r L 4 p e0

0.5

ò 0

x2 + 1

= 2

+ 0.5

ò

0

r L dx

....

2

x +1

... changing limits

x2 + 1

4 p e0

dx

ò

dx x2

2 ´ 1 ´ 10 - 9

z

Solution : Q = 10 -8 C, r = 5 m, h = 5 m. The ring is shown in the Fig. 4.8.

=

= ln é x + x 2 + a 2 ù ûú ëê

é 0.5 + 0.5 2 + 1 ù - ln [1]ü ý úû êë þ

Example 4.6.4

rL =

+a2

0.5 2 rL ì é 2 ùü ln x + x + 1 úûýþ 0 4 p e 0 íî êë

ì ln í 4 p ´ 8.854 ´ 10 - 12 î = 8.649 V =

VA

Fig. 4.7

+ 0.5

VA =

\

z

x2 + 1

4 p e0

x

dx

Q 10 -8 = circumference 2pr

P(0,0,5)

10 -8 10p

z rL

R

= 0.3183 nC m

And

y

O

Consider the differential length dL on the ring. dQ = r L dL \

dL

dL = r d f = 5 d f ... in xy plane

x

r=5m

Fig. 4.8 TM


af


\

4 - 12

dQ = 5 r L df dQ dV = and R = 4pe 0 R

\

dV =

\

V =

5r L df

=

4pe 0 5 2 + 5 2 2p

ò

f= 0

5r L df 4pe 0 50

=


r 2 + z2 5 r L df 4pe 0 5 2 + 5 2 5r L

4pe 0 50

=

´ [f] 20 p =

5 r L df

... r = z = 5

4pe 0 50 5 ´ 0.3183 ´ 10 -9 ´ 2p 4p ´ 8.854 ´ 10 -12 ´ 50

= 12.7102 V

Example 4.7.4 y

Solution : The ring lies in z = 0 i.e. xy plane as shown in the Fig. 4.9. Consider the differential surface area dS at point P at a distance of r from the origin.

dS R+1 r rS

Hence differential charge dQ is dQ = r S dS

O

P x

R

The dS in the xy plane is r dr df \

dQ = r S r dr df

\

dV =

\

V =

r r dr df dQ = S 4 p e0 r 4 p e0 r 2p R + 1

ò ò

f= 0 r = R

= =

z

Fig. 4.9

r S dr df 4 p e0

rS [r]RR + 1 [f]20 p 4 p e0 rS r [R + 1 - R][ 2 p ] = 2 eS V 4 p e0 0

This shows that the potential at the origin due to the ring is independent of the inner radius R. Example 4.7.5 Solution : Note that the charge is distributed over the entire surface of disc so it is a surface charge. The surface charge density is, 40 40 Total charge 3 3 rS = = = = 1.061 nC/m 2 Area p r2 p ´ ( 2) 2 Refer Ex. 4.7.2 and hence

TM



V = \

4 - 13


rs é 2 a + h 2 - h ù with a = 2 m, h = 5 m úû 2e 0 êë

V = 23.0776 V

Example 4.8.1 Solution : Consider a sphere of radius R with a uniform charge density r v . Case 1 : Let point P is outside sphere (r > R).

Gaussian surface

The E is directed radially outwards, along a r direction. ... Normal to a r dS = r 2 sin q dq df a r

×

\ dy = D d S = Dr a r

×

y =

ò

S

Solving,

+ + + + + R + + + ++ + + + + + + + + + + + + + + + ++

... Gauss’s law 2

r sinq dq df

= D r r 2 sin q dq df \

+

×

D dS = Dr =

2p

p

ò ò

(a r

×

a r = 1)

D r r 2 sin q dq df = Q i.e. D =

4pr 2 Q

ar

4pr 2

E =

Now

V = - ò E d L and d L = dr a r

\

V = -

... For r > R

ar

×

r

ò

r= ¥

Q 4pe 0 r

dS

Fig. 4.10

Q

and

4pe 0 r 2

E

Charged sphere

f= 0 q= 0

Q

P

2

ar

×

dr a r = -

r

ò

r= ¥

Q 4pe 0 r 2

dr

Key Point The limits to be taken against the direction of the E i.e. from r = ¥ to r.

\

V = -

Q 4pe 0

At r = ¥, V = 0 i.e. K = 0

r

ò

r= ¥

1 r2

hence

r Q é- 1 ù + K + K =+ êë r úû r = ¥ 4pe 0 r

dr = -

Q 4pe 0

V=

Q 4pe 0 r

Now total charge contained by sphere is, Q = Volume of sphere ´ r v =

\

4 pR 3 r v 3

4 pR 3 r v R 3 r v V = 3 = 4pe 0 r 3re 0

... r > R ... (1)

This is potential outside the spherical shell. TM



4 - 14

Electric Work, Energy and Potential Charged sphere

Case 2 : Let point P is inside sphere (r < R). D = D r a r and dS = r 2 sin q dq df a r

×

\ dy = D d S = D r r 2 sin q dq df \

y = Q =

ò

S 2p

=

p

ò ò

Gaussian surface

r

×

D dS

P R E = Erar D = Drar

D r r 2 sin q dq df = 4p r 2 D r

f= 0 q= 0

\

Dr =

\

Q =

Q

Q

and D =

Fig. 4.11

ar

4pr 2 4pr 2 The charge Q enclosed by radius r < R must be considered.

ò

r v dv =

2p

p

r

ò ò ò

r 2 sin q dr dq df =

f= 0 q= 0 r = 0

v

\

4 3 pr r v rr v ar = ar D = 3 2 3 4pr

\

V = - ò E d L and d L = dr a r

i.e.

E=

×

4 3 pr r v 3

D rr v = a 3e 0 r e0

The limits of r are from r = R to r against the direction of E. \

V = -

r

ò

r= R

r

-rv rr v -rv ér2 ù r 2 - R2 dr = + K1 = 6 e0 3e 0 3e 0 ê 2 ú ë ûR

From equation (1), for r = R, V = \

R 2r v 3 e0

(

) + K1

... (2)

R 2r v 3 e0

= 0 + K1

... (3)

Using in (2), V=-

ù R 2r v rv r é R2 - r 2 r 2 - R2 + = v ê + R2 ú = 6 e0 3 e0 3 e0 2 û ë

(

)

(

r v 3R 2 - r 2 6 e0

Example 4.9.2 Solution : The two line charges are shown in the Fig. 4.12. Now V = 100 V at the origin O (0, 0, 0). TM


)

... r < R ... (4)


4 - 15


Let us obtain potential difference VPO using standard result. x=1,z=2 A(1,y,2)

Case 1 : Line charge 1 \

VPO1 = +

rL ér ù ln ê O1 ú 2pe 0 ë rP1 û

\

rO1 =

(1 - 0) 2

\

rP1 =

(1 - 4) 2 + ( 2 - 3) 2 = 10

\ But \

VPO1

P Line 1

where rO1 and rP1 are perpendicular distances of points O and P from the line 1. The line 1 is parallel to y-axis so do not use y co-ordinates to find rO1 and rP1 . + ( 2 - 0) 2

O (0,0,0)

y x=–1, y=2

x

= 5

B(–1,2,z)

Fig. 4.12

rL é 5 ù = + = - 49.8386 ln ê 2pe 0 ë 10 úû VPO1 = VP1 - VO

Line 2

z

where

VO = 100 V

– 49.8386 = VP1 - 100

... Absolute potential of P due to line charge 1 \ VP1 = 50.16 V Case 2 : Line charge 2, which is parallel to z-axis. Do not consider z co-ordinate to find perpendicular distance. \

rO2 =

( -1 - 0) 2 + ( 2 - 0) 2 = 5

and

rP2 =

( -1 - 4) 2 + ( 2 - 1) 2 = 26 rL é 5 ù = - 118.5417 V ln ê 2pe 0 ë 26 úû

\

VPO2 =

But

VPO2 = VP2 - VO

where VO = 100 V

\ VP2 = – 118.5417 + 100 = – 18.5417 V This is absolute potential of P due to line charge 2 \ VP = VP1 + VP2 = 50.16 – 18.5417 = 31.6183 V Note Students can use the method of using consant C to find absolute potential of P due

to line charge 1 and line charge 2. Adding the two, potential of P can be obtained. The answer remains same. For reference, the constant C 1 = C 2 = 215.721 for both the line charges. Example 4.9.3 Solution : The arrangement is shown in the Fig. 4.13.

TM



4 - 16

Let rL be the infinite line charge at the centre of one of the cylindrical tubes. Then potential difference due to r = 15 cm infinite line charge is given by, r r L VAB = ln B 2p e 0 rA r

\

r

–8

L

= 7.7621 × 10

0.9 m rB A

B

r = 15 cm

C

rA

0. 9 L 2500 = ln 2p e 0 0.15

\


0.15 m

V = 750 V rC = 0.2567 m

Fig. 4.13

C/m

The point at which VAC = 750 V is to be obtained. \

r

VAC =

\

L

2p e 0

ln

rC rA

i.e. 750 =

r 7.7621 ´ 10 -8 ln C 2p e 0 0.15

rC = 0.2567 m

z E1 E 2

Example 4.9.4 Solution : a) The charges are shown in the Fig. 4.14 (a).

rS1

For z > 1.5 m, r E1 = S1 a z due to r S1 2 e0 E2 \

z = 1.5 m

O

r = S2 a z due to r S2 2 e0

E = =

E1

E2

rS2

1 r + r S2 ] a z 2 e 0 [ S1 1 2 e0

z = –0.5 m

[50 e 0 - 50 e 0 ] a z

= 0 V/m

x E1

for z > 1.5 m

E2

Fig. 4.14 (a)

For – 0.5 < z < 1.5, E1 =

r S1 2 e0

(- a z )

\E =

1 2 e0

[ - r S1 + r S2 ] a z

and E2 =

r S2 (a z ) 2 e0

= - 50 a z V/m For z < – 0.5, E1 =

r S1 (- a z ) 2 e0

y

and TM



4 - 17

E2 =

r S2 2 e0

(- a z )

\E =

1 2 e0

[ - r S1 - r S2 ] a z

=

1 2 e0


[ -50 e 0 + 50 e 0 ] a z

= 0 V/m \

E = 0 V/m for z > 1.5 m = - 50 a z V/m for – 0.5 < z < 1.5 = 0 V/m for z < 1.5

b) To find potential at any point on z axis. dL = dx a x + dy a y + dz a z Select any point A (x, y, z) and B (x, y, – 0.5)

×

VAB = - ò E d L = - ò

\

( -50 a z ) (dx a x

+ dy a y + dz a z

)

z= z

= + 50

ò

dz

... z varies from – 0.5 to z for points B to A.

z = - 0.5

= + 50 [z]-z 0.5 = 50 [z + 0.5] V But

VAB = VA - VB

where VA is potential at any point z VB is potential at z = – 0.5 which

V(z)

is zero. \

100 V

VA = 50 [z + 0.5] V = V(z)

The graph of V(z) as a function of z is shown in the Fig. 4.14 (b).

z –0.5

For z > 1.5 and z < – 0.5, the potential is zero as E is zero in that region.

1.5

Fig. 4.14 (b)

Example 4.9.5 Solution : The various charges are shown in the Fig. 4.15. There are three charge configurations. TM



4 - 18


z

x = –3plane

P(5,6,7)

dS Q1 C(3,–1,2)

rS y

rL x

Fig. 4.15

Case 1 : Point charge Q 1 = 200 p e 0 C at C ( 3, - 1, + 2). Q1 VP = + C 1 where C 1 = Constant 4 p e0 R1 R1 =

2 (5 - 3) 2 + [ 6 - ( -1)] + [7 - 2]2

\ R 1 = 78 To find C 1 , V = 0 V at Q (0, 0, 1) Q1 + C1 \ VQ = 4 p e0 R2 where

R2 =

\

0 =

\

VP =

... Distance between P and C

[0 - 3]2 + [ 0 - ( -1)]2 + [1 - 2]2 200 p e 0 4 p e 0 11 200 p e 0

+ C1

4 p e 0 ´ 78

i.e.

= 11

C 1 = – 15.0755

– 15.0755 = – 9.4141 V

Case 2 : Due to line charge along x axis. rQ rL ln VPQ = 2 p e0 rP

... Potential difference

As line charge is along x axis, any point on it (x, 0, 0). \

rQ =

( 0 - 0) 2 + (1 - 0) 2 = 1

and

rP =

( 6 - 0) 2 + (7 - 0) 2 = 85 40 p e 0 ln 2 p e0

\

VPQ =

But

VPQ = VP - VQ

... ^ Distance from line charge ... x not considered

1 = – 44.4265 V 85 and

VQ = 0 V

... Absolute potential of P \ VP = VPQ + VQ = – 44.4265 V Case 3 : Surface charge in the plane x = – 3 i.e. parallel to yz plane. TM



4 - 19


×

A

Note As E due to infinite surface charge is known use V AB = -

ò

E dL.

B

So

E =

rS a 2 e0 x

... a x is normal to yz plane

Point P is infront of plane as x co-ordinate of P is 5 hence + a x . dL = dx a x + dy a y + dz a z rS dx \ E dL = 2 e0

×

\

VPQ = -

P

ò

Q

xQ = 0

and

5

0

... Potential between P and Q

rS 5 rS 5 8e dx = = - ´ 0 = – 20 V 2 e0 2 e0 2 e0

VPQ = VP - VQ

But

VQ = 0 V

and

VP = – 20 V Total VP = – 9.4141 – 44.4265 – 20 = – 73.8406 V

... Absolute potential of P


¶V ¶V ù é ¶V ax + ay + az ú E = -Ñ V = -ê x y z ¶ ¶ ¶ û ë é ¶V ê - ( 2x) = 2 y ( 2x) + 0 - 4 ê ¶x 2 2 êë x + y

(

)

é -2y ¶V ê 2 = 2x + 0 - 4 ê ¶y 2 2 êë x + y

(

)

ù 8x ú = 4xy + 2ú x2 + y 2 úû

(

ù 8y ú = 2x 2 + 2ú x2 + y 2 úû

(

)

)

2

2

¶V = 0 + 20 - 0 = 20 ¶z \

×

x P = 5 hence

VPQ = - ò

\ \

rS dx 2 e0

×

... a x a y = a x a z = 0

ìé 8x ïê E = - í ê 4xy + ïê x2 + y 2 îë

(

)

ù é 8y ú ê a + 2x 2 + 2ú x ê x2 + y 2 úû êë

(

)

ü ù ï ú a + 20 a z ý 2ú y ï úû þ

{ [- 60 + 0.0268]a x + [72 - 0.0112]a y + 20 a z }

\

E at P = -

\

= + 59.9732 a x - 71.9888 a y - 20 a z V/m D at P = E at P ´ e 0 = 0.531 a x - 0.6373 a y - 0.177 a z nC/m 2 TM



4 - 20

Now

rv = Ñ· D

and

D = e0 E

hence


(

)

r v = Ñ · E e0

¶E x ¶E y ¶E z + + ¶x ¶y ¶z

Ñ· E =

é ¶ ê 8x = 4xy + ¶x ê x2 + y 2 êë

(

)

ù é 8y ú ¶ ê 2 2x + 2 ú ¶y ê x2 + y 2 úû êë

(

(

é 2 ì 2 üù é ì 2 2 2 8 - 8x 2 x 2 + y 2 ï ú ê ï x + y ê ï x +y ê ê ïï ( 2x) ïïúú ê ïï = - ê 4y + í 0 +í ý ê ê ú 2 2 4 ï ï x +y ê ï ê ï ïú ê ï ê ú ïî þï û êë ïî ë

(

)

(

(

= - 4y -

8

)

)

)

32 x 2

+

8

-

+

2

3

2

(

)

üù ( 8) - 8y 2 x 2 + y 2 ïú

(x

2

+y

2

)

( 2y) ïïúú

ý –0 ú ïú ïú ïþ úû

4

32 y 2

(x 2 + y 2 ) (x 2 + y 2 ) (x 2 + y 2 ) (x 2 + y 2 ) 2

)

ù ú ¶ - ( 20) 2 ú ¶z ûú

3

At P, x = 6, y = – 2.5 and z = 3. \ \

Ñ · E = 10 - 4.4816 10 -3 + 0.01527 - 4.4816 ´ 10 -3 + 2.651 ´ 10 -3 = 10.00895

[

]

r v at P = e 0 Ñ · E = 8.854 ´ 10 -12 ´ 10.00895 = 88.6193 pC/m 3

Example 4.12.6 Solution : V = x 2 y(z + 3) = x 2 yz + 3x 2 y i)

¶V ¶V é ¶V ù ax + ay + az ú E = - ÑV = - ê ¶y ¶z ë ¶x û = -[(2xyz + 6xy) a x + (x 2 z + 3x 2 ) a y + x 2 y a z ]

... (1)

At (3, 4, – 6) E = - [( -144 + 72) a x + ( -54 + 27) a y + 36 a z ] = 72a x + 27a y - 36a z V m ii) \

D = e0 E ¶E y ¶E z ù é ¶E + Ñ · D = Ñ · e 0 E = e 0 ( Ñ · E) = e 0 ê x + ¶y ¶z úû ë ¶x = e 0 [2yz + 6y + 0 + 0] = e 0 y[2z + 6] = r v

\

Q =

ò

v

r v dv =

1

ò

1

ò

1

ò

e 0 [2yz + 6y] dx dy dz

z= 0 y= 0 x= 0

TM


... Use equation (1)


4 - 21

1

1

ò

=

ò

z= 0 y= 0 1

e 0 [2xyz + 6xy] 10 dy dz

1

ò ò

= e0


z= 0 y= 0

1

é 2y 2 z 6y 2 ù [2yz + 6y] dy dz = e 0 ò ê + dz 2 2 úú û0 z = 0 êë 1

1

1

ù é z2 = e 0 ò (z + 3) dz = e 0 ê + 3zú = 3.5 e 0 = 3.5 ´ 8.854 ´ 10 -12 = 30.989 pC 2 û0 ë z= 0 Example 4.12.7 Solution : The given potential is, V = 10 y (x 3 + 5) = 10 x 3 y + 50 y ¶V ¶V ù é ¶V ax + ay + az ú E = - ÑV = - ê x y z ¶ ¶ ¶ û ë

i)

[

]

= - 30 x 2 y a x + 50 a y = - 30x 2 y a x - 50 a y At y = 0,

E = - 50 a y V m

ii) At y = 0,

¶V ¶V ù é ¶V ax + ay + az ú E = - 50 a y = - ê x y z ¶ ¶ ¶ û ë ¶V = 50 and integrate hence V = 50 y + K ¶y

\

But y = 0, V = K = constant This proves that as potential is constant on y = 0 surface, it is equipotential. iii) For y = 0, E = - 50 a y V m 2 i.e. D = e 0 E = - 50 e 0 a y C m 2 For y = 0, \

×

dS = dx dz a y

D d S = - 50 e 0 dx dz

\

Q =

ò

S

×

D dS =

1

.... a y 2

ò ò

=1

- 50 e 0 dx dz ... charge in the region

Example 4.12.8 i)

y

z = 0 x= 0

= - 50 e 0 [x]20 [z]10 = - 100 e 0 C = - 0.8854 nC Solution :

×a

1 ¶V 1 ¶V ù é¶ V ar + aq + afú E = -Ñ V= - ê q r r sin f f ¶ r ¶ ¶ û ë

1 10 1 10 ù é 20 = - êsin q cos f a r + cos f cos q a q + sin q ( - sin f) a f ú 3 2 2 r r s in q r r û ë r TM



=

20 r3

4 - 22

sin q cos f a r -

10 r3


cos q cos f a q +

10 r3

sin f a f

10 10 ù é 20 sin f a f ú sin q cosf a r cos q cos f a q + D = e0 E = e0 ê 3 3 3 r r û ër p p At point æç 2, , 0ö÷, r = 2, q = è 2 ø 2

and

f=0

\

D = 2.5 e 0 a r = 22.135 ´ 10 -12 a r C / m 2

ii)

W = -Q

final

ò

E × dL

initial final

Now

ò

E × d L = Potential difference between initial and final

initial

Initial point A (1, 30° , 120° ) and final point B ( 4, 90° , 60° ) \

B

ò

E × dL

= VAB = VA - VB

A

VB

= V at B =

VA

= V at A =

\

VAB

\

W

10 r2

sin q cos f

10 r2

= at B

sin q cos f

10

( 4) 2 =

at A

sin ( 90° ) cos ( 60° ) = 0.3125 V

10

(1) 2

sin ( 30° ) cos (120° ) = - 2.5 V

= VA - VB = - 2.5 - 0.3125 = - 2.8125 V = - Q VAB = -10 ´ 10 -6 ´ - 2.8125 = 28.125 mJ

... Same as above

Example 4.12.9 Solution : a) \

æp yö V = E 0 e - x sin ç ÷ è 4 ø ¶V ¶V ù é¶V ax + ay + az ú E = -Ñ V = -ê x y z ¶ ¶ ¶ û ë

¶V æp yö -x = E 0 sin ç ÷ ( -1) e ¶x è 4 ø

... y is constant

¶V æp yö p = E 0 e - x cos ç ÷ ¶y è 4 ø4

... x is constant

¶V = 0 ¶z

... z is absent

TM



\

4 - 23


py py p ù é a x + E 0 e - x cos a y ú V/m E = - ê - E 0 e - x sin 4 4 4 û ë

At P ( 0 , 1 , 1) , E = E 0 [0.7071 a x - 0.555 a y ] V/m b)

V = E 0 r cos q

\

1 ¶V 1 ¶V ù é¶V ar + aq + a E = -Ñ V = -ê r ¶q r sin q ¶ f f úû ë ¶r

¶V ¶V ¶V = E 0 cos q, = - E 0 r sin q, =0 ¶r ¶q ¶f \

E = - E 0 cos q a r + E 0 sin q a q V/m

Convert P ( 0,1, 1) to spherical co-ordinates. x 2 + y 2 + z 2 = 2, f = tan -1

r = \

y p z = , q = cos -1 = 45° x 2 r

E = + E 0 [ - 0.7071 a r + 0.7071 a q ] V/m


V=

100 z2 +1 100

r cos f, r = 3, f = 60º, z = 2 ´ 3 ´ cos 60º = 30 V

i)

V =

ii)

1 ¶V ¶V ù é ¶V E = –Ñ V = – ê ar + a + a r ¶f f ¶z z úû ë ¶r

22 +1

ì 1 100 ï 100 cos f = –í ar + ´ ´ r( –sin f) a f + 100 r cos 2 2 +1 r z + 1 z ï î

é ê –2 z fê 2 êë z + 1

= –10 a r + 17.32 a f + 24 a z V/m iii)

E =

(

)

ù ü ú ï a 2 ú zý úû ï þ

K Using given values

10 2 + 17.32 2 + 24 2 = 31.24 V/m.

iv)

dV = E = 31.24 V/m dN

v)

aN = –

vi)

é1 ¶ 1 ¶E f ¶E z ù r v = Ñ · D = Ñ · e0 E = e0 Ñ · E = e0 ê rEr ) + + ( r ¶f ¶z úû ë r ¶r

E E

= 0.32 a r – 0.55 a f – 0.768 a z

(

)

K (1)

[

]

TM



From (1), Er =

\

rv

–100 cos f z2

+1

4 - 24

, Ef =

100 sin f z2

+1

, Ez =


200 zr cos f

(z 2 + 1)

2

ì é ùü ï 1 ¶ é –100 r cos f ù 1 ¶ é 100 sin f ù ¶ ê 200 z r cos f úï = e0 í ú + ¶z ê ú + r ¶f ê 2 2 2 úý r ¶r êë 2 z + z + 1 1 û û ë ï êë z + 1 úûï î þ

(

(

)

)

ì é 2 ï 1 –100 cos f 1 100 cos f ê z +1 = e0 í ´ + + 200 r cos fê r r z2 + 1 z2 + 1 ï êë î

(

ì 200 r cos f 1 – 3z 2 ï = e 0 í0 + 3 ï z2 + 1 î

(

)

Example 4.12.11 Solution : V = 2x2y – 5z, \

)üï = e ý ï þ

2

(

)

(z 2 + 1)

4

úý úûï þ

ì 200 ´ 3 ´ cos 60º ´( –11)ü 3 ý = –233.75 rC/m 3 5 î þ

0í

P(– 4, 3, 6)

V = 2 ´ ( -4) 2 ´ 3 - 5 ´ 6 = 66 V ¶V ¶V ù é ¶V ax + ay + a z ú = – [4xy a x + 2x 2 a y - 5 a z ] E = -ÑV = - ê ¶ x ¶ y ¶ z û ë

\ At P,

ü

ù (1) – ( z)( 2) z 2 + 1 ( 2z) úï

…(1)

E = +48 a x - 32 a y + 5 a z V / m 2 D at P = E at P ´ e 0 = 0.425 a x – 0.2833 a y + 0.0442 a z nC/m

\

rv = Ñ· D =

¶D x ¶D y ¶D z + + ¶x ¶y ¶z

…(Use equation 1)

3 = e 0 [-4y + 0 + 0] = - 4 e 0 y C/m 3

\ r v at P, r v = – 4 ´ 3 e 0 = – 12 e 0 = – 0.1062 nC/m Example 4.12.12

Solution : Refer example 4.12.11 for the procedure and verify the answers : i) V= 251 V ii) E = 24a x - 30a y - 96a z V/m iii) D = e 0 E = 0.215 a x - 0.2656 a y - 0.85 a z nC / m 2 Example 4.12.13 Solution : Given V is in cylindrical system. ¶V 1 ¶V ù é¶V i) ar + af + az ú E = -Ñ V = – ê r ¶f ¶z û ë ¶r ¶V é ¶ æ1 öù = cos 2f ê ç ÷ ú = cos 2 f ¶r ë ¶ r è r øû

é 1 ù ê- 2 ú ë r û TM



4 - 25


¶V 1é ¶ ù 1 = ( cos 2f) ú = r ( - 2 sin 2f) r êë ¶ f ¶f û

and

¶V =0 ¶z

2 sin 2f 1 -2 sin 2f ù cos 2f é cos 2 f ar + afú = ar + af E = - ê2 2 r r r r r2 û ë

\

At B ( 2, 30° , 1), E = 0.125 a r + 0.433 a f V/m

×

ii) r v = Ñ D = Ñ

×

Ñ E = = \

×(e

0

×]

[

)

E = e0 Ñ E

1 ¶ 1 ¶ Ef ¶ Ez r Er ) + + ( r ¶r r ¶f ¶z 1 ¶ é r cos 2 f ù 1 ¶ é 2 sin 2f ù 1 æ 1 ö 1 2 + 0 = cos 2f ç - ÷ + ´ + r ¶ r êë r 2 úû r ¶ f êë r 2 úû r è r2 ø r r2

[2 cos 2f]

×]

é cos 2f 4 cos 2f ù é 3 cos 2f ù r v = e0 Ñ E = e0 ê+ ú e0 ú=ê r3 r3 û ë r3 û ë

[

At A (0.5, 60°, 1),

rv =

3 ´ cos ( 2 ´ 60° ) e 0

( 0.5)

3

= - 12 e 0 = – 0.106 nC/m 3


¶V ¶V ù é ¶V E = – ÑV = – ê ax + ay + az ú ¶ x ¶ y ¶ z û ë ¶V ¶ = [3x2y + 2yz2 + 3xyz] = 6xy + 0 + 3yz ¶x ¶x ¶V ¶ 2 2 2 2 = [3x y + 2yz + 3xyz] = 3x + 2z + 3xz ¶y ¶y ¶V ¶ = [3x2y + 2yz2 + 3xyz] = 4yz + 3xy ¶z ¶z 2

2

E = – {[6xy + 3yz] ax + [3x + 2z + 3xz] ay + [4yz + 3xy] az} \ At point (1, 2, – 1), x = 1, y = 2, z = – 1 \

E = – 6 ax – 2 ay + 2 az V/m

Example 4.12.15 Solution : i) \

2

2

2

V = 2x – y – z and P(0.5, 1.5, 1) 2

2

2

V = 2(0.5) – (1.5) – (1) = –2.75 V

…Potential

¶V ¶V ù é¶V ax + ay + az ú E = -Ñ V = – ê ¶y ¶z û ë ¶x = - [ 4x a x - 2y a y - 2z a z ] = - 4x a x + 2y a y + 2z a z V/m \

D = e 0 E = e 0 [- 4x a x + 2y a y + 2z a z ] C/m TM

2



4 - 26

×

rv = Ñ D =


¶ Dx ¶ Dy ¶ Dz = - 4e 0 + 2e 0 + 2e 0 = 0 C/m3 + + ¶x ¶y ¶z

ii) V = 6rfz For P, x = 0.5, y = 1.5, z = 1

…r = r in cylindrical system

y = 71.565°, r = x 2 + y 2 = 1.5811 x \ P(r = 1.5811, f = 71.565°, z = 1) p ´ 71.565° = 1.249 rad use f in radians i.e. f = 180° V = 6 ´ 1.5811 ´ 1.249 ´ 1 = 11.8492 V \ In cylindrical system, f = tan -1

Now

¶V 1 ¶V 1 ù é¶V ar + af + a z ú = - [6 f z a r + ´ 6rz a f + 6rf a z ] E = -Ñ V = - ê r r ¶ ¶ f ¶ r z û ë

\

E = - 6 fz a r - 6z a f - 6r f a z

\

D = e 0 E = e 0 [- 6 f z a r - 6z a f - 6r f a z ]

\

rv = Ñ×D =

\

rv =

- 6 f ze 0 1 ¶ 1 ¶ Df ¶ Dz 1 1 = ´ [- 6 f ze 0 ] + ´ ( 0) + ( 0) = ( r Dr) + + r r ¶r r ¶f r r ¶z

- 6 ´ 1.249 ´ 1 ´ 8.854 ´ 10 -12 3 = – 41.96 pC/m 1.5811


Given, E =

10 6 ar r

i.e. | E| =

10 6 r

and | E |2 =

10 12 r2

For free space, e = e 0 \

WE =

=

1 2 1 2

ò

e |E|2 dv

dv = r dr df dz

where

vol

ò

e0 ´

v

10 12 r2

´ r dr df dz =

e0 2

ò

v

10 12 dr df dz r

z = 0 to 50 m, f = 0 to 2 p and r = 0.01 to 0.05 m \

WE =

=

10 12 e 0 2

50

ò

2p

ò

0.05

ò

z = 0 f = 0 r = 0.01

10 12 e 0 1 0.05 2p dr df dz = [z]50 0 [f] 0 [ ln r ] 0.01 r 2

10 12 ´ 8.854 ´ 10 - 12 ´ 50 ´ 2 p ´ [ ln 0.05 - ln 0.01] = 2.2383 kJ 2

Example 4.13.6 Solution : The charges existing at the corners of an equilateral triangle are shown in the Fig. 4.16. TM



4 - 27

Electric Work, Energy and Potential Q1 = 1 C

When Q 1 is placed, no other charge is present, \ WE1 = 0 J When Q 2 is placed, Q 1 is present. ù é Q1 \ WE2 = Q 2 V21 = Q 2 ê ú pe 4 R 0 21 û ë

1m

Q3

2 ù 3C 1m = J ú 1û 4pe 0 Fig. 4.16 both are present, Q1 Q2 ù ù é é + Q3 ê + Q 3 V 32 = Q 3 ê ú ú pe pe 4 R 4 0 31 û 0 R 32 û ë ë

When Q 3

1 é = 2 ´ ê ë 4pe 0 ´ is placed, Q 1 and Q 2

\

WE3 = Q 3 V 31

\

9 1 2 ù ù é é = WE3 = 3 ´ ê + 3 ´ ê J ú ú pe 1 pe 1 4 pe ´ 4 ´ 4 0 0 0 û û ë ë

\

WE = WE1 + WE2 + WE3 = 0 + =

11 4p ´ 8.854 ´ 10 - 12

2 9 + 4pe 0 4pe 0

= 9.8865 ´ 10 10 J

Example 4.13.7 ¶V ¶V ù é ¶V ax + ay + az ú E = -Ñ V = -ê x y z ¶ ¶ ¶ û ë ¶V ¶V ¶V = 1 + y, = – 1 + x, =2 ¶x ¶y ¶z

Solution :

[(1 + y) a x + ( x - 1) a y + 2 a z ] - [(1 + 2) a x + (1 - 1) a y + 2 a z ] = - 3 a x - 2 a z

\

E = -

At (1, 2, 3),

E =

Now

WE = E

\

1 2

WE =

e0 E

2

dv

vol

(1 + y) 2 + ( x - 1) 2 + ( 2) 2

=

| E|2 =

ò

V/m

(1 + y) 2 + ( x - 1) 2 + ( 2) 2

= y 2 + 2y + x 2 - 2x + 6

1 2

and

ò

e0 E

2

dv

dv = dx dy dz

vol

The cube is centered at the origin. Thus x varies from – 1 to + 1, y from – 1 to + 1 and z from – 1 to + 1. \

WE =

e0 2

1

ò

1

1

ò

ò

z = - 1y = -1x = -1

Changing limits from – 1 to 1

to

(y 2 + 2y + x 2 - 2x + 6)

dx dy dz

0 to 1 of each integral, making it twice. TM


1m

Q2 2C


\

WE =

4 - 28

e0 ( 2 ´ 2 ´ 2) 2

= 4 e0

1

1

1

1

ò ò ò

z = 0 y = 0 x= 0


(y 2 + 2y + x 2 - 2x + 6) dx dy dz

1

é 2 ù x3 xy + 2xy + - x 2 + 6xú dy dz ê ò ò 3 û z= 0 y = 0 ë 1

é y3 ù x3 = 4 e 0 ê xz + zxy 2 + yz - x 2 yz + 6 xyzú = 0.2361 nJ 3 3 êë úû 0 Example 4.13.8 Solution : The arrangement is shown in the Fig. 4.17. 2d = 2

R31 = 2 ×

2 d=

2 m = R42 R41

Let Q1 is placed first when all other charges are absent. Hence W1 = 0 J

Ö2d 2

Q4

d 2

d 2 d

é Q1 ù W2 = Q2 V21 = Q2 ê ë 4pe 0 R 21 úû

R43

Fig. 4.17

W4 = Q4 V41 + Q4 V42 + Q4 V43 ù ù é Q3 é Q2 é Q1 ù = Q4 ê + Q4 ê + Q4 ê ú ú ú 4 p e 4 p e 4 p e R R R 0 41 û 0 42 û 0 43 û ë ë ë

But

Q1 = Q2 = Q3 = Q4 = Q = 4 nC

\

W = W 1 + W2 + W3 + W4 = 0+

=

Q2 Q2 Q2 Q2 Q2 Q2 + + + + + 4p e 0 4p e 0 4p e 2 4p e 0 4p e 2 4p e 0 0 0

Q2 4p e 0

( 4 ´ 10 -9 ) 2 2 ù é × 5.4142 = 0.7785 µJ = 4 + êë 2 úû 4p ´ 8 . 854 ´ 10 -12


E=

10 6 r6

ar ,

\

E =

10 6 r6

,

E TM

2

=

R32

d 2

For Q3, W3 = Q3 V32 + Q3 V31 = Q3 ù ù é Q1 é Q2 êë 4p e 0 R 32 úû + Q 3 êë 4p e 0 R 31 úû For Q4,

Q2

Ö2d 2

R21 = R32 = R43 = R41 = 1 m = d

For Q2,

R21

Q1

d = 1m

10 12 r 12


Q3


\

WE =

1 2

4 - 29

ò

2

e E

dv =

vol

1 2

ò

e

10 12

vol


dv

r 12

dv = r dr df dz in cylindrical system z = 0 to 200 mm i.e. 0 to 0.2 m, f = 0 to 2 p and r = 0.05 to 0.1 m. 0.2 2p

0.1

\

WE =

1 e 10 12 2

\

WE =

1 2p f (10 e 0 ) 10 12 [z] 0.2 0 [ ]0 2

ò ò

ò

1

r dr df dz

12 z = 0 f = 0 r = 0.05 r

0.1

ò

r = 0.05

1 r 11

... e = 10 e 0

dr

0.1

é r -10 ù 10 13 = e 0 ´ 0.2 ´ 2 p ´ ê 2 -10 ú ë û 0.05 é ù 1 1 = 55.6313 ´ ê + ú = 5.691 ´ 10 13 J 10 10 10 0.1 10 0.05 êë ( ) ( ) úû Example 4.13.10 Solution : V = r 2 z sin f

... cylindrical system

¶V ¶V ù é¶ V ar + af + az ú E = -Ñ V= -ê r¶f ¶z û ë¶r

\

1 = - é 2 r z sin f a r + r 2 z cos f a f + r 2 sin f a z ù êë úû r WE =

1 2

WE =

e 0 | E|2 dv

vol

4 r 2 z 2 sin 2 f + r 2 z 2 cos 2 f + r 4 sin 2 f

| E| = \

ò

e0 2

[4 r

ò

vol

2

]

z 2 sin 2 f + r 2 z 2 cos 2 f + r 4 sin 2 f dv

dv = r dr df dz \ WE =

e0 2

e = 0 2

=

e0 2

p/ 3

2

ò

4

ò ò

z= - 2 f= 0 r= 1 2

ò

z= - 2 2

ò

[

]

r 3 4 z 2 sin 2 f + z 2 cos 2 f + r 2 sin 2 f dr df dz

p / 3ì

4 4 2 ü ér4 ù ér4 ù ér6 ù ï 2 2 2 ï 2 2 + 4 z sin z f cos f + sin f í ý df dz ê 4 ú ê 4 ú ê 6 ú ò ë û ë û ë û ïþ 1 1 1 f = 0ï î p/ 3

ò

z= - 2 f= 0

[255 z

2

] df dz

sin 2 f + 63.75 z 2 cos 2 f + 682.5 sin 2 f

TM



e = 0 2

=

=

e0 2 e0 2

4 - 30


p/ 3 ì

2 2 ü éz3 ù éz3 ù ï 2 ï 2 2 2 ò í255 sin f ê 3 ú + 63.75 ê 3 ú cos f + 682.5 sin f [z]-2 ý df û -2 ë û -2 ë ï f = 0î þï p/ 3

[1360 sin

2

[4090 sin

2f+

ò

f= 0 p/ 3

ò

]

f + 340 cos 2 f + 2730 sin 2 f df

]

340 cos 2 f df

f= 0

p/ 3 p/ 3 ìï 1 - cos 2 f 1+ cos 2 f üï 4090 d dfý f + 340 í ò ò 2 2 ïþ ïî f= 0 f= 0

=

e0 2

=

e 0 ìï 4090 2 íï 2 î

p/ 3

sin 2f ù 340 é + êë f - 2 úû 2 f= 0

sin 2 f ù p / 3 ïü é + f êë 2 úû f = 0 ýï þ

e e0 ì 340 é p p ü 2045 é - 0.433ù + + 0.433ùý = 0 í êë 3 2 2 î 2 êë 3 ûú ûúþ = 6.6735 nJ

=

{( 2045 ´ 0.6141) + (170 ´ 1.48019)}

Example 4.14.3 Solution : i) The dipole moment is given by, p = Qd

where d = d a z

Here d = Distance between charges = 2 mm \

p = 3 ´ 10 -6 ´ 2 ´ 10 -3 a z = 6 a z nCm

ii) In spherical system, E is given by, E =

=

Qd 4p e 0 r 3

[2 cos q a r + sin q a q ]

3 ´ 10 -6 ´ 2 ´ 10 -3 4p ´ 8.854 ´ 10 -12 ´ (2) 3

[2 cos 40º a r + sin 40º a q ] = 10.3275 a r + 4.333 a q V m

Example 4.14.4 Solution : p = 3 a x – 2 a y + a z nCm at origin (0, 0, 0). i) P(2, 3, 4) \

r = 2 a x + 3 a y + 4 a z , |r| = V =

p· a r 4p e 0 r 2

=

p· r 4p e 0 r 3

=

29, a r =

r r

[( 3 ´ 2) + ( -2 ´ 3) + (1 ´ 4)] ´ 10 - 9 4p ´ 8.854 ´ 10 - 12 ´ ( 29 ) 3

ii) r = 2.5, q = 30º, f = 40º TM


= 0.2302 V


4 - 31


Converting given spherical to cartesian co-ordinates x = r sin q cos f = 0.9575, y = r sin q sin f = 0.8034, z = r cos q = 2.165 \ \

r = 0.9575 a x + 0.8034 a y + 2.165 a z , |r| = 2.5 V =

p· a r 4p e 0 r 2

=

[(3 ´ 0.9575) + ( -2 ´ 0.8034) + (1 ´ 2.165)] ´ 10 - 9 z

Example 4.14.5 Solution : The dipole is shown in the Fig. 4.18.

(0, 0, 0.1)

Given P (0.3, 0, 0.4) in cartesian co-ordinates. \

x = 0.3, y = 0, z = 0.4

\

r =

q

x 2 + y 2 + z 2 = 0.5

x

Vp =

P(r, q, f)

r1

r

r2 y

d = 0.2 m (0, 0, – 0.1)

z q = cos - 1 é ù = 36.8698º êë r úû y f = tan - 1 = 0º x \

= 1.9734 V

4p ´ 8.854 ´ 10 -12 ´ (2.5) 3

Fig. 4.18

Q é d cos q ù 1.5 ´ 10 - 9 = 4 p e 0 êë r 2 úû 4p ´ 8.854 ´ 10 - 12

é 0.2 ´ cos (36.8698º ) ù ê ú = 8.6282 V (0.5) 2 êë úû

qqq

TM


5

Conductors, Dielectrics and Capacitance Solutions of Examples for Practice

Example 5.2.6 I = ò J · dS

Solution :

Now J is in the direction of a r and dS in the direction of a r in cylindrical system is r dr df. 3 – 400 r r dr df J · dS = 10 e

\

2 p 0 . 002

\

ò

I =

ò

3

10 r e

and r = 2 mm = 2 × 10

– 400 r

f= 0 r= 0

2p

dr df = 10 3 [f]0

0 . 002

òre

–3

m

- 400 r

dr

r=0

Use integration by parts, ò uv = u ò v - ò u ¢ ò v Select u = r 0 . 002

\

òre

and

- 400 r

0

v = e - 400 r

[

]

= r ò e - 400 r dr - ò 1 ´ ò e - 400 r dr dr 0 . 002

=

é r e - 400 r r e - 400 r e - 400 r e - 400 r ù -ò dr = ê ú - 400 - 400 - 400 160 ´ 10 3 û 0 ë

=

0.002 ´ 0.4493 1 + - 2.808 ´ 10 -6 = 1.1955 ´ 10 -6 3 - 400 160 ´ 10

I = 10 3 ´ 2 p ´ 1.1955 ´ 10 -6 = 7.5115 mA

\ Example 5.2.7 Solution :

From continuity equation, I =

ò

S

×

J dS

For y = 0 plane, the normal vector is a y and hence dS normal to a y is dx dz a y . \

0.002

I =

ò

0.1

ò

×

10 2 | x| a y dx dz a y =

z = - 0.002 x = - 0.1

(5 - 1) TM

0.002

ò

0.1

ò

z = - 0.002 x = - 0.1


10 2 | x| dx dz


= 10

2

5-2

[z] 0.002 - 0.002

Conductors, Dielectrics and Capacitance 0.1

é| x |2 ù 2 ò |x | dx = 2 ´ 100 ´ [ 0.002 - ( - 0.002) ] ´ ê ú êë 2 úû 0 x= 0 0.1

= 200 ´ 4 ´ 10 -3 ´

(0.1) 2 = 4 mA 2

Example 5.2.8 Solution : J = 10 r 2 z a r - 4 r cos 2 f a f I =

ò J · dS = ò (Ñ · J) dv S

vol

1 ¶ 1 ¶J f ¶ J z , J r = 10 r 2 z, J f = - 4r cos 2 f (r J r ) + Ñ· J = + r ¶r r ¶f ¶z 1 ¶ 1 ¶ = (10 r 3 z) + - 4r cos 2 f = 30 r z + 8 sin f cos f r ¶r r ¶f

[

\

I =

ò

( 30 r z + 8 sin f cos f) r dr df dz

vol 2. 8

ò

=

]

2p

ò

3

ò

[ 30 r

2

]

z + 4 r sin 2f dr df dz

z = 2 f = 0 r= 0 2p

2. 8

ò

=

ò

3

é 30 r 3 z 4 r 2 sin 2f ù + df dz ê ú 3 2 ê ú ë û 0 r=0

z= 2 f = 2. 8 2 p

ò [270 z + 18 sin 2f]df dz

ò

=

z = 2f= 0 2. 8

2p

ò [270 z f + 9 cos 2f]f = 0

=

z =2

2. 8

2. 8

é z2 ù dz = ò [540 p z dz] = 540 p ê ú z ë ûz=2 z= 2

= 3257.203 mA = 3.257 A Example 5.2.9 Solution :

b)

a) J at r = 3, q = 0° and f = p is, 2 cos ( 0° ) a r + 20 e - 6 sin 0° a f - 2 sin 0° cos p a f = 0.222 a r A/m 2 J = ( 9) I =

ò

J · dS

S

dS in a r direction is r 2 sin q dq df a r \

I =

ò

S

[

J · r 2 sin q dq df a r

]= ò

S

2 r2

[

cos q r 2 sin q dq df

]

... a r · a r = 1

aq · ar = af · ar = 0

TM


Electromagnetic Field Theory 20°

2p

ò

=

5-3

ò

Conductors, Dielectrics and Capacitance

2 cos q sin q dq df

f= 0 q= 0 20°

20°

2p

ò

=

f= 0

é - cos 2q ù sin 2 q dq df = ê ú 2 û0 ë q= 0

ò

[f]20 p

= 0.735 A


J =

4 r

cos q a r + 20 e - 2r sin q a q - r sin q cos f a f

2

Given

r = 3,

\

J =

\

=

J

q=0

4 32

and

f=p

cos 0 a r + 20 e - 2 ´ 3 sin 0 a q - 3 sin 0 cos p a f

… sin 0 = 0

4 2 a A/m 9 r

Example 5.2.11 Solution : For z = 0,

x = 1, from 3x + z = 3

For x = 0, z = 3, from 3x + z = 3 I = ò J · dS

z z=3

S

For x = 0,

dS = dy dz ( - a x )

For x = 1,

dS = dy dz ( + a x )

For y = 0,

dS = dx dz ( - a y )

For y = 2,

dS = dx dz ( + a y )

For z = 0,

dS = dx dy ( - a z )

For z = 3,

dS = dx dy ( + a z )

\

3

2

ò ò

I =

+

3

y=0 y y=2 x

1

ò ò

z= 0 x= 0

x=1

z=0

Fig. 5.1

- 3x dydz +

z= 0 y= 0

x=0

(x = 0)

3

2

ò ò

z= 0 y= 0 2

ò

(y - 3)dxdz + (y = 2)

3

3x dydz + (x = 1)

1

ò

y= 0 x= 0

1

ò ò

- (y - 3) dxdz (y = 0)

z= 0 x= 0

- (z + 2)dxdy + (z = 0)

2

ò

1

ò

y= 0 x= 0

(z + 2)dxdy (z = 3)

= 0 + 3[y] 20 [z] 03 + 3 [x] 10 [z] 03 - [x] 10 [z] 03 - 2 [x] 10 [y] 20 + 5 [x] 10 [y] 20 = 0 + 18 + 9 – 3 – 4 + 10 = 30 A Example 5.2.12 Solution : The current is given by, I =

ò

S

×

J dS TM



5-4


Assuming J given in a z direction, dS = r dr df a z \

×

J dS =

\

I =

10 4 ´ r dr df = 10 4 dr df r 2p

4 ´ 10 - 3

f= 0

r= 0

ò

ò

10 4 dr df = 10 4 [r]40 ´ 10

-3

[f]20 p

= 10 4 ´ 4 ´ 10 - 3 ´ 2 ´ p = 80 p A Example 5.4.3 Solution :

d = 0.8 mm, L = 2 cm, I = 20 A I I 20 = 39.788 ´ 10 6 A/m 2 = = | J| = S p 2 p 3 2 d ´ [0.8 ´ 10 ] 4 4

Now

| J| = s|E|

\

| E| =

And

|J| 39.788 ´ 10 6 = = 0.686 V/m s 5.8 ´ 107

V = EL = 0.686 ´ 2 ´ 10 -2 = 0.0137 V R =

V 0.0137 = = 6.86 ´ 10 -4 W I 20


2 electrons/atom, 0.09375 ´ 10 26 atoms/kg

Density = 9000 kg/m 3 , e = Charge on 1 electron = 1.6 ´ 10 - 19 C n = Concentration of electrons/m 3 = 2 ´ 0.09375 ´ 10 26 ´ 9000 = 1.6875 ´ 10 29 m 3 \

r = Charge concentration = n e = 1.6875 ´ 10 29 ´ 1.6 ´ 10 - 19 = 2.7 ´ 10 10 C m 3

i)

r = 1 mm, A = p r 2 = p ´ (1 ´ 10 - 3 ) 2 = 3.1416 ´ 10 - 6 m 2

\ Mobile charge per unit length = r ´ A = 2.7 ´ 10 10 ´ 3.1416 ´ 10 - 6 = 84.823 ´ 103 C / m ii)

I = 1A

\

J =

A = 3.1416 ´ 10 - 6 m 2

I 1 = = 318.30914 ´ 10 3 A m 2 A 3.1416 ´ 10 - 6

v = mE Now

and

and

J=sE

s = r m = neµ

… r = ne TM



5-5

\

J = n em E = n e v

\

v =


J J 318.30914 ´ 10 3 = = = 1.178 ´ 10 - 5 m / s 10 ne r 2.7 ´ 10

Example 5.6.3 Solution : V = – 6000 z, e r = 3.6 1) E = – ÑV = –

¶V a z = 6000 a z V/m ¶z

2) P = c e e 0 E \

…

¶V ¶V = =0 ¶x ¶y

c e = e r – 1 = 2.6

where

P = 2.6 ´ 8.854 ´ 10 –12 ´ 6000 a z = 138.122 a z nC / m 2

3) r S = P · a n and a n = a z as space is between z = 0 and z = 1 \

r S = (138.122 a z ) · a z = 138.122 nC/m 2

Example 5.6.4 P = ce e 0 E

Solution :

eR = c e + 1

i.e. c e = e R - 1 = 2.7 – 1 = 1.7

–6 \ [– 0.2 a x + 0.7 a y + 0.3 a z ] × 10 = 1 . 7 ´ 8 . 854 ´ 10 -12 E

i) \

E = - 13 . 2874 a x + 46.5060 a y + 19.9311 a z kV/m

ii)

2 D = e 0 e R E = - 0 . 3176 a x + 1 . 111 a y + 0 . 4764 a z mC/m

iii)

-Ñ V = E

where Ñ V = Voltage gradient

\ Magnitude of voltage gradient = |E| \

|E| =

(13 . 2874) 2 + ( 46 . 506) 2 + (19 . 9311) 2 = 52.3125 kV/m


|E| = 10 kV/m,

e R = 255

a)

|D| = e 0 e R |E |= 8.854 ´ 10 -12 ´ 255 ´ 10 ´ 10 3 = 2.2577 ´ 10 -5 C/m 2

b)

|P| = c e e 0 |E |

where c e = e R - 1 = 254

= 254 ´ 8.854 ´ 10 -12 ´ 10 ´ 10 3 = 2.2489 ´ 10 -5 C/m 2 c)

V = |E|d = 10 ´ 10 3 ´ 1.5 ´ 10 - 3 = 15 V


c e = 0.12,

D = 1.6 n C m 2

P = polarisation = c e e 0 E TM



= c e e0

5-6

D e0 eR

where


eR = c e + 1

\

P =

ce D 0.12 ´ 1.6 ´ 10 - 9 = 0.1714 nC / m 2 = 1.12 ce +1

\

E =

D 1.6 ´ 10 - 9 = 161.3475 V/m = e0 eR 8.854 ´ 10 - 12 ´ 1.12


z

i) M(4, –2, 1) is shown in the Fig. 5.2. Conductor M

As point M is inside the conducter E = D = 0.

N

ii) N(–3, 1, 4) is shown in the Fig. 5.2. It is located inside the dielectric. D N = r S i.e. DN = r S a y = 4 a y nC/m 2 EN =

y

rS 4 ´ 10 –9 = 150.59 V/m = e 0 e r1 8.854 ´ 10 –12 ´ 3

Fig. 5.2

K As per boundary conditions

E tan = D tan = 0 \

Dielectric er1 = 3

E = EN = 150.59 a y V/m, D = DN = 4 a y nC/m 2

Example 5.8.5 Solution : E = 60 a x + 20 a y – 30 a z

[

D = e 0 E = 8.854 ´ 10 –12 60 a x + 20 a y – 30 a z \

]

D = 0.531 a x + 0.177 a y – 0.265 a z nC/m 2 DN = D as Dtan = 0 as per boundary conditions

\

r S = DN = D = ( 0.531) 2 + ( 0.177 ) 2 + ( 0.265) 2 = 0.619 nC/m 2

Example 5.9.3 Solution : The two media are separated by z = 0 plane and ± a z are the directions of normal to the surface. D1 = 2 a x + 5 a y - 3 a z nC/m 2 D1 = DN1 + Dtan 1 Normal direction to the surface is ± a z hence the part of D1 in the direction of ± a z is DN1 . \ DN1 = -3 a z nC/m 2

TM



5-7


Dtan 1 = D - DN1 = 2 a x + 5 a y nC/m 2

\

According to boundary conditions, DN1 = DN2 = - 3 a z while \

Dtan 1 Dtan 2

=

Dtan 2 =

e e1 = r1 e2 e r2 2 5

2ax +5ay

i.e.

Dtan 2

(2 a x + 5 a y ) = 0.8 a x + 2 a y z

=

5 2

nC/m 2 D1

z>0 er1 = 5

Normal az

q1 Surface

z=0 plane

q2

–az

z<0 er2 = 2 D2

Fig. 5.3

D2 = DN2 + Dtan 2 = 0.8 a x + 2 a y - 3 a z nC/m 2

\

Energy density WE1

1 D1 = 2 e1

2

2

1 D1 = 2 e 0 e r1 2

=

and

WE2 =

1 ´ 2

(

æç ( 2) 2 + (5) 2 + ( -3) 2 ö÷ ´ 10 -9 ø è

)

8.854 ´ 10 -12 ´ 5

1 D2 2 e2

2

2

= 0.4291 mJ/m 3

é 0.8 2 + 2 2 + -3 2 ´ 10 -9 ù ( ) ( ) ( ) úû 1 êë = 2 8.854 ´ 10 -12 ´ 2

2

= 0.3851 mJ/m 3

To find angle of D2 with z axis i.e. - a z is to be obtained by dot product. \

D2 · ( - a z ) =

D2

a z cos q 2

\

[0.8 a x + 2 a y - 3 a z ]· ( - a z ) =

\

+ 3 = 3.6932 cos q 2

( 0.8) 2 + ( 2) 2 + ( - 3) 2 cos q 2 i.e.

q 2 = 35.678° TM


... a z

= 1


Alternatively, tan q 2 = Now

\

5-8

D tan 2 ( 0.8) 2 + ( 2) 2 = D N2 3

i.e.

D2 =

( 0.8) 2 + ( 2) 2 + ( - 3) 2 = 3.6932

D1 =

( 2) 2 + (5) 2 + ( -3) 2 = 6.1644

D2 D1

=

= c e e0 E = c e e0

As

D

=

( c e + 1) e 0 E

\

P

=

(c e ) D ( c e + 1)

\

P

=

P2 P1

=

q 2 = 35.678°

3.6932 = 0.599 6.1644

P

\


( e R - 1) eR

D

( c e + 1) e 0

But

eR = c e + 1

D

e R2 - 1 e 1 ´ D2 ´ R1 ´ e R2 e R1 - 1 D 1

= 0.599 ´

( 2 - 1) 2

´

5 = 0.3743 (5 - 1)


E1 = 100 a x + 80 a y + 60 a z

At boundary,

E1 = Etan 1 + EN1

Now EN1 is projection of E1 in the normal direction, given by the dot product. \

EN1 =

[ E1 ×a N12 ] a N 12

é = ê 100 a x + 80 a y + 60 a z ë

(

= \ \

)×æçè - 72 a x + 73 a y + 76 a z ö÷ø ùúû a N12 …

( - 28.57 + 34.285 + 51.428 ) a N12

= 57.143 a N12

EN1 = – 16.326 a x + 24.489 a y + 48.979 a z V/m Etan 1 = E1 - EN1 = 116.326 a x + 55.511 a y + 11.021 a z V/m

At the boundary, Etan 2 = Etan 1

and

e EN1 = 2 e1 EN2

TM



\ \

EN2 =

5-9


3 e0 e1 = 1.5 EN1 EN1 = E e2 2e 0 N1

E2 = Etan 2 + EN2 = Etan 1 + 1.5 EN1 = 91.837 a x + 92.245 a y + 84.489 a z V/m

Example 5.9.5 Solution : As shown, z axis is normal to the surface. So part of E1 which is in the direction of a z is normal component of E1 . \ EN1 = 5 a z V m And \

E1 = EN1 + Etan 1 Etan 1 = E1 - EN1 = 2 a x - 3 a y V/m

... (1)

At the boundary of perfect dielectrics,

Now

Etan 1 = Etan 2 = 2 a x - 3 a y V/m

... (2)

Dtan 2 = e 2 Etan 2 = e 0 e r2 Etan 2

... (3)

and

DN1 = e 1 EN1 = e 0 e r1 EN1

... (4)

But

DN2 = DN1 = e 0 e r1 EN1

... (5)

and

D2 = DN2 + Dtan 2 = e 0 e r1 EN1 + e 0 e r2 Etan2

[(

)

= e 0 5 2 a x - 3 a y + 2 (5 a z ) \

] = 8.854 ´ 10 [10 a -12

x

- 15 a y + 10 a z

]

D2 = 88.54 a x - 132.81 a y + 88.54 a z pC/m 3

As D N1 , E N1 are in same direction and D 1 , E 1 are in same direction, D N1 = D 1 cos q 1¢ i.e. E N1 = E 1 cos q 1¢ where q ¢1 is angle measured w.r.t. normal. E N1 = 5 and E 1 = \

cos q 1¢ =

E N1 5 = E1 6.1644

2 ( 2) 2 + ( - 3) + (5) 2 = 6.1644

i.e. q 1¢ = 35.795°

This q 1¢ is angle made by E1 with the normal while q 1 is shown with respect to horizontal. \ q 1 = 90 - q 1¢ = 90 – 35.795 = 54.205° Similarly if q ¢2 is angle made by E2 with the normal then, cos q ¢2 =

e 0 e r1 EN1 E N2 D N2 D N1 = = = E2 D2 D2 D 2

= \

e0 ´ 2 ´ 5 e 0 ´ 10 2 + ( - 15) + (10) 2 2

q 2¢ = 60.982°

=

10 = 0.485 20.6155

i.e. q 2 = 90 - q 2¢ = 29.017° TM


... D N2 = D N1


5 - 10


Example 5.9.6

z

Solution : The arrangement is shown in the Fig. 5.4. E1 = 5 a x – 2 a y + 3 a z kV/m i) The normal directions to the plane z = 0 are ± a z . \ EN1 = 3 a z , Etan 1 = 5 a x – 2 a y From boundary conditions,

z>0 er1 = 4

z<0 er2 = 3

a1 q2

E2

i.e.

EN2

E q1 E1 N1

z=0

a2 EN2 Etan1 – az

Etan 1 = Etan 2 = 5 a x – 2 a y E N1 e 3 = r2 = E N2 e r1 4

az

Etan2

4 = EN1 = 4 a z 3

Fig. 5.4

\ E2 = EN2 + Etan 2

ii)

= 5 a x – 2 a y + 4 a z kV/m E N1 3 = tan q 1 = E tan1 25 + 4 tan q 2 =

iii)

WE1 =

E N2 = E tan2

4 25 + 4

i.e.

q 1 = 29.12º

…with interface

i.e.

q 2 = 36.6º

…with interface

1 1 e | E |2 = ´ 4 ´ 8.854 ´ 10 -12 ´ ( 25 + 4 + 9) 2 ´ 10 6 2 1 1 2 3

WE2

= 672.904 µJ/m 1 1 = e | E |2 = ´ 3 ´ 8.854 ´ 10 -12 ´ ( 25 + 4 + 16) 2 ´ 10 6 2 2 2 2 3

= 597.645 µJ/m

iv) Cube is at (3, 4, – 5) i.e. z = – 5 hence it is in the region z < 0 with e r1 = 3. Volume of cube = 2 ´ 2 ´ 2 = 8 m3 \ Energy in cube = WE2 ´ volume = 597.645 ´ 8 = 4.781 mJ Example 5.9.7

Jtan2

Solution : The arrangement is shown in the Fig. 5.5.

J2 (E2)

The electric field intensities are E 1 and E 2 in media 1 and 2 respectively.

JN2 Medium 2 s2

q2

As per boundary conditions,

Boundary

E tan1 = E tan2 While J 1 and J 2 are current densities in the two media. Similar to flux densities D the boundary condition for current densities J states that J N1 = J N2 . E tan1 E tan2 = … (1) \ J N1 J N2 TM


q1

Medium 1 s1

JN1 J1 (E1) Jtan1

Fig. 5.5


5 - 11


From Fig. 5.5, J N1 = J 1 cos q 1 and J N2 = J 2 cos q 2

… (2)

From the point form of Ohm's law, J = sE \

E tan1 =

J tan1 s1

and

E tan2 =

From Fig. 5.5, J tan1 = J 1 sin q 1 \

E tan1 =

and

J 1 sin q 1 s1

J tan2 s2

J tan2 = J 2 sin q 2

and E tan2 =

J 2 sin q 2 s2

… (3)

Using equations (2) and (3) in equation (1), æ J 1 sin q 1 ö æ J 2 sin q 2 ö ç s ÷ ç s ÷ è ø è ø 1 2 = J 1 cos q 1 J 2 cos q 2 \

i.e.

tan q 1 tan q 2 = s1 s2

tan q 1 s = 1 tan q 2 s2

… Proved

Example 5.9.8 Solution : The normal direction to the y = 0, plane is a y hence out of E 2 , 12 a y is the normal component of E 2 . \ E N2 = 12 a y V/m But \

E 2 = E tan 2 + E N2

E1 er1 = 4 y=0 er2 = 1

E tan 2 = 5 a x + a z V m

At the boundary of perfect dielectrics, E2 = 5ax + 12ay + az

E tan 1 = E tan 2 = 5 a x + a z V m and

E N1 e = r2 E N2 e r1

\

E N1 = 3 a y

\

i.e.

1 E N1 = 12 a y 4

Fig. 5.6

E1 = E tan 1 + E N1 = 5 a x + 3 a y + a z V m

Example 5.13.4 Solution : r S = 2 mC / m 2 , A = 1 m 2 , Gradient = 105 V/m, d = 1 mm Q = r S ´ A = 2 ´ 10 –6 C

\

Gradient =

V V i.e. 105 = d 1 ´ 10 –3

C =

Q 2 ´ 10 –6 = = 20 nF V 100

i.e. V = 100 V

TM



5 - 12



2

A = 100 cm , d = 0.03 mm, e r = 3 C =

e e A eA 8.854 ´ 10 -12 ´ 3 ´ 100 ´ 10 -4 = 0 r = d d 0.03 ´ 10 -3

= 8.854 ´ 10 -9 F E = 200 kV/cm = \

…Capacitor 200 ´ 10 3 10 -2

V/m = 2 ´ 107 V/m

V = E ´ d = 2 ´ 107 ´ 0.03 ´ 10 -3 = 600 V

…Voltage rating

Example 5.13.6 Solution : The arrangement is shown in the Fig. 5.7. The e r varies linearily from e r1 to e r2 . The equation for this linear behaviour is, er = K x + A At x = 0,

e r = e r1

d x=0

er varies linearily er1

\ A = e r1

x=d

er2

Fig. 5.7

At x = d,

e r = e r2

\

e r2 = K d + e r1

i.e.

K=

er 2 - er 1 d

é e - er 1 ù er = ê r 2 ú x + er 1 d û ë Let the plate at x = 0 carries positive charge. +rS \ ax E1 = 2e \

while \

E2 =

... (1)

... Due to plate at x = 0

-rS (- a x ) 2e

E = E1 + E2 =

×

+

rS a e x

V = - ò E dL = -

= -

x = 0

ò

x = d

... Due to plate at x = d ... Between the plates

x= 0

ò

x= d

rS a e x

× dx a

rS é æ e r2 - e r1 ö ù ÷ x + e r1 ú e 0 ê çè d ø ë û

x

dx

x= 0

= -

rS ì æ e - e r1 ö ü d ´ ln í ç r2 ÷ x + e r1 ý ( d e e0 è ø r2 - e r1 ) î þx = d TM


... e = e 0 e r

... a x

×a

x

=1


= =

5 - 13


rS d ln {e r1} - ln {e r2 - e r1 + e r1} ] e 0 ( e r2 - e r1 ) [

-rS d rS d ée ù ée ù ln r1 = ln ê r2 ú e 0 [e r2 - e r1 ] êë e r2 úû e 0 ( e r2 - e r1 ) ë e r1 û

And

Q = rS A

\

C=

Q = V

rS A rS d æe ö ln ç r2 ÷ e 0 ( e r2 - e r1 ) è e r1 ø

i.e.

C =

e 0 ( e r2 - e r1 ) A F æe ö d ln ç r2 ÷ è e r1 ø

Example 5.13.7 Solution : A = 1 cm 2 = 1 ´ 10 - 4 m 2 , d = 1 cm = 1 ´ 10 - 2 m, e r = 6, e 0 = 8.854 ´ 10 - 12 For parallel plate capacitor, e e A eA 6 ´ 8.854 ´ 10 - 12 ´ 1 ´ 10 - 4 = 0.5312 pF C = = r 0 = d d 1 ´ 10 - 2 Example 5.13.8 Solution :

A = 0.8 m 2 ,

E = 10 6 V m

d = 0.1 mm, er = 1000,

C =

e e A eA 8.854 ´ 10 –12 ´ 1000 ´ 0.8 = 0 r = = 70.832 mF d d 0.1 ´ 10 –3

E =

V d

i.e.

10 6 =

V 0.1 ´ 10 –3

i.e. V = 100 V


The capacitance of a co-axial cable is, C =

a = As ratio \

2peL b ln æç ö÷ èaø

where

a = inner radius,

0.0295 = 0.01475 inches 2

and

b is to be used, no need to convert to metres. a C =

(

2p ´ 8.854 ´ 10 -12 ´ 2.26 ´ 20 ´ 10 -2 0.058 ö ln æç ÷ è 0.01475 ø

b=

)

b = outer radius

0.116 = 0.058 inches 2

= 18.365 pF

Example 5.14.4 Solution : The D field between the plates in cylindrical co-ordinates is of the form D = D f a f where D f depends only on r. TM



5 - 14


Let plate at f = 0º is V = 0 and plate at f = aº is V = Vo. a æ Df ö Vo = – ò E · d L = – ò ç a f ÷ · (rdf a f ) e è ø f= 0

=

–D f r e

a

ò df =

f= 0

–D f ra e

Where e = e 0 e r

eVo ra The charge density on the plate f = a is eVo r S = D n = –D f = ra

\

Df = –

The total charge on the plate is, Q =

\

C =

ò r S ds =

L

ò

r2

ò

z = 0 r = r1

e Vo eVo L r2 dr dz = ln ra a r1

e L r2 Q = ln a r1 Vo

Example 5.15.3 Solution : The sphere is shown in the Fig. 5.8. d a = Radius of sphere = = 1 cm 2

3 cm

r1 = a + Thickness = 1 + 3 = 4 cm

a

This forms two capacitors in series. C 1 = Capacitor due to spherical arrangement of

r1

concentric spheres 4pe = æ1 - 1ö ç ÷ è a bø

Fig. 5.8

Here a = 1 cm and b = r1 = 4 cm \

C1 =

4p ´ 2.26 ´ e 0 éæ 1 1 ö æ öù ÷-ç ÷ú ê çè 2 2 ë 1 ´ 10 ø è 4 ´ 10 ø û

= 3.3527 pF

And C 2 = Due to isolated sphere of r1 = 4 cm = 4 p e r1 But between this sphere of radius r1 and sphere at ¥, the dielectric is free space e 0 . \

C 2 = 4pe 0 r1 = 4p ´ 8.854 ´ 10 -12 ´ 4 ´ 10 -2 = 4.4505 pF TM



\


1 1 1 + C1 C2

C eq =

=

5 - 15

... C 1 and C 2 in series

C1C2 3.3527 ´ 10 -12 ´ 4.4505 ´ 10 -12 = C1 + C2 10 -12 [3.3527 + 4.4505]

= 1.9121 pF Example 5.15.4 Solution : For spherical capacitor, 4pe C = and é1 - 1ù êë a b úû \

C =

a = 1 cm,

b = 2 cm and e r = 2

4p ´ 2 ´ 8.854 ´ 10 -12 = 4.4505 pF ù é 1 1 ú ê ë 1 ´ 10 -2 2 ´ 10 -2 û

Example 5.15.5 Solution : i) The capacitance of a single isolated sphere is, C = 4pea

... e = e 0 , a =

1.5 = 0.75 m 2

= 4 p ´ 8.854 ´ 10 - 12 ´ 0.75 = 83.447 pF ii) For co-axial cable, C =

2p e L where, L = 1.5 m, e r = 2.26, a = 0.6 mm, b = 3.5 mm b ln é ù ëê a ûú

2p ´ 8.854 ´ 10 - 12 ´ 2.26 ´ 1.5 = 106.935 pF 3.5 ù ln é êë 0.6 úû iii) Consider the cylindrical conductor suspended above conducting plane as shown in the Fig. 5.9. C =

The capacitance of this arrangement is given by, C = L

2pe cosh -1

the

b = 1.5 mm

h = 15 m Plane

æhö ç ÷ è bø

Fig. 5.9

e = e 0 , h = 15 m, b = 1.5 ´ 10 - 3 m

TM



C = L

\

5 - 16


2p ´ 8.854 ´ 10 - 12 = 5.6173 pF m 15 æ ö cosh - 1 ç ÷ è 1.5 ´ 10 - 3 ø

Example 5.16.5 Solution : This is composite capacitor with, d1 = 6 mm = 6 ´ 10 -3 m, e r1 = 4 d2 = 4 mm = 4 ´ 10 -3 m, e r2 = 1 and A = 500 × 500 mm = 0.25 m 2

and

C =

1 d1 d + 2 e1 A e2 A

=

A d1 d2 + e1 e2

=

0 . 25

=

6 ´ 10 -3 4 ´ 10 -3 + 4 ´ e0 1 ´ e0

2

0 . 25 ´ 8 . 854 ´ 10 -12 1. 5 ´ 10 -3 + 4 ´ 10 -3

= 402.4545 pF Example 5.16.6 Solution : The arrangement is shown in the Fig. 5.10. d 2 = 0.002 m er2 = 6

d = d 1 + d 2 = 0.01 m \

d 1 = 0.01 – 0.002 = 0.008 m e r2 = 6 e r1 = 6

C2 =

e2 A d2

A = 1 m2 e A C1 = 1 d1

Wood

d2

Air

d1 er1 = 1

and

Fig. 5.10

and two are in series.

\

C1 =

e 0 e r1 A 8.854 ´ 10 - 12 ´ 1 ´ 1 = = 1.1067 ´ 10 - 9 F d1 0.008

\

C2 =

e 0 e r2 A 8.854 ´ 10 - 12 ´ 6 ´ 1 = = 26.562 ´ 10 - 9 F d2 0.002

\

C eq =

C1 C2 1.1067 ´ 10 - 9 ´ 26.562 ´ 10 - 9 = C1 + C2 1.1067 ´ 10 - 9 + 26.562 ´ 10 - 9

= 1.0624 ´ 10 - 9 F = 1.0624 nF Example 5.16.7 Solution : In this case, the dielectric interface is normal to the plates. Hence two capacitors formed are in parallel. Now d = 2 mm same for both the dielectrics A 1 = 1.2 ´ 2 = 2.4 m 2

and TM

A 2 = 0.8 ´ 2 = 1.6 m 2



\

5 - 17


C1 =

e 0 e R1 A 1 8.854 ´ 10 -12 e R1 ´ 2.4 = = 1.062 ´ 10 -8 e R1 F d 2 ´ 10 -3

C2 =

e 0 e R2 A 2 8.854 ´ 10 -12 ´ 2.5 ´ 1.6 = = 1.7708 ´ 10 -8 F d 2 ´ 10 -3

\

C eq = C 1 + C 2 = 1.7708 ´ 10 -8 + 1.062 ´ 10 -8 e R1

But

C eq = 60 nF

\

60 ´ 10 -9 = 1.7708 ´ 10 -8 + 1.062 ´ 10 -8 e R1

\

i.e. 1.062 ´ 10 -8 e R1 = 4.2292 ´ 10 -8

e R1 = 3.9823


V = 100 V, 44.21 mJ/unit area is energy.

For a parallel plate capacitor, C =

e0 A d

\

E =

1 e0 A 2 V 2 d

But

E = Energy per unit area A

\ 44.21 ´ 10 - 6 = \

and

E= i.e.

1 C V2 2 E 1 e0 V2 = A 2 d

1 8.854 ´ 10 - 12 ´ ´ (100) 2 2 d

d = 1.0013 mm

... Seperation between plates

Example 5.17.6 C 1 = 2 mF, V = 100 V 1 1 E = C V 2 = ´ 2 ´ 10 –6 ´ 100 2 = 0.01 J 2 1 2 The total energy when two capacitors are connected in parallel must remain same as before. 2 2 1 1 E = C 1 Veq + C 2 Veq 2 2 2 2 1 1 0.01 = ´ 2 ´ 10 –6 Veq + ´ 2 ´ 10 –6 ´ Veq \ 2 2 Solution :

2

\

0.01 = 2 ´ 10 –6 Veq

\

Veq = 70.7106 V

…Voltage across each capacitor


The arrangement is shown in the Fig. 5.11. TM



5 - 18


a) The capacitor is not a function of voltage V or E, it depends on dielectric and physical dimensions. For coaxial conductors, A = 50 mm 2p e L C = b b ln æç ö÷ èaø =

2p (10 e 0 ) ´200 ´ 10 -3

é 100 ´ 10 -3 ù ln ê -3 ú ë 50 ´ 10 û b) E is function of r hence using,

×

+

V = - ò E dL = -

= -

r= a

ò

r= b

r= a

ò

r= b

= 160.518 pF B = 100 mm

Fig. 5.11

10 6 a r r

×[dr a ] r

10 6 r= 50 mm dr = - 10 6 [ ln r ]r = 100 mm r

[

= - 10 6 ln 50 ´ 10 -3 - ln 100 ´ 10 -3 c)

WE =

a

(

] = 693.1471 kV

1 1 C V 2 = ´160.518 ´ 10 -12 ´ 693.147 ´ 10 3 2 2

)

2

= 38.5606 J

Example 5.17.8 Solution : A = 50 ´ 50 cm2, V = 250 V, d 1 = 1 mm, e r = 1 \ \

C1 =

e0 A 8.854 ´ 10 –12 ´ 50 ´ 50 ´ 10 –4 = 2.2135 ´ 10 –9 F = d1 1 ´ 10 –3

Q = C 1 V = 2.2135 ´ 10 –9 ´ 250 = 5.5337 ´ 10 –7 C E 1 = Energy stored =

1 C V 2 = 6.9171 ´ 10 –5 J 2 1

When separation is increased to 3 mm i.e. d2 = 3 mm \

C2 =

e0 A 8.854 ´ 10 –12 ´ 50 ´ 50 ´ 10 –4 = = 7.3783 ´ 10 –10 F –3 d2 3 ´ 10

The charge remains same Q 5.5337 ´ 10 –7 = = 750 V \ V¢ = C2 7.3783 ´ 10 –10 E 2 = energy stored = \

1 C ( V 1 ) 2 = 2.075 ´ 10 –4 J 2 2

Work done = E 2 – E 1 = 1.3834 ´ 10 –4 J = 138.34 mJ

Example 5.17.9 2

Solution : d = 5 mm, S = 80 cm , e r = 10 TM


2 L=

00

mm


i)

C =

ii)

C = E = WE =

5 - 19


e S 8.854 ´ 10 -12 ´ 10 ´ 80 ´ 10 -4 = 141.664 pF = d 5 ´ 10 -3 Q V

i.e. Q = CV = 141.664 ´ 10

V 50 = = 10 kV/m, d 5 ´ 10 -3

– 12

´ 50 = 7.0832 nC 2

D = e 0 e r E = 0.8854 µC/m

1 1 2 – 12 2 CV = ´ 141.664 ´ 10 ´ (50) = 0.17708 µJ 2 2

iii) Though source is disconnected and the dielectric is removed, Q on the surface remains same \

Q = 7.0832 nC, D = r S = E =

WE = But

C =

\

WE =

iv)

V =

Q 7.0832 ´ 10 -9 2 = 0.8854 µC/m = -4 S 80 ´ 10

D 0.8854 ´ 10 -6 = = 100 kV/m e0 8.854 ´ 10 -12

… e r = 1 as dielectric removed

1 1 1 Q2 Q 2 CV2 = C é ù = 2 2 êë C úû 2 C e 0 S 8.854 ´ 10 -12 ´ 80 ´ 10 -4 = 14.1664 pF = d 5 ´ 10 -3 1 (7.0832 ´ 10 -9 ) 2 = 1.7708 µJ 2 14.1664 ´ 10 -12 Q 7.0832 ´ 10 -9 = 500 V = C 14.1664 ´ 10 -12

Example 5.17.10 Solution : A = 1 m 2 , d = 1 mm, e r = 25, C =

V = 1000 V

e e A eA 8.854 ´ 10 - 12 ´ 25 ´ 1 = 221.35 nF = 0 r = d d 1 ´ 10 - 3

Q = CV = 221.35 ´ 10 - 9 ´ 1000 = 2.2135 ´ 10 - 4 C eA For the plate separation 'x', the capacitor is C = . x For the fixed voltage V across the plates,

\

V

¶(CV) ¶C ¶Q = V = V2 ¶x ¶x ¶x

While the energy stored in a capacitor =

1 CV 2 2

TM



5 - 20


¶WE ¶C ¶ é1 1 = CV 2 ù = V 2 úû 2 ¶x ¶x ¶x êë 2

\

The force between the plates is given by, F = =

\

¶WE 1 ¶Q ¶C ¶C + V = - V2 + V2 2 ¶x ¶x ¶x ¶x

1 2 ¶C V 2 ¶x

F = -

but C =

eA ¶C - eA hence = x ¶x x2

(1000) 2 ´ 8.854 ´ 10 - 12 ´ 25 ´ 1 1 2 eA == – 110.675 N V 2 x2 2 ´ (1 ´ 10 - 3 ) 2

... x = d = 1 mm

Example 5.17.11 Solution : A = 30 ´ 30 = 900 cm 2 , V = 1000 V, d = 5 mm, e r = 1 \

C =

e0 er A 8.854 ´ 10 –12 ´ 1 ´ 900 ´ 10 –4 = 159.372 pF = d 5 ´ 10 –3

WE =

1 1 CV 2 = ´ 159. 372 ´ 10 –12 ´ (1000) 2 = 79.686 mJ 2 2

| E| =

V 1000 = 200 ´ 10 3 V m = d 5 ´ 10 –3

\ Energy density =

1 1 e e | E|2 = ´ 8.854 ´ 10 –12 ´ ( 200 ´ 10 3 ) 2 2 0 r 2

= 0.17708 J m 3

qqq

TM


6

Poisson’s and Laplace’s Equations Solutions of Examples for Practice

Example 6.2.5 Solution : V1 is function of r alone, V1 = \

Ñ 2 V1 =

while V2 = 3 hence

1 r2

¶ ¶r

6 r 1

æ 2 ¶ V1 ö ¶ ÷ = 2 çr ¶ ¶r ø r è r

¶ V2 =0 ¶r

6 ö 1 ¶ æ 2 ( - 6) = 0 çr ´ - 2 ÷ = 2 è ø r r ¶r

Ñ 2 V2 = 0

hence

i) Both V1 and V2 satisfy Laplace's equation. ii) V1 at r = 2 is V1 =

6 = 3, 2

V2 at r = 2 is V2 = 3

Example 6.2.6 Solution : Find Ñ 2 V Ñ2V =

¶ Vù ¶ é ¶2V 1 ¶ é 2 ¶ Vù 1 1 sin q + + r ú ê ú ê ¶ r û r 2 sin q ¶ q ë ¶ q û r 2 sin 2 q ¶ f 2 r2 ¶ r ë

=

( -2) ù 1 ¶ é 2 1 50 ¶ é ù + sin q ´ ´ cos q ú + 0 r ´ 50 sin q 2 ¶r ê 3 ú 2 sin ¶ q ê 2 r r û r q r û ë ë

=

1 ¶ é -100 sin q ù 1 ¶ é 50 sin q cos q ù + ú 2 ¶ r ëê 2 sin ¶ q ê ú r û r r q r2 û ë

= = =

1 r2

1 1 ¶ é 50 æ 1 ö ( -100 sin q) ç - ÷ + ´ ´ sin 2q ù 2 2 úû è r ø r sin q r 2 ¶ q êë 2

+100 sin q r4 100 sin q r4

+

+

1 r4

sin q

50 r4

sin q

-

´ 25 ´ 2 cos 2q = 100 sin q r4

=

100 sin q

50 r4

sin q

r4

+

¹0

Hence given potential field does not satisfy Laplace’s equation. (6 - 1) TM


1 r4

sin q

´ 50 [1 - 2 sin 2 q]


6-2

Poisson's and Laplace's Equations


The potential due to dipole is given by, Q d cos q K cos q V = = 2 4pe 0 r r2

…K=

Qd = Constant 4pe 0

According to Laplace's equation in spherical system, 1 ¶ æ 2 2 ö 1 ¶ æ K ö Ñ2V = ç r ´ K cos q ´ - 3 ÷ + 2 ç sin q ´ 2 ´ sin q ÷ + 0 2 ø r ¶r è r ø r sin q ¶q è r = = =

¶ é - 2 K cos q ù 1 ¶ æ K sin 2 q ö + ÷ ç ê ú r ¶r ë û r 2 sin q ¶q è r 2 ø

1 r2 1

1

´ - 2 K cos q ´ -

r2

2 K cos q r4

-

r2

2 K cos q

+

1 r 2 sin

q

´

K r2

´ 2 sin q ( - cos q)

= 0

r4

This shows that potential due to electric dipole satisfies Laplace's equation. Example 6.4.7 Solution :

As r v is not zero, use Poisson's equation. - - 10 6 e 0 r = 10 6 Ñ2V = - v = e e0

[

]

... e = e 0 as free space

As V is the function of x only and not of y and z, ¶2V Ñ2V = = 10 6 ¶x 2 Integrating,

¶V = ¶x

ò 10

Integrating,

V =

ò (10

6

dx + C 1 = 10 6 x + C 1 6x+

)

C 1 dx + C 2 =

10 6 x 2 + C1x + C2 2

At x = 0, V = 0 V hence, 0 = 0 + 0 + C 2 , C 2 = 0 At x = 1 mm = 1 ´ 10 -3 m, V = 2 V 10 6 1 ´ 10 -3 2

(

)

2

(

+ C 1 1 ´ 10 -3

\

2 =

\

V = 0.5 ´ 10 6 x 2 + 1500 x

)

i.e. C 1 = 1500

V

At x = 0.5 mm = 0.5 ´ 10 -3 m,

[

V = 0.5 ´ 10 6 0.5 ´ 10 -3

]

2

(

)

+ 1500 ´ 0.5 ´ 10 -3 = 0.875 V

TM



From V,

E = - ÑV = -

6-3


[

]

[

]

¶V ¶ ax = 0.5 ´10 6 x 2 + 1500 x a x = -1 ´ 10 6 x - 1500 a x ¶x ¶x

E x = - 1 ´ 10 6 x - 1500 V/m

\

At x = 1 mm = 1 ´ 10 -3 m, E x = -1 ´ 10 6 ´ 1 ´ 10 -3 - 1500 = - 2500 V/m Example 6.4.8 Solution : The field intensity is a function of r only and not of q and f. From Poisson's equation. Using spherical co-ordinate system, r rv 1 ¶ æ 2 ¶V ö i.e. ... (1) Ñ2V = - v çr ÷=2 e e ¶r ø r ¶r è It is known, E = - ÑV E is a function of r only and not of q and f. ¶V ¶V \ a = E r a r i.e. E r = E = ¶r r ¶r

... (2)

Case 1 : For r < a, E r = Ar 4 Ar +4 = -

\

Using in equation (1),

¶V ¶r

¶V = - Ar 4 ¶r

i.e.

(

1 ¶ - Ar 6 r 2 ¶r

) = - rev

r2

i.e. i.e.

1 r2

¶V = - Ar 6 ¶r

[- A6r ] = - re 5

r v = 6Aer 3 C/m 3

\

v

.. For r £ a

Case 2 : For r > a, E r = Ar -2 Ar -2 = -

\

Using in equation (1), \

¶V ¶r

i.e.

r2

¶V = - Ar -2 ´ r 2 = - A ¶r

r 1 ¶ ( - A) = - v 2 e ¶ r r

i.e.

0=-

rv e

r v = 0 C/m 2

.. For r > a

Example 6.4.9 Solution : The free space satisfies Laplace's equation as charge free. \ Ñ2V = 0 In given spherical shells, V is the function of r only and not of q and f. Hence in Ñ 2 V, ¶ V ¶ q and ¶ V ¶ f are zero. 1 ¶ æ 2 ¶ Vö \ Ñ2V = ÷=0 çr ¶r ø r2 ¶ r è

¶ æ 2 ¶ Vö ÷=0 çr ¶r è ¶r ø

i.e.

TM


... (1)


Integrating, r 2

6-4

¶V = 0 + C1 ¶r

where


C 1 = Constant of integration

... (2)

C1 ¶V = ¶r r2

\ Integrating,

V =

C1

ò r2

... (3) 1 dr + C 2 = C 1 æç - ö÷ + C 2 è rø

... (4)

C 2 = Constant of integration

where

At r = 0.1 m,

V=0

1 ö 0 = C 1 æç ÷ + C2 è 0.1 ø

i.e.

\ - 10 C 1 + C 2 = 0 At r = 2 m,

... (5)

V = 100 V

i.e

1 100 = C 1 æç - ö÷ + C 2 è 2ø

\ - 0.5 C 1 + C 2 = 100

... (6)

Solving (5) and (6), C 1 = 10.5263 and C 2 = 105.2631 10.5263 + 105.2631 r

\

V = -

Now

E = -ÑV = = -

¶V a ¶r r

... (7)

(other components zero)

¶ é 10.5263 10.5263 ù + 105.2631 a r = a r V/m ú ê r ¶r ë û r2

93.1998 ù é 10.5263 ar ú = a r pC / m 2 D = e0 E = e0 ê2 2 r r û ë

and

... free space

Example 6.4.10 Solution : Let the field varies along x direction only and the mean value of volume charge density is r V . According to Poisson's equation. r dV - r V d2 V d2 V = ... (1) = – V i.e. ò = x+A e e dx dx 2 dx 2 \

ò

- r V x2 dV = V= + Ax + B dx e 2 E = -

But \

E

... (2)

¶V ù ù é-r V é+r V ax = – ê x+ Aú ax = ê x - A úa x e e ¶x û û ë ë

= 100 V m at x = 0 m

E =

i.e. 100 = 0 – A i.e. A = – 100

rVx + 100 e

... (3)

TM



6-5


At x = 1550 m, | E| = 25 V/m and e = e 0 r ´ 1550 25 = v + 100 i.e. r v = - 0.4284 pC m 3 \ e0 4 p R 3 with 3

i) The volume of the atmosphere between the altitudes 0 to 1500 m is R=

1500 = 750 m. 2

\ Q (mean space charge) = r v ´

4 p R3 3

= - 0.4284 ´ 10 -12 ´

4 ´ p ´ (750) 3 = - 757.045 mC 3

ii) D = e 0 E, on the surface of earth x = 0 m D = e 0 ´ (E at x = 0) = e 0 ´ 100 C m 2

\

The D acts in the normal directions as per boundary conditions hence D = DN . \

r S on earth = | DN | = | 100 ´ e 0 | = 885.4 pC / m 2


r = 0.2m

The spherical shells are shown in the Fig. 6.1.

The E is in radial direction and hence V is also the function of r r = 0.1m alone and independent of q and f. 1 ¶ æ 2 ¶V ö ... Laplace's equation V = 100 V \Ñ2V = çr ÷=0 ¶r ø r 2 ¶r è ¶ æ 2 ¶V ö çr ÷ = 0 ¶r è ¶r ø ¶V Integrating, r 2 = ò 0 + C1 = C1 ¶r

\

\

ò

V =

At r = 0.1 m,

V = 0

\

0 = -

\ Hence

Fig. 6.1

... (1)

C ¶V = 1 ¶r r2

Integrating,

Solving,

e0

C1 r2

dr + C 2 = -

C1 + C2 r

... (2)

and r = 0.2 m, V = 100 V C1 + C2 0.1

and

100 = -

C1 + C2 0.2

C1 = 20, C2 = 200 20 + 200 V r ì 20 ¶V ¶ æ 1 öü a r = - é+ 200ù a r -í -20 ç - ÷ý a r E = -ÑV = ú ê r ¶r ¶r ë è r 2 øþ û î

V = -

TM



\

E = -

6-6

20 r2


a r V/m - 20 e 0

- 177.08

a r pC/m 2 r2 r Note that as outer shell is at higher potential, E is directed from outer to inner shell and hence in - a r direction.

\

D = e0 E =

2

ar =

Example 6.4.12 Solution : The potential is changing with respect to y only hence, Ñ2 V =

¶2V ¶ y2

=0

... Laplace's equation

Integrating twice, V = C 1 y + C 2 \

¶V ¶V ù é¶ V ax + ay + a z ú = - C1 a y E = -Ñ V = - ê x y z ¶ ¶ ¶ û ë

\

D = e 0 E = - C 1 e 0 a y C/m

But given

D = 253 a y N C/m C1 = -

\

253 ´ 10 - 9 8.854 ´ 10 - 12

2

2

253 ´ 10 -9 = - C 1 e 0

i.e

= - 28.574 ´ 10 -3

At y = 0.01 m, V = 0 i.e. 0 = - 28.574 ´ 10 3 ´ 0.01 + C 2 \

C 2 = 285.746 V = - 28.574 ´ 10 3 y + 285.746

\

So voltage at y = 0 m is, V = 285.746 V and voltage at y = 0.02 m is, V = – 285.746 V Example 6.4.13 Solution :

The planes are shown in the Fig. 6.2.

The potential is constant with respect to r and z so the Laplace’s equation for variable f is, 1 ¶2V = 0 r ¶ f2 Integrating,

¶V = C 1 and ¶f

Integrating, V = C 1 f + C 2

At f = 0°, V = 0 hence C 2 = 0 \

V = C1 f

TM



6-7

At f = a°, V = V0 hence C 1 =

V0 a

z

V0 f a

\

V =

Now

E = -Ñ V

V=0

Plane 1

In cylindrical co-ordinates, ÑV=


¶V 1 ¶V ¶V a + a + a ¶ r r r ¶f f ¶ z z

f = 0º

ì 1 ¶ é V0 ù ü \ E = - í0 + êë a fúû a f + 0ý r f ¶ î þ =-

V = V0

V0 a V/m ar f

f = aº

Fig. 6.2


i) Poisson's equation states that, r Ñ2V = - v e0 Ñ2V = -

\

200 e 0

r 2.4

e0

From the conditions given q and f. 1 ¶ \ Ñ2V = r2 ¶ r \

¶ ¶r

=-

... As free space e = e 0

200 r 2.4

it is clear that V is a function of r only and not the function of é 2 ¶ Vù êr ¶ r ú û ë

i.e.

1 ¶ r2 ¶ r

200 é 2 ¶ Vù ê r ¶ r ú = - 2.4 r û ë

é 2 ¶ Vù - 0.4 ê r ¶ r ú = - 200 r û ë

Integrate, r 2

¶V 200 r 0.6 = - ò 200 r - 0.4 dr + C 1 = + C 1 = - 333.33 r 0.6 + C 1 0.6 ¶r

As E is the function of r only we can write, E = - Ñ V = and \

Plane 2

a

Er = -

¶V a = Er a r ¶r r

¶V ¶r

... (2)

- r 2 E r = – 333.33 r 0.6 + C 1

... (3)

But as r ® 0, r 2 E r ® 0 \

0 = 0 + C1

Using in equation (1), r 2

... (1)

.... (Given) i.e.

C1 = 0

... (4)

¶V = – 333.33 r 0.6 ¶r

i.e. TM

¶V = – 333.33 r - 1.4 ¶r



Integrate,

6-8

ò

V = – 333.33

833.325

( ¥ ) 0.4 833.325

V =

\

r - 1.4 dr + C 2

r - 0.4 833.325 + C2 = + C2 ( - 0.4) ( r ) 0.4

= – 333.33 Use V ® 0 as r ® ¥, 0 =


( r) 0.4

+ C2 = 0 + C2

... (5)

i.e. C 2 = 0

V

... (6)

ii) Let us verify this using Gauss's law.

× Ñ× E =

Ñ D = rv \

×

i.e. Ñ e 0 E = r v

+ 200 e 0 rv 200 where E = E r a r and no other component exist. = = e0 r 2.4 e 0 r 2.4

Consider the radial component of E in spherical co-ordinate system and hence divergence of E is, 1 ¶ 200 ¶ 200 i.e. r 2 Er = r 2 Er = 2 ¶r 2.4 ¶r r r r 0.4

(

)

(

Integrate, r 2 E r = 200

)

r - 0.4 + 1 + C 1 = 333.33 r 0.6 + C 1 0.6

But r 2 E r ® 0 as r ® 0 hence C 1 = 0 \

r 2 E r = 333.33 r 0.6

\

E = E r a r = 333.33 r - 1.4 a r V/m

Now

V = - ò E dL

\

V = - ò 333.33 r -1.4 a r dr a r = - 333.33 ò r - 1.4 dr

×

= - 333.33

where dL = dr a r in radial direction

×

r - 0.4 833.33 + C2 = + C2 - 0.4 ( r ) 0.4

But V = 0 as r ® ¥ hence C 2 = 0 \

V =

833.33

( r) 0.4

V

This is same as obtained above using Poisson's equation.

qqq TM


7

Magnetostatics Solutions of Examples for Practice

Example 7.3.3 Solution : Fig. 7.1.

The points are shown in the Point 1 A (1,2,4) dL

The direction of dL is from A to B. So let us obtain unit vector in the direction from B (–2,–1,3) A to B. RAB a AB = | RAB| =

=

O

y R12

( -2 - 1) a x + ( -1 - 2) a y + ( 3 - 4) a z ( -3) 2 + ( -3) 2 + ( -1) 2

x

Point 2 C (3,1,–2)

-3 a x - 3 a y - a z

Fig. 7.1

19

\ dL = dL a AB =

Now, a R12 =

\

aR12

R12 R12

dL ´ a R12 =

[

10 -4 -3 a x - 3 a y - a z

]

19 =

( 3 - 1) a x + (1 - 2) a y + ( -2 - 4) a z ( 2) 2 + ( -1) 2 + ( -6) 2

=

2ax -ay - 6az 41

ax ay az - 3 - 3 -1 = 17 a x - 20 a y + 9 a z 3 -1 -6 ... (without considering

\

I dL ´ a R12 =

... From A to C

[

6 ´ 10 -4 17 a x - 20 a y + 9 a z 19 ´ 41

] = 2.1497 ´ 10 -5

(7 - 1) TM


1 and 19

1 ) 41

[17 a x - 20 a y + 9 a z ]


\

dH =

7-2

I dL ´ a R12 4p (R 12 )

2

Magnetostatics

[

2.1497 ´ 10 -5 17 a x - 20 a y + 9 a z

=

4p ´ ( 41 )

[

]

2

]

= 4.172 ´ 10 -8 17 a x - 20 a y + 9 a z

= 0.7093 a x - 0.8344 a y + 0.3755 a z mA/m Example 7.3.4 Solution : P(0, 1, 0) is a point at which H is to be obtained. Case 1 : - ¥ < z < ¥ hence general point on the element is (2, – 4, z). R12 = ( 0 – 2) a x + 1 – ( – 4) a y + ( 0 - z) a z = – 2 a x + 5 a y - za z

[

4 + 25 + z 2 = z 2 + 29, a R12 =

| R12| =

dH P =

\

ax

ay

az

0 -2

0 5

0.4 -z dHP

]

I dL ´ a R 12 4p R 212

=

R12 | R12|

(

0.4dz a z ´ –2 a x + 5 a y – z a z

(

)

4 p z 2 + 29

)

z 2 + 29

= – 0.8 a y – 2 a x é ù ê - 0.8 a y - 2 a x ú = ê dz 3/ 2 ú 2 êë 4p z + 29 úû

Putting z = 29 tan q,

(

¥

i.e.

)

¥

dz

ò

z = –¥

(z 2 + 29)

3/ 2

ò

HP =

z = –¥

=

1 29

+90 º

ò

- 0.8 a y - 2 a x

(

)

4p z 2 + 29

cos q dq =

q = - 90 º

3/ 2

2 29

Changing limits of integration from z = 29 tan q. \ z = – ¥, q = –90º and z = +¥, q = 90º 2 – 0.8 a y – 2 a x = - 0.0591 a x – 0.0236 a y A/m \ HP = 4p ´ 29

[

]

Case 2 : -3 < z < 3 Only change is the limits of integration. 3

ò

\

HP =

1 29

=

+5.906º

ò cos q d q =

0.2058 29

–5.906º (z 2 + 29) 0.2058 [– 0.8 a y – 2 a x ] = – 0.00608 a x – 0.00243 a y A/m 4p ´ 29 z = –3

\

dz 3/ 2

Case 3 : 0 < z < ¥

TM


dz


7-3 ¥

ò

\

z= 0

\

HP =

dz

(z 2 + 29)

3/ 2

1 29

=

(

1 – 0.8 a y – 2 a x 4p ´ 29

Magnetostatics 90 º

ò

1 29

cos q dq =

0º

) = – 0.0295 a x – 0.0118 a y

Example 7.3.5 Solution : The small wire is shown in the Fig. 7.2. a RQP =

A/m

z

RQP | RQP|

RQP = (0 - 1) a x + (2 - 0) a y + (2 - 0) a z = -ax + 2ay + 2az |RQP | = \

a RQP =

P(0, 2, 2)

1+4+4 = 3 -ax + 2ay + 2az 3

ax 2A

I dL = 2 a x \

dH =

I dL ´ a RQP 4p R

I dL ´ a RQP =

y

Q (1, 0, 0)

x

2

ax 2 1 3

RQP

O

Fig. 7.2

ay 0 2 3

az 4 4 0 = az - ay 3 3 2 3

4 4 az - ay 3 3 = - 0. 0117 a y + 0 . 0117 a z A/m dH = 4 p ´ ( 3) 2

\

Example 7.3.6 Solution : The square is placed in the xy plane as shown in the Fig. 7.3. z

–yay Point 2

–2.5

C

P B –xax

R12 –2.5

P R12

D x

2.5

A

2.5

y

dx Point 1

Fig. 7.3 TM


dx


7-4

Magnetostatics

Consider differential element dx along AB of the square. dL = dx a x \ The R12 joining differential element to point P is, R12 = – x a x – y a y i.e.

\

dL ´ a R12 =

ax

ay

az

dx -x

0 -y

0 0

R12 =

x 2 + y 2 i.e. a R12 =

-x a x - y a y x2 + y 2

= - y dx a z

According to Biot-Savart law, dH =

=

I dL ´ a R12

(

- 2.5

ò

x = 2.5

Put x = 2.5 tan q,

H = 0°

ò

q = 45°

æç x 2 + y 2 ö÷ è ø

)

... y = 2.5 for segment AB

3/ 2

- 25 dx a z

(

... Considering R12

2

4 p x 2 + 2.5 2

)

= 2

3/ 2

0

ò

x = 2.5

- 25 dx a z

(

4 p x 2 + 2.5 2

)

3/ 2

dx = 2.5 sec 2 q dq

Limits, x = 2.5, q = 45°

= – 0.6366

+ y2

10 ´ ( - 2.5) dx a z

H =

\

x2

4p

4 p x 2 + 2.5 2 \

I ( - y dx) a z

=

4 p R 212

and

25 ´ 2 4p

0°

ò

x = 0, q = 0° 2.5 sec 2 q dq a z

q = 45° ( 2.5 )

3

(1 + tan 2 q)

3/ 2

= - 0.6366

0°

ò

q = 45°

1 dq a z sec q

0°

cos q dq a z = – 0.6366 [ sin q ]45° a z = – 0.6366 [ 0 - sin 45° ]a z

= 0.4501 a z A/m This H is due to the segment AB of the square. All sides will produce same H at point P. \ Htotal = 4 H = 4 ´ 0.4501 a z = 1.8 a z A/m Example 7.3.7 R12

Solution :

IdL P(3,4,5)

R12 = -3 a x - 4 a y - 5 a z \

a R12

-3 a x - 4 a y - 5 a z R12 = = |R12| 32 + 42 +52 = -0.4242 a x - 0.5656 a y - 0.7071 a z TM


O (0,0,0)

Fig. 7.4


\

dH2 =

7-5

Magnetostatics

I 1 dL1 ´ a R12 2

4p R 12 I 1 dL1 = I 1 dL1 ´ a R12 =

I 1 dL(given) = 3p (a x + 2a y + 3a z ) mAm ax 3p

ay 6p

az 9p

-0.4242 -0.5656 -0.7071 = -13.328 a x - 12 a y - 5.33 a z + 8 a z + 6.664 a y + 16 a x = 2.672 a x - 5.336 a y + 2.67 a z \

dH2 =

2.672 a x - 5.336 a y + 2.67 a z 4p ´ ( 50 ) 2

= 4.252 a x - 8.4925 a y + 4.252 a z nA m

Example 7.5.4 Solution : i) P (2, 2, 0) : Consider the four sides separately as shown in the Fig. 7.5. I I sin a 2 – sin a 1 ] + sin a 2 – sin a 1 ] \ HP = [ 4pr1 4pr2 [ +

I I sin a 2 – sin a 1 ] + sin a 1 – sin a 2 ] [ 4pr 3 4pr4 [ r1 = 2

r2 = 2 P(2,2,0)

2 2 a2 a1

I = 10 A

a2

P(2,2,0)

2

2

a1

6 I = 10 A

a2 = 45º a1 = –45º

a2 = 45º 6 a1 = –tan–1 –— = –71.56º 2 (b)

(a) r3 = 6 P(2,2,0)

I = 10 A

2

I = 10 A

6

2

a1 a2

6

r4 = 2

2

a1

a2

P(2,2,0) 2 a2 = tan–1 –— = 18.43º 6

6 a2 = tan–1 –— = 71.56º 2 –1 2 a1 = –tan –— = – 45º 2

2 a1 = –tan–1 –— = –18.43º 6 (c)

(d)

Fig. 7.5 TM



Hp =

Magnetostatics

I I I [sin a 2 – sin a 1 ] + [sin a 2 – sin a 1 ] + [sin a 2 – sin a 1 ] 4pr1 4pr2 4pr 3 +

=

7-6

I [sin a 1 – sin a 2 ] 4pr4

10 ì sin 45 – sin ( –45) sin 45 – sin ( –71.56) + 4p íî 2 2 sin (18.43) – sin ( –18.43) sin (71.56) – sin ( –45)ü + + ý 6 2 þ

10 {0.7071 + 0.8278 + 0.1053 + 0.8278} = 1.964 a z A m 4p Use the above procedure for the remaining points and verify the answers : =

ii) 1.78 a z A m iii) –0.1178 a z A m Iv) –0.3457 a x – 0.3165 a y + 0.1798 a z A m Example 7.5.5 Solution : For a conductor in the form of regular polygon of n sides inscribed in a circle of radius R, the flux density B at the centre is given by, m nI p B = 0 tan æç ö÷ … Refer Ex. 7.5.3 è 2pR nø For given conductor, n = 6, R = 1 m, I = 5 A \

B =

4p ´ 10 -7 ´ 6 ´ 5 p tan æç ö÷ = 3.4641 mWb m 2 è6ø 2p ´ 1

Example 7.5.6 Solution : Consider the various sections of the B circuits. x Section I : Section AB is shown in the Fig. 7.6 (a) , PM is the perpendicular on AB. 3m M r = l (PM) \

2.5 a2 a1

r=2 m

5m P 2.5

The triangle ABC is right angled triangle. \

x = tan -1

4 = 53.13° 3

A

Fig. 7.6 (a)

\ a 2 = 90 - x = 36.8698° From the symmetry of the circuit, a 1 = a 2 = 36.8698° But as A is below point P, a 1 = - 36.8698° r = l (PM) = BP cos a 2 = 25 cos (36.8698°) = 2 m \

H1 =

4m

I sin a 2 - sin a 1 ] a N 4p r[

TM


C


=

7-7

Magnetostatics

10 sin 36.86°- sin ( - 36.86° ) ] a N = 0.4774 a N A/m 4 p ´ 2[

Section II : Section B to C along a semicircle as shown in the Fig. 7.6 (b).

B

R=2.5

The H at the centre of circular conductor is given by (I/2R) a N .

P

\ Due to semicircular portion, 1 I 10 ´ a = a = a N A/m H2 = 2 2R N 4 ´ 2.5 N Section III : Section C to A is shown in the Fig. 7.6 (c).

C

Fig. 7.6 (b) B

PM is perpendicular on AC. 2.5

As triangle ABC is right angled triangle, x = tan -1 \

P

3 = 36.8698° 4

3m

a 1 = 90 – x = 53.13°

2.5

a1

a2

x A

From the symmetry of the circuit,

2m

C

2m

M 4m

a 2 = a 1 = 53.13°

Fig. 7.6 (c)

But a 1 is negative as point C is below point P, a 1 = – 53.13° r = l (PM) = PC cos a 1 = 2.5 cos (53.13°) = 1.5 \ H3 =

I 10 sin 53.13°- sin ( - 53.13° ) ] a N = 0.8488 a N A/m sin a 2 - sin a 1 ] a N = [ 4p r 4 p ´ 1.5 [

Hence total H at point P is, H = H1 + H2 + H3 = 2. 3262 a N A/m Example 7.5.7 Solution : Consider the sections of given loop to calculate H at P. Section I : The portion AB of the circuit, as shown in the Fig. 7.7, PM is the perpendicular on AB from P. Note that a 1 and a 2 are to be measured from perpendicular line from P to the conductor. And r = l (PM)

B M

r a2 0.5 m a1 A

The triangle ABP is a right angled triangle hence, x = tan -1 \

x

AP 1 = tan -1 = 63.43° PB 0.5

a 2 = 90 - x = 26.565°

P

1m

Fig. 7.7

... from D ABP ... from D PMB

TM



And

7-8

Magnetostatics

a 1 = 90 - a 2 = 63.43°

But a 1 is negative as point A is below point P. r = l (PM) = BP cos a 2 = 0.5 cos (26.565°) = 0.4472 m \

H1 = =

I sin a 2 - sin a 1 ] a N 4p r[ 5 sin 26.565 - sin ( - 63.43° ) ] a N 4 p ´0.4472 [ B

= 1.1936 a N A/m where

a N = Unit vector normal to the plane is which the circuit is placed Section II : The semicircle B to C.

R=0.5 m

P

I

The H at the centre of circular conductor is (I/2R) a N where R is radius of the conductor. Hence H2 due to semicircular loop is, H2

C

1 I 5 = a = a = 2.5 a N A/m 2 2R N 4 ´ 0.5 N

Fig. 7.7 (a)

Section III : The portion C to A is shown in the Fig. 7.7 (b). A

PM is perpendicular on AC. x = tan -1

1m

P

x

a1

0.5 = 26.565° as triangle 1

r a2 0.5 m M

APC is right angled triangle \

90º C

Fig. 7.7 (b)

a 2 = 90 – x = 63.43°

And a 1 = 90 – a 2 = 26.565° But a 1 is negative as point C is below point P. r = l (PM) = PC cos a 1 = 0.5 cos (26.565°) = 0.4472 m \

H3 = =

I sin a 2 - sin a 1 ] a N 4p r[ 5 sin 63.43°- sin ( - 26.565° ) ] a N 4 p ´0.4472 [

L a 1 = –26. 56º

= 1.1936 a N A/m Hence the total H at point P is H = H1 + H2 + H3 = [1.1936 + 2.5 + 1.1936] a N = 4.8873 a N A/m TM



7-9

Magnetostatics

z

Example 7.5.8 Solution : Consider the arrangement as shown in the P (0, 0, 5) Fig. 7.8. a 1 = 0 and a 2 = - tan -1 H = where

2 = –21.801° 5

a2

r=5

I [sin a 2 - sin a 1 ] a N 4p r

0

I = 10 A 2

aN = ay

Side 1 Fig. 7.8

a 1 and a 2 are negative as both the ends of the conductors are below point P. \

H =

10 [sin( -21.801° ) - sin(0° )] (a y ) = – 0.0591 a y A/m 4p ´ 5


The arrangements are shown in the Fig. 7.9. z

z ¥

5m a r 2 3m a 1

¥ P

P(1,2,3) 3m

a1

y

y x

a2

5m

r

P(1,2,3) y

O x

x

z = –¥ (a)

(b)

(c)

Fig. 7.9

Case a : It is infinitely long straight conductor. I P (1, 2, 3), I = 10 A a , H = 2pr f Now \

r = H =

x2 + y 2 = 1 + 4 = 5 m

and

f = tan -1

y = tan -1 2 = 63.43º x

10 a f = 0.7117 a f A/m 2p´ 5

To find x component, take dot product with a x . \

H x = H · a x = 0.7117 a f · a x = – 0.7117 sin f

Similarly

H y = H · a y = 0.7117 a f · a y = + 0.7117 cos f

\

H x = – 0.6365,

and a f · a z = 0

H y = 0.3183 TM


x


\

7 - 10

Magnetostatics

H = – 0.6365 a x + 0.3183 a y A/m

Case b : It is a finite length conductor with z 1 = 0 and z 2 = 5 m. [Refer Fig. 7.9 (b)] y r = x 2 + y 2 = 1 + 4 = 5 m , f = tan -1 = 63.43º at point P x 3 = 53.3° but negative as that end is below point P. a 1 = tan -1 5 \ H=

a 1 = – 53.3°

and

a 2 = tan -1

2 = 41.81° 5

I 10 sin a 2 - sin a 1 ] a f = [sin 41.81°- sin ( - 53.3° )]a f = 0.5225 a f 4pr [ 4p´ 5

\

H x = H · a x = 0.5225 ( a f · a x ) = 0.5225 ( - sin f)

and

H y = H · a y = 0.5225 ( a f · a y ) = 0.5225 ( cos f)

\

H x = – 0.4673, H y = 0.2337 i.e. H = – 0.4673 a x + 0.2337 a y A/m

Case c : It is a conductor from z = 5 to z = ¥. [Refer Fig. 7.9 (c)] r =

x 2 + y 2 = 1 + 4 = 5 m, f = 63.43º

a 1 = tan -1

¥ 2 2 = tan -1 = 41.81° and a 2 = tan -1 = 90° r r 5

Both a 1 and a 2 are positive as above point P. \H =

I 10 sin a 2 - sin a 1 ]a f = [sin 90 - sin 41.81] a f = 0.1186 a f 4pr [ 4p´ 5

(

) = 0.1186 ( - sin f)

(

) = 0.1186

\

H x = H · a x = 0.1186 a f · a x

and

H y = H · a y = 0.1186 a f · a y

\

H x = – 0.106, H y = 0.053 i.e. H = – 0.106 a x + 0.053 a y A/m

Example 7.7.2

( cos f)


Example 7.7.3 Solution : For H on the axis of a circular loop given by, H =

I r2 2

2 (r + z 2 ) 3

2

a z A/m

In this example, r = 3 from x 2 + y 2 = (3) 2 , I = 10 A, z = ± 5. For (0, 0, 5), H = 0.227 a z A/m For (0, 0, – 5), H = – 0.227 a z A/m TM


... f = 63.43º


7 - 11

Magnetostatics

Example 7.7.4 Solution : We know that, H =

Ir 2 2 (r 2 + z 2 ) 3/ 2

a z A/m

In this example, r = 5 cm and I = 50 mA. The ring is in z = 1 plane and point P (0, 0, – 1) hence z = 2 cm. 50 ´ 10 -3 ´ (5 ´ 10 -2 ) 2 a z = 0.4 a z A/m H = \ 2[(5 ´ 10 -2 ) 2 + ( 2 ´ 10 -2 ) 2 ] 3 / 2 Example 7.7.5 Solution : The H at the centre of the circular coil with N turns is given by, NI ... R = r1 = 5 cm H = 2R 10 ´ 1

=

= 100 A/m

2 ´ 5 ´ 10 -2 The H at the centre of coil A i.e. point P due to coil B is, Ir 2 H =

(

2 r 2 + z2

)

P

10 cm

Q

3 2

Coil A N = 10 r1 = 5 cm I1 = 1A Coil B N=1 r2 = 7.5 cm I2 = ?

Fig. 7.10

where r = r2 = 7.5 cm, I = I2 and z = Distance between point P and coil = 10 cm \

(

I 2 7.5 ´ 10 -2

H =

(

é 2 ê 7.5 ´ 10 -2 ë

)

2

) + (10 ´ 10 -2 ) 2

3 2ù2

= 1.44 I2

úû

The total H at P = 100 + 1.44 I2 which must be zero, \

I2 =

-100 = – 69.44 A 1.44

The negative sign indicates direction of I2 is opposite to that of I1. Example 7.7.6 Solution : The coil is shown in the Fig. 7.11 placed in xy plane with z-axis as its axis. z = 100 m at which H is to be obtained. r = radius of coil =

TM

d = 25 m, I = 28 ´ 10 4 A 2



7 - 12

Magnetostatics z

Hence H at the point on the axis of a circular loop is, I r2

H = 2

(

r2

+ z2

)

az

3/ 2

28 ´ 10 4 ´ ( 25) 2

=

[

2 25 2 + 100 2

]

P

z = 100 m

az

3/ 2

d = 50 m O

y

= 79.894 a z A/m 4

I = 2.8×10 A x

Example 7.9.5

Fig. 7.11

Solution : The planes are shown in the Fig. 7.12. For sheet in z = 0 plane, aN = az

z

… towards P

For sheet in z = 4 plane, aN = – az

K = 18 ax

z=4

… towards P

P (1,1,1)

z=0

1 1 H1 = K ´ a N = ´ ( - 8 a x ) ´ a z = 4 a y 2 2 1 1 K ´ a N = ´ (18 a x ) ´ -a z = 9 a y 2 2 \ H at P = H1 + H2 = 13 a y A/m. H2 =

y K = – 8 ax

x

Fig. 7.12

Example 7.9.6 Solution : The current from current density is given by, I =

ò

×

J dS

dS = r dr df a z normal to a z as J is in a z 2p

\

I =

r

×

4.5 e -2r a z r dr df a z = 4.5

ò ò

f = 0r = 0

2p

r

ò ò

r e -2r dr df

f= 0 r = 0

Using integration by parts, 2p

= 4.5

ò

f= 0

df

{r ò

e -2r

dr - ò 1 ò

e -2r

}

dr dr

r 0

TM

r

ì r e -2r ü e -2r = 4.5 ( 2 p) í -ò drý -2 -2 î þ0



7 - 13

Magnetostatics

r

ì - r e -2r 1 -2r 1 ü 9 p ì r e -2r 1 e -2r ü = 9 pí = 9 pí - e + ý= + ý 4 -2 2 4 4 2 -2 î î þ þ0

{1 - 2 r e

-2r

- e -2r

}A

For r = 0.5, I = 7.068 [1 – 0.3678 – 0.3678] = 1.8676 A Consider a closed path with r ³ 0.5 such that the enclosed current I is 1.8676 A. According to Ampere's circuital law, ò H dL = I

×

\

2p

ò

H f r df = I

... H = H f a f , d L = r df a f

f= 0

\

2 p r H f = 1.8676

\

H =

Hf =

i.e.

0.2972 a f A/m r

1.8676 0.2972 = 2pr r for r ³ 0.5 m


For a given point P (0.01, 0, 0). r =

( 0.01) 2 + ( 0) 2 = 0.01 m = 10 mm

Thus P is in the region b < r < c. \

H =

I é c2 - r 2 ù a 2 p r ê c 2 - b2 ú f û ë

=

é (0.011) 2 - (0.01) 2 ù 6 = 50.113 a f A/m 2p ´ 0.01 êê (0.011) 2 - (0.009) 2 úú ë û


Due to infinite long conductor along z-axis. H1 =

20p ´ 10 -3 10 ´ 10 -3 I af = af = a f A/m r 2p r 2p r

At r = 0.5 cm, no current sheet is enclosed. H = H1 =

10 ´ 10 -3

a f = 2 a f A/m 0.5 ´ 10 -2 At r = 1.5 cm, current sheet at r1 = 1 cm is getting enclosed. It carries current in z direction. K1 = 400 ´ 10 - 3 a z A/m \

\

I enc = K1 ´ 2p r1 = 400 ´ 10 -3 ´ 2p ´ 1 ´ 10 -2 = 0.02513 A

... r1 = 1 cm for sheet

According to Ampere's circuital law,

ò

×

H d L = I enc

2p

i.e.

ò

H f r df = I enc

... H = H f a f and dL = r df a f

f= 0 TM



\

H f (2pr) = I enc

\

H2 =

7 - 14

i.e.

Hf =

4 ´ 10 -3 a f A/m r

I enc 0.02513 = 2pr 2pr and

14 ´ 10 -3

So at r = 1.5 cm, H = H1 + H2 =

Magnetostatics

1.5 ´ 10 - 2

H1 =

10 ´ 10 -3 a f A/m r

a f = 0.933 a f A/m

At r = 2.5 cm, second sheet also gets enclosed for which, K2 = - 250 ´ 10 -3 a z A/m \

I enc = K2 ´ 2 p r2 = - 250 ´ 10 -3 ´ 2p ´ 2 ´ 10 -2 = – 0.03141 A

... r2 = 2 cm for sheet

According to Ampere's circuital law,

ò \

×

2p

H dL = I enc

i.e.

ò

H f r df = I enc

f= 0

Hf =

I enc - 0.03141 = 2pr 2pr

i.e.

H3 =

- 5 ´ 10 -3 a f A/m r

4 10 5 H = H1 + H2 + H3 = æç + - ö÷ ´ 10 -3 a f è r r rø

So at r = 2.5 cm, =

9 ´ 10 -3 9 ´ 10 -3 a f = a f = 0.36 a f A/m r 2.5 ´10 -2

Example 7.9.9 Solution : The planes are shown in the Fig. 7.13. i) P(1, 1, 1) For z = 0 plane, a N = +a z at P 1 1 \ H1 = K ´ a N = [( –10 a x ) ´ a z ] 2 2 –10 = – a y = +5 a y A/m 2 For z = 4 plane, a N = – a z at P as P is below z = 4 plane. 1 1 \ H2 = K ´ a N = [( –10 a x ) ´ ( –a z )] 2 2

Q (0,–3,10)

[ ]

= \

[ ( )] = +5 a

10 – –a y 2

y

z

K = +10 ax

P(1,1,1) y

x

K = –10 ax z = 0

A/m

Fig. 7.13

H = H1 + H2 = 10 a y A/m at P(1, 1, 1) ii) Q(0, –3, 10)

For z = 0 plane, a N = +a z at Q hence H1 = +5 a y A/m. TM

z=4



7 - 15

Magnetostatics

For z = 4 plane, a N = +a z at Q as Q is above the plane 1 1 \ K ´ a N = [(10 a x ) ´ a z ]K a x ´ a z = –a y H2 = 2 2 = \

[ ]

10 – a y = –5 a y A/m 2

H = H1 + H2 = 0 A/m at Q (0, –3, 10)

Example 7.9.10 Solution : The sheet is located at y = 1 on which K is in a z direction. The sheet is infinite and is shown in the Fig. 7.14.

z y=1 plane

The H will be in x direction. a) Point A ( 0, 0, 0) a N = - a y normal to current sheet at point A \ Now

H =

[

1 1 K´ aN= 40 a z ´ - a y 2 2

]

H =

K = 40 az

x

az ´ ay = -ax

\

y

Fig. 7.14

1 + 40] a x = 20 a x A/m 2[

b) Point B (1, 5, - 2) This is to the right of the plane as y = 5 for B. \

a N = a y normal to sheet at point B

\

H =

[

1 1 K áN = 40 a z ´ a y 2 2

] = - 20 a x

A/m

Example 7.9.11 Solution : The arrangement is shown in the Fig. 7.15. r = 0.5 cm

r = 0.5 cm P

A

B P

0.5 m

0.5 m 1m

Right hand rule

(a)

(b)

Fig. 7.15

TM



7 - 16

Magnetostatics

The current carrying conductors are seperated by 1 m. The two currents are in opposite directions hence according to right hand thumb rule, the field produced at P is in the same direction due to both the conductors as shown in the Fig. 7.15 (b). I1 I H at P = H1 + H2 = \ af + 2 af 2p d 1 2p d 2 Now \

d1 = d2 = 0.5 m, I1 = I2 = 100 A H at P =

2 ´ 100 a = 63.6619 a f A/m 2p ´ 0 . 5 f

Now at the surface of conductor A, H1 = H2 =

\

I a 2p r f

where

r = Radius of conductor

I 2p ´ (1 - 0 . 5 ´ 10 -2 )

a f is due to other conductor

é ù 100 100 + a = 3199.09 a f A/m HA = ê -2 2p ´ 0 . 995 úú f êë 2p ´ 0 . 5 ´ 10 û

Same is the value of H on the surface of conductor B but in opposite direction. \

HB = – 3199.09 a f A/m

Example 7.9.12 Solution : Consider the conductor as shown in the Fig. 7.16. Region 1 : r < a Ienc = 0 \

as conductor is hollow

H = 0 A/m i.e. B = 0 Wb/m

2

…r
Region 2 : a < r < b Consider the cross-section upto radius r as shown in the Fig. 7.16. The total current I flows through the area p [b 2 - a 2 ]. Hence the current enclosed by the closed path is, p [r 2 - a 2 ] [r 2 - a 2 ] I’ = I= I p [b 2 - a 2 ] [b 2 - a 2 ]

b

Fig. 7.16 Hollow aluminium conductor

The H is only in a f direction hence H = H f a f Consider dL in a f direction i.e. dL = r df a f \

H · dL = H f r df TM



7 - 17

Magnetostatics

According to Ampere’s circuital law,

ò \

2p

ò

H · dL = Ienc = I’ H f r df =

f= 0

\

H f r [f]

2p 0

=

[r 2 - a 2 ] [b 2 - a 2 ] [r 2 - a 2 ] [b 2 - a 2 ]

i.e.

H =

\

B = m0 H =

i.e.

I

2pr[b 2 - a 2 ]

\

B =

H f r df =

[r 2 - a 2 ] [b 2 - a 2 ]

I

I

[r 2 - a 2 ] I

\

ò

Hf =

[r 2 - a 2 ] I 2pr[b 2 - a 2 ]

a f A/m

…a
4p ´ 10 -7 ´ [r 2 - (1 ´ 10 -2 ) 2 ] ´ 1000 2p r[( 2 ´ 10 -2 ) 2 - (1 ´ 10 -2 ) 2 ]

af

0.666[r 2 - 1 ´ 10 -4 ] 2 a f Wb/m r

…a < r < b

Region 3 : r > b The entire current gets enclosed i.e. Ienc = 1000 A \

\

H =

I 1000 159.154 a f A/m a = a = 2pr f 2p r f 4

2 ´ 10 -4 2 a f Wb/m B = m0 H = r

…r > b

The variation of B against r is shown in the Fig. 7.17.

B in 2 Wb/m 0.01 Wb/m 2

0 Axis of conductor

Fig. 7.17


a

From the point form of Ampere's circuital law, Ñ´H = J

In the cartesian system, ax ¶ Ñ´H = ¶x y cos (a x)

ay ¶ ¶y 0

az ¶ ¶z y + ex

TM


b

r


7 - 18

(

é ¶ y cos (a x) ¶ y + e x ù é ¶ x =ê y+ e úax + ê ¶z ¶x ê û ë¶y ë

(

)

(

)ùú a ú û

Magnetostatics

y

ù é ¶ + êy cos (a x) ú a z û ë ¶y

)

= (1) a x + 0 - e x a y + ( - cos a x) a z On yz plane, x = 0 \ J on yz plane = a x - e 0 a y - cos 0 a z = a x - a y - a z A / m 2 Example 7.10.4 Solution : In cylindrical co-ordinates Ñ ´ A is given by, é 1 ¶ (r A f ) 1 ¶ A r ù é1 ¶ A z ¶ A f ù é¶A r ¶A z ù ar + ê af + ê Ñ Á = ê úa ú ú r ¶f ú z ¶r ¶r û ¶z û ë ¶z êë r ër ¶f û Now A r = 0,

A f = sin 2f and A z = 0 ù é 1 ¶ ( r sin 2f) ¶ sin 2f ù é - 0ú a z \ Ñ ´ A = ê0 ú a r + [0 - 0] a f + ê r ¶ ¶ z r û ë û ë sin 2f sin 2f = [0 - 0] a r + 0 a f + az = az r r p p At æç 2, , 0ö÷ , r = 2, f = , z = 0 è 4 ø 4 p p sin æç 2 ´ ö÷ sin æç ö÷ è è2ø 4ø \ Ñ Á = az = a z = 0.5 a z 2 2 Example 7.10.5 In the spherical co-ordinates, curl H is given by,

Solution : Ñ´H =

é ¶ sin q H f ¶ H q ù 1 1 é 1 ¶ H r ¶ (r H f ) ù ar + ê úa ê ú r sin q ë r ê sin q ¶ f ¶q ¶f û ¶r ú q ë û +

1 é ¶ (r Hq ) ¶ Hr ù a r ê ¶r ¶ q úû f ë

Now H r = 0, \ Ñ´H =

H q = 2.5,

Hf = 5

¶ (5 r ) ù é ¶ 5 sin q ¶ 2.5 ù 1 1 é 1 é ¶ ( r 2.5) ù a + a + ê 0- 0ú a f r sin q êë ¶ q r ë ¶r ¶ r úû q ¶ f úû r r êë û

=

1 1 1 5 cos q - 0] a r + [-5] a q + [2.5] a f [ r sin q r r

=

5 5 2.5 cot q a r - a q + af r r r

TM



p At æç 2, , 0ö÷ , è 6 ø

r = 2, q =

Ñ´H =

\

7 - 19

Magnetostatics

p , f = 0° 6

5 p 5 2.5 cot ar - aq + a = 4.33 a r - 2.5 a q + 1.25 a f 2 6 2 2 f

Example 7.11.4 Solution : The portion of the cylinder is shown in the Fig. 7.18. The flux crossing the given surface is given by, f = ò B · dS S

z=1

dS

1m

dS normal to a r direction is, dS = r df dz a r \ f = ò m 0 H · dS

... B = m 0 H

z=0

S

= m0

1

p/4

ò ò

2.39 ´ 10 6 cos f df dz

f=0

z = 0 f= 0

f = p/4

Fig. 7.18 p/4

= 2.39 ´ 10 6 m 0 [ sin f] 0

[z]10

p \ f = 2.39 ´ 10 6 ´ 4 p ´ 10 -7 ´ é sin - sin 0ù [1 - 0] = 2.1236 Wb êë úû 4 Example 7.11.5 z

Solution : The arrangement is shown in the Fig. 7.19. B = m0 H = = f =

2.239 ´ 10 6 ´ 4p ´ 10 -7 cos f a r r dS

2.8136 cos f a r Wb m 2 r

ò

B · dS

s

p 4

=

ò

1

ò

f= - p 4 z= 0

2.8136 cos f ´ r df dz r p 4

= 2.8136 [sin f] - p

4

– p/4

p/4

Fig. 7.19

´ [z] 10

= 2.8136 ´ [0.7071 - ( - 0.7071)] = 3.9789 Wb

TM



7 - 20

Magnetostatics

Example 7.12.5 Solution : a)

The A is a vector magnetic potential. ¶A x ¶A y ¶A z + + Ñ· A = ¶x ¶y ¶z =

b)

¶ ¶ ¶ [ 3y - z]+ ¶ y [2 xz]+ ¶ z [0] = 0 ¶x

... Proved.

B = ÑÁ ÑÁ =

ax ay az ¶ ¶ ¶ ¶x ¶y ¶z 3y - z 2 xz 0

= a x [0 - 2x] + a y [ - 1 + 0] + a z [2z - 3] \

B = Ñ ´ A = - 2x a x - a y + (2z - 3) a z

At P (2, – 1, 3), \

x = 2, y = – 1, z = 3

B = - 4 a x - a y + 3 a z Wb/m 2 A at P = - 6 a x + 12 a y Wb/m H at P =

Now

J = Ñ´H H =

\

1 B = m0 m0

1 m0

[- 4 a x - a y + 3 a z ]

A/m

where

{- 2x a x - a y + (2z - 3) a z }

ax ay az 1 ¶ ¶ ¶ J = m 0 ¶x ¶y ¶z - 2x - 1 2z - 3 =

1 m0

{a x [0 - 0] + a y [0 - 0] + a z [0 - 0]}

= 0 A/m 2 Example 7.12.6 Solution : For a current sheet of current density K, the magnetic field intensity is given by, 1 K áN H = 2 For the given sheet, a N = a z and K = Ky a y TM



KáN =

\ \

H =

7 - 21

ax

ay

0

Ky

0

0

Magnetostatics

az 0 = Ky a x 1

1 K a A/m, 2 y x

B = m0 H =

m0 K a 2 y x

Wb/m 2

Now ÑÁ = B But Ñ ´ A in cartesian system is, ax ay az ¶ ¶ ¶ ÑÁ = ¶x ¶y ¶z Ax Ay Az As B has only a x component, consider a x of Ñ ´ A and equate. é ¶A z ¶A y ù m 0 m ¶A z ¶A y = = 0 Ky Ky a x i.e. \ ax ê ú 2 2 ¶z û ¶y ¶z ë ¶y The vector magnetic potential A must be independent of x and y. Thus -

\

¶A y ¶z

=

Integrating, A y =

m0 K 2 y

ò

Let at z = z 0 , A y = 0 0 = -

-

... As A y can be function of z.

m0 m K dz + C 1 = - 0 Ky z + C 1 2 y 2

m0 Ky z 0 + C 1 2

\

Ay =

m0 Ky 2

( z 0 - z)

\

A =

m0 Ky 2

( z 0 - z) a y

Ky a y = K

But

¶A z = 0. ¶y

hence

A =

i.e.

C1 =

m0 Ky z 0 2 … For z > 0

Wb/m m0 2

K ( z 0 - z ) Wb/m

Example 7.12.7 Solution : The B from vector magnetic potential is given by, ax ay az ¶ ¶ ¶ B = ÑÁ= ¶x ¶y ¶z Ax Ay Az \

B =

ax ay az ¶ ¶ ¶ = a x [3 - 3] + a y [2 - 2] + a z [5 - 3] = 2 a z ¶x ¶y ¶z 4x + 3y + 2z 5x+ 6y + 3z 2x+ 3y + 5z

TM



\

7 - 22

Magnetostatics

B = 2 a z Wb/m 2

Its magnitude is constant in the direction a z . It is a conservative field. Example 7.12.8 Solution :

10

A =

ax ¶ ¶x 10

B =Ñ Á =

x2 + y 2 + z2 =

x2

+ y2 + z2

ay ¶ ¶y

az ¶ ¶z

0

0

- 20 z (x 2

+ y2

+ z2)2

=

ay +

ax

ù ù ¶ é 10 ¶ é 10 ay ê 2 ú ê úa ¶z ê x + y 2 + z 2 ú ¶y ê x 2 + y 2 + z 2 ú z ë û ë û

20 y (x 2

+ y2 + z2)

az

qqq

TM


8

Magnetic Forces, Materials and Inductance Solutions of Examples for Practice

Example 8.2.4 Solution : Q = Charge = 5 ´ 10 – 18 C B = – 0. 4 a x + 0.2 a y – 0.1 a z T v = (2 a x – 3 a y + 6 a z ) ´ 105 m / s a) By definition, F = Q ( E + v ´ B) At

t = 0,

\

0 = 5 ´ 10 – 18 [E + (2 a x – 3 a y + 6 a z )105 ´ (– 0.4 a x + 0.2 a y – 0.1 a z )]

\

E = – [(2 a x – 3 a y + 6 a z )105 ´ (– 0.4 a x + 0.2 a y – 0.1 a z )]

\

E = – 105

\

E = – 105 [(+0.3 – 1.2) a x – (– 0.2 + 2.4) a y + (0. 4 – 1.2) a z ]

\

E = 0.9 ´ 105 a x + 2.2 ´ 105 a y + 0.8 ´ 105 a z

\

E = (0.9 a x + 2.2 a y + 0.8 a z )10 5 V / m

F= 0

ax

ay

az

2 –3 6 – 0.4 0.2 – 0.1

b) Let E field be in x-direction only. Then we can write, E = Ex a x The force is given by, F = Q ( E + v + B) \

F = 5 ´ 10 – 18 [(E x a x ) + {(2 a x – 3 a y + 6 a z )105 ´ (– 0.4 a x + 0.2 a y – 0.1 a z }]

\

F = 5 ´ 10 – 18 [(E x a x – (0.9 ´ 105 a x + 2.2 ´ 105 a y + 0.8 ´ 105 a z )] (8 - 1) TM



8-2

Magnetic Forces, Materials and Inductance

F = 5 ´ 10 – 18 [(E x – 90000) a x – 22000 a y – 80000 a z ]

\

But|F| = 2 ´ 10 –12 N. Hence finding magnitudes on both the sides and equating, \

2 ´ 10 – 12 = 5 ´ 10 – 18 (E x – 90000) 2 + (– 220000) 2 + (– 80000) 2

\ (E x – 90000) 2 + (– 220000) 2 + (– 80000) 2 = (400000) 2 Solving for E x , we get \ (E x – 90000) 2 = 1.052 ´ 10 11 \ Taking square root we get, E x – 90000 = ± 324345.5 E x = 414. 355 ´ 103 V / m or E x = – 234.3955 V / m

\


Given : Q = 18 nC = 18 ´ 10 - 9 C v = v a v = 5 ´ 10 6 [0.6 a x + 0.75 a y + 0.3 a z ] m/s i) The electric force exerted by E on charge Q is given by, Fe = Q E = 18 ´ 10 - 9 [( - 3 a x + 4 a y + 6 a z ) ´ 10 3 ] = - 54 ´ 10 - 6 a x + 72 ´ 10 - 6 a y + 108 ´ 10 - 6 a z = ( - 54 a x + 72 a y + 108 a z ) mN The magnitude of the force exerted on the charge is given by, | Fe| =

( -54 ´ 10 -6 ) 2 + (72 ´ 10 -6 ) 2 + (108 ´ 10 -6 ) 2 = 140.5844 mN » 140.6 mN

ii) The magnetic force exerted by B on the charge Q is given by Fm = Q v ´ B = 18 ´ 10 - 9 [5 ´ 10 6 (0.6 a x + 0.75 a y + 0.3 a z )

´ (- 3 a x

+ 4 a y + 6 a z ) 10 - 3 ]

= [(0.054 a x + 0.0675 a y + 0.027) ´ ( - 3 a x + 4 a y + 6 a z ) ´ 10 - 3 )]

=

ax

ay

az

0.054 - 3 ´ 10 - 3

0.0675 4 ´ 10 - 3

0.027 6 ´ 10 - 3

= [( 405 - 108) ´ 10 - 6 ] a x - [( 324 + 81) ´ 10 - 6 ] a y + [(216 + 202.5) ´ 10 - 6 ] a z = (297 a x - 405 a y + 418.5 a z ) ´ 10 - 6 N Hence the magnitude of the force exerted on charge is given by, TM



|Fm | =

8-3


(297 ´ 10 - 6 ) 2 + ( - 405 ´ 10 - 6 ) 2 + (418.5 ´ 10 - 6 ) 2 = 653.74 mN

iii) Total force exerted on charge both B and E acting together is given by, F = Fe + Fm

= [- 54 ´ 10 - 6 a x + 72 ´ 10 - 6 a y + 108 ´ 10 - 6 a z ] + [297 ´ 10 - 6 a x - 405 ´ 10 - 6 a y + 418.5 ´ 10 - 6 a z ]

= (243 a x - 333 a y + 526.5 a z ) mN Hence (243 ´ 10 - 6 ) 2 + ( - 333 ´ 10 - 6 ) 2 + (526.5 ´ 10 - 6 ) 2 = 668.6855 mN ~ 668.7 mN

|F| = Example 8.2.6

Solution : i) Let the position of the charge is given by P(x, y, z). The force exerted on charge by E is given by, F = QE According to Newton's second law,

… (1)

dv d 2 z … (2) = dt dt 2 Equating equations (1) and (2) we can write, d2 z = Q × E = (– 0.3 ´ 10 – 6 ´ 30 a z ) … (3) m dt 2 The initial velocity is constant and it is in x-direction so no force is applied in that direction. Rewritting equation (3), we get, QE d2 z … (4) = 2 m dt F = ma =m

Integrating once equation (4) by separating variables, we get QE dz = vz = t + k1 dt m where k1 is constant of integration.

… (5)

To find k1 : At t = 0, initial velocity in z-direction is zero. Substituting values in equation (5), we get 0 = 0 + k1 i.e. k1 = 0 Thus equation (5) becomes, QE dz = v = t dt m Integrating equation (6) with respect to corresponding variables we get, Q E æt2 ö z = ç ÷ + k2 m è 2ø where k2 constant of integration. To find k2 : At t = 0, charge is at origin. Substituting values in equation (7) we get, TM


…(6)

… (7)


0 =

8-4


Q E æ 0ö ç ÷ + k 2 i.e. k2 = 0 m è 2ø

Hence solution of the equation (3) is given by, Q E 2 – 0.3 ´ 10 – 6 ´ 30 a z 2 z = t = t 2m 2 ´ 3 ´ 10 – 16

… (8)

At t = 3 msec, z =

– 0.3 ´ 10 – 6 ´ 30 2 ´ 3 ´ 10

– 16

´ ( 3 ´ 10 – 6 ) 2 = – 0.135 m

Let us consider initial constant velocity in x-direction, the charge attains x co-ordinate of, x = vt = ( 3 ´ 105 )( 3 ´ 10 – 6 ) = 0.9 m Hence at t = 3 msec, the position of charge is given by, P(x, y, z) = (0.9, 0, – 0.135) m ii) To find velocity at t = 3 m sec using equation (6), we get, v =

QE (– 0.3 ´ 10 – 6 ´ 30 a z ) t = ( 3 ´ 10 – 6 ) – 16 m 3 ´ 10 4

= – 9 ´ 10 a z m/sec The actual velocity of charge can be obtained by including initial constant velocity in x-direction as, v = (3 ´ 10 5 a x – 9 ´ 10 4 a z ) m / sec iii) The kinetic energy of the charge is given by, 2 1 1 K.E. = m |v|2 = ´ 3 ´ 10 – 16 ´ é ( 3 ´ 105 ) 2 + (–9 ´ 10 4 ) 2 ù êë úû 2 2 \

K.E. = 1. 4715 ´ 10 – 5 J

Example 8.2.7 Solution : The magnitude of velocity is given as v = 6 ´ 10 6 m/s. The direction of this velocity is specified by an unit vector. Thus we can write, v = v a v = 6 ´ 10 6 [- 0.48 a x - 0.6 a y + 0.64 a z ] m/s The force experience by a moving charge in a steady magnetic field B is given by, F= Q v ´ B = - 60 ´ 10 -9 [( 6 ´ 10 6 ) ( - 0.48 a x - 0.6 a y + 0.64 a z ) ´ ( 2 a x - 6 a y + 5 a z ) (1 ´ 10 -3 )] = ( - 3.6 ´ 10

-4

ax

ay

az

) - 0.48 - 0.6 0.64 = ( - 3.6 ´ 10 -4 ) [0.84 a x + 3.68 a y + 4.08 a z ] 2 5 -6

TM



8-5


= ( - 0.3024 a x - 1.3248 a y - 1.4688 a z ) ´ 10 -3 N Thus the magnitude of the force on a moving charge is given by, F = ( - 0.3024 ´ 10 -3 ) 2 + ( -1.3248 ´ 10 -3 ) 2 + ( - 1.4688 ´ 10 -3 ) 2 = 2.0009 mN Example 8.2.8 Solution : Given : H =

0.01 a A m, Q = 1 pC = 1 ´ 10 - 12 C, v = 10 6 a y m sec m0 x

The magnetic force exerted on charge Q is given by, Fm = Q v ´ B But

… assuming free space for which m r = 1

B = m H = m0 mr H = m0 H

\

Fm = Q v ´ (m 0 H)

\

é æ 0.01 ö ù Fm = Q v ´ êm 0 ç ÷ax ú m ë è 0 ø û

\

Fm = 1 ´ 10 - 12 10 6 a y ´ ( 0.01 a x )

\

Fm = 1 ´ 10 - 8 ( - a z ) N

(

)

… (Q a y ´ a x = - a z )

Example 8.3.4 Solution : A force exerted on a current element in a magnetic field is given by, F = I dL ´ B But current element is 4 cm long i.e. 0.04 m long. It carriers current of 10 mA in y-direction. The magnetic field is given by, 5 A/m a H = m x But

5 B = m H = m é a x ù = 5 a x T. êë m úû

Hence, force exerted is given by, F = {10 ´ 10 -3 [0.04 a y ]} ´ {5 a x } \

F = -2 ´ 10 -3 a z N = -2 a z mN

Example 8.3.5 Solution : A force exerted a current carrying conductor in the magnetic field B is given by, F = IdL ´ B ... (1) From given data, I = 10 A, dL = 4 a y and B = 0.05 a x T TM



8-6


Substituting values in equation (1), we get, F = [10 (4 a y ) ´ (0.05 a x )] = [10 ´ 4 ´ 0.05) (a y ´ a x ) … ay ´ ax = – az

= 2 (– a z ) N Example 8.4.5 Solution : For air, m = m 0 m r = m 0 = 4p ´ 10 - 7 H/m d = Distance of separation = 10 cm = 10 ´ 10 - 2 m I 1 = I 2 = I = Current = 100 A In general, the force between two parallel long wires is given by, m I1 I2l m I I l F = = 0 1 2 2p d 2p d Hence force per meter length is given by, m I I F 4 ´ p ´ 10 - 7 ´ 100 ´ 100 = 0.02 N/m = 0 1 2 = 2p d l 2 ´ p ´ 10 ´ 10 - 2

As two parallel conductors carry equal current of 100 A but directions are opposite. Hence they will repel each other. Thus the nature of the force is repulsive force. Example 8.4.6 Solution : Force between two parallel conductors is given by, mI1 I2 l F = 2p d For free space, m = m 0 m r = m 0 = 4 p ´ 10 -7 H/m d = 10 cm = 10 ´ 10 -2 m I 1 = I 2 = 10 A Hence force per unit length is given by, F 4p ´ 10 -7 ´ 10 ´ 10 = = 0.2 mN/m l 2 ´ p ´ 10 ´ 10 -2 Example 8.4.7 Solution : Assume that parallel conductors are in air. \

m = m 0 m r = m 0 = 4p ´ 10 –7 H/m

Two conductors are separated by distance d = 10 m. The force between two infinitely long straight conductors is given by, F =

m I 1I 2 l m 0 I 1I 2 = l 2pd 2pd TM



8-7


Hence force per meter is given by, F 4p ´ 10 –7 ´ 10 ´ 10 = 2 mN = 2p ´ 10 l

\

Example 8.4.8 Solution : A rectangular loop with width a and height b is placed parallel to infinitely long conductor as shown in the Fig. 8.1. Due to infinitely long wire the magnetic flux density is given by m0 I1 a T B1 = 2p r 0 f Force exerted on side AB is given by, F1 = - I 2 ò B 1 ´ d L2 = -I2

=

ù é m 0I 1 êë 2p r 0 a f úû z=0

´

´

ò

-m0 I1 I2 2 pr0

... (Qa f =

b

Infinitely long wire B

I1

I2

O

r0 A

Rectangular loop C

b

D

a

(dz a z )

Fig. 8.1

b

ò

dz a r

z=0

a z = + ar )

- m 0 I 1I 2 b ar N 2p r 0

Force exerted on side BC is given by, F2 = -I 2 ò B 1

=

-m 0 I 1I 2 2p

= +

´ d L2

= -I 2

r = r0 + a

ò

r =r0

r = r0 + a

ù é m 0I 1 êë 2p r a f úû

´ (dr a r )

dr (- a z ) r

ò

r =r0

... (Q a f ´ a r = -a z )

m 0 I 1I 2 ì é 1 ù é 1 ùü - ln ê ú ý a z N ln 2 p íî êë r 0 + a úû ër 0 û þ

Force exerted on side CD is given by F3 = -I 2 ò B 1 =

´ d L2

-m0 I1 I2 2p (r 0 + a)

= -I 2

ù é m 0I 1 êë 2p(r 0 + a) a f úû ´ [dz a z ] z= b

0

ò

b

0

dz a r =

ò

- m 0 I 1I 2 + m 0 I 1I 2 b ( - b) a r = a N 2p (r 0 + a) 2p (r 0 + a) r TM



8-8

Force exerted on side DA is given by, F4 = -I 2 ò B 1

=

=

-m 0 I 1I 2 2p -m 0 I 1I 2 2p

´ d L2

r0

= -I 2

ò

r0 + a

ù é m 0I 1 êë 2p r a f úû ´ [dr a r ] +a

ò

r0 r0


-m 0 I 1I 2 dr ( -a z ) = r 2p

ì é 1 ù é 1 ùü í ln ê r ú - ln ê r + a úý ( -a z ) ûþ ë 0 î ë 0û

ì é 1 ù é 1 ùü í l n ê r + a ú - ln ê r ú ý a z N û ë 0 ûþ î ë 0

... Adjusting negative sign

Now total force on a loop is given by, F = F1 + F2 + F3 + F4 \

F =

m 0 I 1I 2 b 2p r 0

\

F =

-m 0 I 1I 2 b é 1 1 ù ar ê 2p ë r 0 + a r 0 úû

ar +

m 0 I 1I 2 b a 2p (r 0 + a) r

Rearrange terms inside bracket and taking negative sign out of it we get, F =

m 0I 2I 2 b é 1 1 ù ( -a r ) N êë r 0 2p r 0 + a ûú

This indicates force of attraction and infinitely long wire tries to attract a rectangular loop towards it. Example 8.5.8 Solution : For N turns loop, the maximum value of magnetic torque is given by, Tmax = N B I S, where S = Area of a square loop Now for a square loop, each side is 15 cm \

–2

Tmax = 200 ´ 1 ´ 5 (15 ´ 10

i.e.

15 ´ 10– 2 m

´ 15 ´ 10

–2

) = 22.5 Nm

Example 8.5.9

I = 100 mA = 100 ´ 10 -3 A Solution : Given B = 0.2 a x - 0.1 a y + 0.2 a z T i) The force exerted on segment AB is given by, FAB = IdL ´ B Now,

dL = 0.6 a x

\

FAB = (100 ´ 10 -3 ) 0.6 a x ´ (0.2 a x - 0.1 a y + 0.2 a z )

\

FAB = -12 a y - 6 a z mN

TM



8-9


ii) Now the torque on a loop is given by, T = m ´ B = IS ´ B Now area of triangle shaped loop placed in x-y plane can be written in vector form as, S =

1 [0.6 a x ´ (0.4 a x + a y )] = 0.3 a z 2

Hence

T = 100 ´ 10 -3 (0.3 a z ) ´ (0.2 a x - 0.1 a y + 0.2 a z )

\

T = 3 a x + 6 a y mN. m

z

Example 8.5.10 Solution : Consider a circular loop in z = 0 plane as shown in the Fig. 8.2. Current is in a f as shown in the Fig. 8.2. The given magnetic field is uniform given by æa + a z ö B = B0 ç x ÷ T è ø 2

O

r y I

x B

The magnetic dipole moment of a planar circular loop is given by, m = (I S) a n where S is the area of the circular loop.

Fig. 8.2

Note that the loop is laying in z = 0 plane. Thus the direction of unit normal a n must be decided by the right hand thumb rule. Let the fingures point in the direction of current (in a f direction), then the right thumb gives the direction of a n which is clearly a z . \

( ) a z = (p r 2I) a z

m = I p r2

The total torque is given by T = m ´ B =

=

(

)

p r 2I a z ´

p r 2 B0 I 2

æ p r 2 B0 = çç 2 è

B0 2

(a x

+az ) =

ax

ay

az

0 1

0 0

1 1

=

p r 2 B0 I 2

p r 2 B0 I 2

[a z ´ ( a x + a z ) ]

[- (- a )] y

Iö ÷ a y N-m ÷ ø

Example 8.5.11 Solution : a) The field is uniform and hence it will not produce any translation of the loop. TM



8 - 10


Hence we can write, the magnetic torque is given by, T = I S´B From the given Fig. 8.5.11, the square loop has sides of 4m each. Also it is placed in x - y plane. Hence the area is given by, dS = ( 4 a x ) ´ ( 4 a y ) = 16 a z m 2 Hence the magnetic torque at origin is given by, T = 0.6(16 a z ) ´ (100 a y ) ´ 10 -3 = - 0.96 a x N × m b) A(0, 0, 0) and B = 200 a x + 100 a y mT : Again the field is uniform and hence it will not produce any translation of the loop. Hence similar to part (a), we can write, T = I S ´ B = 0.6(16 a z ) ´ ( 200 a x + 100 a y ) ´ 10 -3 \

… (Q a z ´ a x = +a y a z ´ a y = –a x )

T = - 0.96 a x + 1.92 a y N × m

Example 8.7.4 Solution : i) The relative permeability m r can be obtained as, mr =

1.8 ´ 10 –5 m = m0 4 ´ p ´ 10 –7

= 14.3239

The magnetic field intensity and magnetization are related to each other as, M = cm ×H =

(m r – 1) H

M 120 = 9 A/m = 14. 3239 – 1 mr – 1

\

H =

ii)

n = Number of atoms = 8. 3 ´ 10 28 atoms/m3 m = 4. 5 ´ 10 –27 A × m

2

The magnetization is given by,

(

M = (n) (m) = 8. 3 ´ 10 28

)(4.5 ´ 10 –27 )

= 373.5 A/m

Now the magnetic field intensity is given by, H = iii)

373. 5 M = 17.7857 A/m = 22 – 1 mr – 1

The magnetic flux density is given by, B = m H = m 0 m r H = m 0 (1 + c m ) H

TM


... m r = ( c m + 1)


\

H =

8 - 11


300 ´ 10 –6 B = 14.9207 A/m = m 0 (1 + c m ) 4 ´ p ´ 10 –7 ´ (1 + 15)

Example 8.7.5 Solution : But

a) J = Ñ ´ H B B B = = H = m m 0 m r m 0 ( c m + 1)

... (1)

Putting value of H in equation (1) \

\

J = Ñ´

(

)

1 B = Ñ´B m 0 ( c m + 1) m 0 ( c m + 1)

Ñ´B =

ax ¶ ¶x 0.005 y 2

=

[0 - 0] a x

ay ¶ ¶y 0

az ¶ ¶z 0

ù ù é é ¶ ¶ (0.005 y 2 ) ú a y + ê 0 (0.005 y 2 ) ú a z - ê0 ¶ z y ¶ û û ë ë

= – (0.01 y) a z \

J =

\

J =

1 [- 0.01 y a z ] m 0 ( c m + 1) - 0.01 y 4 ´ p ´ 10 -7 ( 6 + 1)

az

Calculating value of J at y = 0.4 m. \

J = – 454.7284 a z A/m 2

The magnitude of J is 454.7284 A/m 2 b)

Jb = Ñ ´ M

But

M = cm H =

... (2) cm B cm B cm B = = m m 0 m r m 0 ( c m + 1)

Putting value of M in equation (2) \ \ \

Jb = Ñ ´ M = Ñ ´

cm B cm = Ñ´B m 0 ( c m + 1) m 0 ( c m + 1)

(

)

Ñ ´ B = – (0.01) y a z 6 Jb = ( - 0.01 y a z ) 4 ´ p ´ 10 -7 ´ 7

Calculating value of J at y = 0.4 m. TM



\

8 - 12


J b = – 2728.37 a z A/m 2

The magnitude of J b is 2728.37 A/m 2 c)

JT = Ñ ´

B m0

We can also write, J T = J b + J = (– 2728.37 a z ) + (– 454.7284 a z ) = – 3183.09 a z A/m 2 Thus the magnitude of J T is 3183.09 A/m 2 . Alternative method : JT = Ñ ´

1 B 1 = Ñ´B = m0 m 0 4 ´ p ´ 10 -7

[

]

( - 0.01 y a z ) = – 7957.74 y

a z A/m 2

Calculating value of J T at y = 0.4 m \

J T = – 3183.09 a z A/m 2

Hence the magnitude of J T is 3183.09 A/m 2 . Example 8.8.6 Solution region 2 gradient direction

But \

: Let f (x, y) = 2x – 5y. Thus f (x, y) increases from region 2 to region 1 as is defined by 2x – 5y < 0 while region 1 by 2x – 5y > 0. Now if we calculate of f (x, y) then it represents a vector with a magnitude and direction. The of that vector is in the direction of increasing value of f(x,y). ¶f ¶f ¶f a + a + a Ñf = ¶x x ¶y y ¶z z f = 2x – 5y Ñf = 2 a x - 5 a y

Now the unit vector normal to the plane is given by, a n21 = \

2ax -5ay 2ax -5ay Ñf = = Ñf 4 + 25 29

a n21 = 0.3714 a x - 0.9284 a y

a) The magnitude of the normal component of B 1 is given by,

( )×(a n21 )]×a n12

B N1 = [ B 1 But

B 1 = m 1 H 1 = m 0 m r1 H 1

= ( 4 ´ p ´ 10 -7 ´ 3) ( 30 a x ) = 113.0973 ´ 10 -6 a x T TM



\ b)

B1

8 - 13


= 113.0973 m T

× ×a

B N1 = ( B 1 a 21 )

= [113.0973 ´ 10

21 -6

ax

× (0.3714 a

x

)

- 0.9284 a y ]

(0.3714 a x - 0.9284 a y )

= 4.2 ´ 10 -5 ( 0.3714 a x - 0.9284 a y ) = 15.59 a x - 38.99 a y m T \

B N1

=

(15.59 ´ 10 -6 ) 2 + ( 38.99 ´ 10 -6 ) 2 = 41.99 m T

c) From the symmetry, B 1 = B tan1 + B N1 \

B tan1 = B 1 - B N1

(

)

= [113.0973 a x - 15.59 a x - 38.99 a y ] ´ 10 -6 = 97.5073 a x + 38.99 a y m T B tan1 = = m 0 m r1

(97.5073 a x

\

H tan1

\

H tan1 = 25.86 a x + 10.34 a y

\

H tan1

=

)

+ 38.99 a y ´ 10 -6

4 ´ p ´ 10 -7

´3

( 25.86) 2 + (10.34) 2 = 27.8505 A/m

d) According to the boundary conditions, B N2 = B N1 = 15.59 a x - 38.99 a y m T B tan2 =

m2 B tan1 m1

\

B tan2 =

4 [97.5073 a x + 38.99 a y ] ´ 10 -6 3

\

B tan2 = 130 a x + 51.98 a y m T

\

B 2 = B tan2 + B N2 B 2 = [(130 a x + 51.98 a y ) + (15.59 a x - 38.99 a y )] ´ 10 -6

\

B 2 = 145.59 a x + 12.99 a y m T

\

H2 =

(

)

145.59 a x + 12.99 a y ´ 10 -6 B2 = m 0 m r2 4 ´ p ´ 10 -7 ´ 4 TM



8 - 14

\

H 2 = 28.96 a x + 2.5842 a y

\

H2

=


( 28.96) 2 + ( 2.5842) 2 = 29.075 A/m

Example 8.8.7 Solution : Given : B 2 = 5 a x + 8 a z mWb m 2 ,

K=

1 a mA m. m0 y

Now the normal component of B 2 is along a z so that the normal component of B 1 is also along a z being continuous at boundary. By definition, B 1n = B 2n = 8 a z

i.e.

Bz = 8

Now for a current at boundary,

(H1 – H2 ) ´ a n12

= K

\

æ B1 B2 ö 1 – ay ç ÷ áz = m m m è 1 0 2 ø

\

æ B1 B2 ö 1 – a ç ÷ áz = m0 y è m 0m 1 m 0m 2 ø

\

æ B1 B2 ö – ç ÷ áz = ay è m1 m2 ø \

(

é Bxa x + By a y + Bza z ê 6 ê ë

) – (5a z + 8a z )ùú ´ a 4

ú û

z

= ay

\

éæ Bx 5 ö ù æ Bz 8 ö ê çè 6 – 4 ÷ø a x + B y – 0 a y + çè 6 – 4 ÷ø a z ú ´ a z = a y ë û

\

æ B x – 5 ö –a ç ÷ y + By a x = a y è 6 4ø

(

)

(

Equation components,

)

By = 0

Bx 5 = –1 – 6 4 \

Bx 5 1 = –1 + = 6 4 4

\

Bx =

Hence

B1 = Bxa x + By a y + Bz a z

\

B 1 = 1.5 a x + 0 a y + 8 a z = (1.5 a x + 8 a z ) mWb m 2

6 = 1.5 4

TM



8 - 15


But

H1 =

B1 1 1 = 1.5 a x + 8 a z ] = 1.5 a x + 8 a z ] m1 6 m0 [ m 0m 1 [

\

H1 =

1 [0.25 a x + 1.333 a z ] mA m m0

Example 8.8.8 Solution : z-axis is normal to the boundary. The normal component is given by, K (1) B N1 = ( B 1 × a N12 ) × a N12 Here below z = 0, there exists medium 2 while above z = 0, medium 1 exists. The field vector travels from medium 1 to 2. a N12 = –a z \ \

B N1 =

[(2a

x

– 3a y + 2a z

) · (–a z )]· (–a z )

= [–2]( –a z )

K Because B 1 is expressed in militesla

= 2 a z MT The tangential component of B 1 is given by,

(

)

B tan1 = B 1 – B N1 = 2 a x – 3 a y + 2 a z – ( 2 a z ) = 2 a x – 3 a y mT. According to boundary conditions, B N2 = B N1 = 2 a z mT Now we can write

(Htan1 – Htan2 )

= a N12 ´ K

K (2)

(

)

2 a x - 3 a y ´ 10 –3 B tan1 B = tan1 = m 0 mr1 m1 4 ´ 4p ´ 10 –7

But

Htan1 =

\

Htan1 = 198.94 2 a x – 3 a y

(

(

)

) A/m

Putting value of Htan1 in equation (2), we get,

[198.94(2a

\

x

)

– 3a y – Htan2

(397.88 a x – 596.82 a y ) – Htan2

\ But

] = (–a

z

) ´ ( 60 a x )

= –60 a y

Htan2 = 397.88 a x – 536.82 a y A/m

[

B tan2 = m 2 Htan2 = 7 ´ 4 ´ p ´ 10 –7 397.88 a x – 536.82 a y

\

B tan2 = 3.4999 ´ 10 –3 a x – 4.7222 ´ 10 –3 a y

\

B tan2 = 3.5 a x - 4.7222 a y mT

Thus

B 2 = B tan2 + B N2 =

{[3.5 a

x

]

} mT

– 4.7222 a y + [2 a z ] TM


]


\

8 - 16


B 2 = 3.5 a x – 4.7222 a y + 2 a z MT

Example 8.8.9 Solution : The surface separating two regions can be defined as, f(x, y, z) = 3x - 2y + 5z The unit vector normal to the plane is given by, 3 ax - 2 ay +5 az Ñ f 3 ax - 2 ay +5 az an = = = Ñf 9 + 4 + 25 38 Then the normal component of H is given by, é æ 3 a x - 2 a y + 5 a z öù é 3 a x - 2 a y + 5 a z ù H1n = ( H · a n ) a n = ê(4 a x + 6 a y - 3 a z ) ç ÷ú ê ú 38 38 è øû ë û ë 1 \ H1n = [12 + 0 + 0 + 0 - 12 + 0 + 0 + 0 - 15][ 3 a x - 2 a y + 5 a z ] 38 -15 (3 a x - 2 a y + 5 a z ) 38

\

H1n =

\

H1n = ( - 0. 3947)(3 a x - 2 a y + 5 a z ) = - 1.1841 a x + 0.7894 a y - 1.9735 a z

But

H1 = H1n + H1t H1 - H1n = (4 a x + 6 a y - 3 a z ) - ( -1.1841 a x + 0.7894 a y - 1.9735 a z )

\

H1t =

\

H1t = 5.1841 a x + 5.2106 a y - 1.0265 a z A m

Now

H2t = H1t = 5.1841 a x + 5.210 a y - 1.0265 a z A m

Also

B 2n = B 1n

\

H2n = =

i.e. m 2 H2n = m 1 H1n

m1 H m 2 1n 2m 0 [-1.1841 a x + 0.7894 a y - 1.9735 a z ] 5m 0

= - 0.4736 a x + 0.3157 a y - 0.7894 a z Now

H2 = H2n + H2t = ( - 0.4736 a x + 0.3157 a y - 0.7894 a z ) + 5.1841 a x + 5.2106 a y - 1.0265 a z ) H2 = 4.7105 a x + 5.5263 a y - 1.8159 a z A m

Now

B 2 = m 2 H2 = 5 m 0 H2 = 5 ´ 4 ´ p ´ 10 -7 H2

\

B 2 = 20 p ´ 10 -7 (4.7105 a x + 5.5263 a y - 1.8159 a z )

\

B 2 = 29.5965 a x + 34.7222 a y - 11.4094 a z m Wb m 2

Example 8.9.5 Solution : The magnetic field intensity is given by, TM



H =

8 - 17

I I = 2pr C 0.3


where C = Circumference

\

H =

But

B = mH = m 0 m r H

\

B = 4 ´ p ´ 10 - 7 ´ 1500 ´ 2 = 3.7699 ´ 10 - 3 T

15 ´ 10 - 2

=2 A m

The flux induced in one turn of a iron ring is given by, f = B×S

where S = Area of cross-section of iron ring

= (3.7699 ´ 10 - 3 ) (3 ´ 10 -4 ) = 1.13097 ´ 10 - 6 Wb Hence the total flux established in the ring is given by, f Total = N× f

where N = Number of turns

= 250 × (1.13097 ´ 10 - 6 ) = 0.2827 mWb Example 8.9.6 Solution : The flux in air gap is also flux in core. f 0.141 ´ 10 -3 = = 0.35 T Si 4 ´ 10 -4 0 . 35 Bi Bi = 849.15 A/m = = = m m 0m r 4 ´ p ´ 10 -7 ´ 328

Bi = Hence Then

Hi

(Hi) (li) = (849.15) (0.44) = 373.626 A

For air gap, cross-sectional area is given by, Sg = (0.02+0.002)2 = 4.84×10– 4 m2 Hence

Hg × lg =

f 0.141 ´ 10 -3 lg = ´ 2 ´ 10 -3 = 463.65 A -7 -4 m 0 Sg 4 ´ p ´ 10 ´ 4.84 ´ 10

Hence

f = H i l i + Hg lg = 373.626 + 463.65 = 837.27 A

Hence

f = N ×I f 837.27 I = = = 2.0932 A N 400

\

Example 8.10.11 Solution : For a given solenoid in air, m = m 0 = 4p ´ 10 -7 Wb/A.m N = 200 d = 6 cm = 6 ´ 10 -2 m

hence r = TM

d = 3 ´ 10 -2 m 2



8 - 18


l = 60 cm = 60 ´ 10 -2 m The inductance of a solenoid is given by, m N2 A l

L =

(

-7 2 -2 m 0 N 2 ( pr 2 ) 4 ´ p ´ 10 ´ ( 200) ´ p ´ 3 ´ 10 = = l 60 ´ 10 -2

)

2

= 2.3687 ´ 10 -4 H = 0.2368 mH Example 8.10.12 Solution : A solenoid is in air, hence m r = 1 Given

N = 400, d = diameter = 10 cm = 10 ´ 10 -2 m, l = 50 cm = 50 ´ 10 -2 m

The inductance of a solenoid is given by

(

)

2 2 m N 2 A (m 0 m r ) N p d 4 = L = l l

where

d A = Area of cross-section = p r 2 = p æç ö÷ è2ø

(

4 ´ p ´ 10

\

L=

-7

é

)

2ê

´ 1 ( 400) ê êë 50 ´ 10 -2

2

(10 ´ 10 - 2 ) p

2

4

=

p d2 4

ù ú ú úû = 3.1583 ´ 10 -3 H = 3.1583 mH

Example 8.10.13 Solution : The inductance of the solenoid is given by, mN 2 A = 20 mH l where l = Length of the solenoid, A = Area = p r 2 , N = Number of turns L =

r Now length is made 2l while the radius is made æç ö÷ . Then the inductance is given by è 2ø

Lnew \

é r 2ù m N 2 ê( p ) æç ö÷ ú è 2 ø ú m N 2 pr 2 êë û = = 8l 2 l ( )

Lnew =

( )

1 æ mN 2 A ö 1 20 ´ 10 -3 ç ÷= 8è l ø 8

(

) = 2.5 mH

Example 8.10.14 Solutions : For inner solenoid : m r = 75, l = 50 cm = 50 ´ 10 - 2 m, N = 1500 TM



8 - 19


d = 2 cm, hence r = 1 cm = 1 ´ 10 - 2 m So the inductance of solenoid is given by, L in =

m m N 2 (p r 2 ) 4 ´ p ´ 10 - 7 ´ 75 ´ (1500) 2 ´ ( p) ´ (1 ´ 10 - 2 ) m N2 A = 0 r = l l 50 ´ 10 - 2

\ L in = 0.1332 H For outer solenoid : m r = 1 (in air), l = 50 cm = 50 ´ 10 -2 m, N = 1200 d = 3 cm hence r =

d = 1.5 cm = 1.5 ´ 10 -2 m 2

So the inductance of outer solenoid is given by, L out = \

m m N 2 (p r 2 ) 4 ´ p ´ 10 - 7 ´ (1200) 2 ´ p ´ (1.5 ´ 10 -2 ) 2 m N2 A = 0 r = l l 50 ´ 10 -2 L out = 2.5582 mH

Example 8.10.15 Solution : N = 700, h = height = 1.5 cm = 1.5 ´ 10 - 2 m r1 = Inner radius = 1 cm = 1 ´ 10 - 2 m, r2 = Outer radius = 2 cm = 2 ´ 10 - 2 m 1) In general, inductance of a toroid of square cross section is given by, L =

=

m 0 N2 h ær ö ln ç 2 ÷ 2p è r1 ø æ 2 ´ 10 -2 ö 4 ´ p ´ 10 -7 ´ (700) 2 ´ 1.5 ´ 10 -2 ln ç ÷ = 1.0189 mH 2 ´ p è 1 ´ 10 -2 ø

2) By general approximate formula for toroid, the inductance is given by, m N2 A L = 0 2p R where A = Area of square cross-section = (1 cm) (1.5 cm) = 1.5 ´ 10 - 4 m 2

\

r = Mean radius = 1.5 cm = 1.5 ´ 10 -2 m 4 ´ p ´ 10 -7 ´ (700) 2 ´ (1.5 ´ 10 - 4 ) L = = 0.98 mH 2 ´ p ´ 1.5 ´ 10 -2

Thus for toroid, with radius larger than the cross-section, the inductance obtained by both formulae are approximately same. Example 8.10.16 Solution : For a solenoid with large length as compared to small cross section, the magnetic field intensity inside the coil can be assumed to be constant and zero for points just outside the solenoid. Let the current flowing through the coil be I 1 . TM



8 - 20


Then the magnetic field intensity is given by, H1 =

N1 I 1 ( 2000) I 1 = = ( 2000) I 1 A/m l1 100 ´ 10 -2

The magnetic flux density is given by,

(

B 1 = m H 1 = m 0 m r H 1 = 4 ´ p ´ 10 -7

) ( 2000 I 1) = (2.5132 ´ 10 -3 ) I 1

Wb / m 2

Total flux produced is given by,

\

(

f1 =

( B1) ( A 1)

f1 =

(3.1582 ´ 10 -6 ) I 1

= 2.5132 ´ 10 -3 ´ I 1

) éêë p (2 ´ 10 -2 )

2ù

úû

Wb

The flux calculated above can only link with the second coil as H 1 and B 1 are zero outside the coil 1. The mutual inductance between two coils is given by M12 \

(

( 4000) 3.1582 ´ 10 -6 I 1 N2 f1 = = I1 I1

)

M12 = 12.633 mH

Example 8.12.3 Solution : Given : N = 1000, f = 10 mWb = 10 ´ 10 –3 Wb, R = 4 W, V = 40 V The current in the coil at steady state is given by, I =

V 40 = = 10 A R 4

Hence the self inductance of a coil is given by, L =

Nf 1000 ´ 10 ´ 10 –3 =1H = I 10

Thus the energy stored in a magnetic field is given by, Wm =

1 2 1 LI = (1) (10) 2 = 50 W 2 2

Example 8.12.4 Solution : The inductance of toroid with N turns and A as area of a toroidal ring cross section, is given by, L=

( )

2 2 mN 2 A (m 0 m r ) N pr = , where R = Mean radius of a toroid, For air, m r = 1 2pR 2pR

TM



\

L =

8 - 21

(4 ´ p ´ 10 -7 )( 400) 2 éêë p ´ (4 ´ 10 -3 ) 2 ´ p ´ (40 ´ 10 -3 )

Magnetic Forces, Materials and Inductance 2ù

úû = 40.2123 mH

The current flowing through an air core toroid is I = 10 A. Then the energy stored by toroid is given by, 1 2 1 Wm = LI = 40.2123 ´ 10 -6 (10) 2 = 2.0106 mW 2 2

(

)

Example 8.12.5 Solution : The energy density in free space in a magnetic field is given by, 1 1 mH 2 = m 0 m r H 2 wm = 2 2 But for free space m r = 1 1 1 \ wm = m H 2 = ´ 4 ´ p ´ 10 - 7 ´ (1000) 2 = 0.6283 J m 3 2 0 2

qqq

TM


9

Time Varying Fields and Maxwell’s Equations Solutions of Examples for Practice

Example 9.2.7 Solution : Here the magnetic flux is constant while the path is rotating with speed of 1500 r.p.m. The field intensity is given by, where v = Linear velocity E = v´B In one minute there are 1500 revolutions which corresponds to 25 revolutions in one second. The distance covered in one second is ( 2pr ) meter. Hence in 25 revolutions the distance travelled is (50pr ) meter. The conductor rotates in f-direction. Hence linear velocity is given by,

(

)

v = (50pr ) a f = 50p 25 ´ 10 -2 a f = 39.27 a f m/s Hence the electric field intensity is given by, E = v ´ B = ( 39.27 a f ) ´ ( 0.5 a r ) = 19.635 ( - a z )

... a f ´ a r = - a z

Induced e.m.f. is given by, e =

ò E · dL

As the conductor is parallel to z axis, dL = (dz) a z \

z= 1

ò 19.635(-a z )· ( dz) a z

e =

z= 0

= -19.635 [z]10 = – 19.635 V

Example 9.2.8 Solution : The circular loop conductor is in X-Y plane. B is in a z direction which is

perpendicular to X-Y plane. Hence, we can write, dS = ( r dr df) a z Total flux is given by f =

ò

B · dS

S

(9 - 1) TM


Electromagnetic Field Theory 2p

ò

=

9-2 0. 15

ò [(0.5 sin 10

f = 0 r= 0

=

(0.5 sin 10 t) 3

[f]20 p

3

Time Varying Fields and Maxwell's Equations

) ]

t a z · [( r dr df) a z ] 0.15

ér2 ù ê 2 ú ë û0

(

é

2

) [2p] êê (0.15) 2

= 0.5 sin 10 3 t

ë

ù ú úû

= 35.3429 sin 10 3 t mWb Now induced e.m.f. is given by, e = -

df d =dt dt

(

[ 35.3429 ´ 10

= - 35.3429 ´ 10 -3

-3

]

sin 10 3 t

)(10 3 ) cos 10 3 t = – 35.3429 cos 10 3 t

V

Hence current in the conductor is given by, i =

e - 35.3429 cos 10 3 t v = – 1.7671 cos 103 t A = R 20

Example 9.2.9

z

Solution : The conductor is placed along x-axis with one end at origin as shown in the Fig. 9.1. Given : B = 0.04 a y T

v

v = Velocity = 2.5 sin 1000 t a z m/sec

y

By definition, the motional electric field intensity is given by, Em = v ´ B \

B x

0.2 m

Em = (2.5 sin 1000 t a z ) ´ (0.04 a y )

\ Em = 0.1 sin 1000 t ( - a x ) V/m Now the induced e.m.f. in the conductor is given by, e = \ \

e =

ò

E m · dL =

ò

Fig. 9.1

… (Q a z ´ a y = - a x )

( v ´ B) · d L

x = 0.2

0.2

x = 0

x = 0

ò

[0.1 sin 1000 t ( - a x )] · (dx a x ) = – 0.1 sin 1000 t

= – 0.02 sin 1000 t V e = - 0.1 sin 1000 t [x] 0.2 0

Example 9.2.10 Solution : The induced e.m.f. is given by, e = B l v sin q = (1.1) (0.5) (30) sin

p = 16.5 V 2

As field and direction of motion are perpendicular to each other, q = TM


p 2

ò

dx


9-3

Time Varying Fields and Maxwell's Equations z

Example 9.2.11 Solution : Here filamentary conductor is fixed and it is placed in z = 0 plane. It encloses area of 0.65 m 2 . \ dS = dS a z Induced e.m.f. according to Faraday's is law is given by, ¶B · dS ¶t

e = -ò

S

= -ò

S

= -ò

B dS

x

¶ é 3 ê 0.05 cos 10 ¶t ê ë

æa y + a z t çç 2 è

öù ÷÷ ú · ( dS a z ) ø úû

Fig. 9.2

( )(- sin 10 3 t) dS

0.05 10 3

2

S

é ù = + 35.355 sin 10 3 t ê ò dSú êë S úû But

ò

y

... a y · a z = 0 az ·az = 1

dS is given as 0.65 m 2 .

S

\

e = 35.355 sin 10 3 t (0.65) = 22.98 sin 103 t V

Example 9.3.8 Solution : The conduction current density is given by

[

]

J C = s E = 5 100 sin 10 10 t = 500 sin 10 10 t A/m 2 The displacement current density is given by JD = =

¶D ¶E ¶ ¶ = ( e E) = ¶ t ( e 0 e r E) = e 0 e r ¶ t ¶t ¶t

(8.854 ´ 10 -12 ´ 1) ¶¶t [100 sin 10 10 t]

= 8.854 ´ 10 -12 ´ 10 10 ´ 100 ´ cos 10 10 t = 8.854 cos 10 10 t A/m 2 For the two current densities, the condition for the equal magnitudes is given by JC s = =1 ew JD \

w =

s 5 = = 5.647 ´ 10 11 e 8.854 ´ 10 -12 ´ 1 TM



But

w = 2p f

\

f =

9-4


w 5.647 ´ 10 11 = 89.877 GHz = 2p 2p

Example 9.3.9 Solution : i) The condition for both the current densities to have equal magnitude is JC

=

JD

s =1 we s e 2 ´ 10 -8

\

w =

\

w =

But

w = 2p f

\

f =

10 -8 36 p

= 226.194 rad/sec

w 226.194 = 36 Hz = 2p 2p

ii) The displacement current density at f = 36 Hz is given by JD = =

d dE 10 -8 é d = e E) = e ( ( 200 sin w t) ùú dt dt 36 p êë dt û 10 -8 10 -8 ´ 200 ´ w ´ cos w t = ´ 200 ´ ( 2 ´ p ´ 36) ´ cos ( 2 ´ p ´ 36) t 36 p 36 p

= 400 ´ 10 -8 cos 72 p t A/m 2 = 4 cos 72p t

m A/m 2

dE dt So from the expressions of J C and J D it is clear that both are always at right angles to each other. So the phase angle between the current densities is 90°. iii)

J C = s E and J D = e


D = eE = e

V d

Hence the displacement current density is given by, JD =

¶D e dV ¶ æ Vö = çe ÷ = è ø d dt d ¶t ¶t

Hence the displacement current is given by e dV ö iD = JD × Area = æç ÷ (A) è d dt ø

… Plate area = A

TM



\

iD =

9-5


eA dV dV =C d dt dt

This current is same as conduction current. dQ dD dV dE eA dV =A = eA =C = dt dt d dt dt dt Hence the conduction current and displacement current is same. The displacement current is given by

\

iC =

iD = =

eA dV ( 2 e 0 ) ( A ) dV 2 ´ 8 . 854 ´ 10 -12 ´ 5 ´ 10 -4 d = = 50 sin 10 3 t 3 d d dt dt dt 3 ´ 10

(

2 ´ 8.854 ´ 10 -12 ´ 5 ´ 10 -4 ´ 50 ´ 10 3 3 ´ 10 -3

)

cos 103t = 0.1475 cos 103t µA

Example 9.3.11 2

Solution :

Area of plate = A = 10 cm = 10 × 10 Distance of separation = d = 2 mm = 2 × 10

–4

–3

m

2

m 6

Dielectric is air \ e r = 1, Applied voltage V = 300 sin 10 t volts Hence displacement current is given by, ( e 0 e r ) A dV eA dV iD = = d dt d dt =

=

8.854 ´ 10

–12

´ 1 ´ 10 ´ 10

2 ´ 10 8.854 ´ 10

–12

–3

´ 10 ´ 10

2 ´ 10

–4

–3

–4

d 300 sin 10 6 t dt 6

´ 300 ´ 10 6 cos 10 6 t = 1.3281 cos 10 t mA


According to condition, =

s we

JC

=

iC i and J D = D A A

\

iC A iD A

=

\

iD

=

JC JD But

\

s w (e 0 e r ) w (e 0 e r ) i C 2pf( e 0 e r ) i C 2 ´ p ´ 1 ´ 10 9 ´ ( 8.854 ´ 10 -12 )( 3) = = s s 5.8 ´ 107

iD = 2.87748 nA

TM



9-6


Example 9.5.9 Solution : The point form of Maxwell's second equation is, Ñ´H = J+

¶D ¶t

But as fields are time invariant, we can write, ¶D = 0 ¶t \

Ñ´H = J

\

J =

ax ¶ ¶x 0

ay az ¶ ¶ ¶y ¶z 0 ( 3 x cos b + 6y sin a )

\

J =

¶ ¶ [ 3 x cos b + 6y sin a ] a x - ¶ x [ 3 x cos b + 6y sin a ] a y ¶y

\

J = 6 sin a a x - 3 cos b a y A / m 2

Example 9.5.10 Solution : Consider Maxwell's equation for static fields, ÑÉ = 0 Consider L.H.S. of equation (1),

... (1)

[

L.H.S. = Ñ ´ E = Ñ ´ x a x + x a y 2

]

ax ay az = ¶/¶x ¶/¶y ¶/¶z x2 x 0

( )

( )

ù é ¶ ù é ¶ ù é ¶ ¶ ¶ ¶ = ê x2 ú a y + ê x2 ú a z ( 0) - ( x)ú a x - ê ( 0) ( x) y z x z ¶ x ¶ y ¶ ¶ ¶ ¶ û ë û ë û ë =

[0] - [0] + (1) a z

= az

But R.H.S. = 0. That means L.H.S. ¹ R.H.S. Thus we have Ñ ´ E ¹ 0 which indicates that the given electric field E is not static. But we can have a static field only if the charge distribution is static. From above calculation it is clear that E is not static implies this electric field cannot arise from static distribution of charge. Example 9.5.11 Solution : Given : E = 20 cos ( wt – 50x) a y V m,

mr = er = 1

i) By definition, D = eE = ( e 0 e r ) E = e 0 E

TM


...For free space


\

9-7

[

D = e 0 20 cos ( wt – 50x) a y

] = 20 e 0[cos ( wt – 50x) a y ]

The current density J D is given by, ¶ ¶D = 20 e 0 cos ( wt – 50x) a y JD = Jd = ¶t ¶t

{

\


} = 20

...(1)

e 0 [ – sin ( wt – 50x)]( w) a y

J D = J d = –20 w e 0 sin ( wt – 50x ) a y A m 2

...(2)

ii) By Maxwell's equation, for free space, – ¶B ÑÉ = ¶t \

ax ay – ¶B = ¶ ¶x ¶ ¶y ¶t 0 20 cos ( w t – 50x)

az ¶ ¶z 0

¶ ü ì ü ì¶ = í0 – 20cos ( wt – 50x)ýa x – {0 – 0} a y + í 20cos ( wt – 50x) – 0ýa z ¶z þ î þ î ¶x \ –

¶B = 20[ – sin ( wt – 50x)( –50)]a z = 1000 sin ( wt – 50x) a z ¶t

¶B = –1000 sin ( wt – 50x) a z ¶t Separating variables, ¶B = [ –1000 sin ( wt – 50x) a z ] dt Integrating both sides é cos ( wt – 50x) ù 1000 B = –1000 ê – ú a z = w cos ( wt – 50x) a z T w û ë By definition, 1000 B B B = = = cos ( wt – 50x ) a z A m H = wm 0 m m 0m r m0

\ –

...(3)

...(4)

Now to find value of w, let us use Maxwell's equation as follows, ¶D = JD Ñ´H = ¶t ax

ay

Ñ ´ H = ¶ ¶x ¶ ¶y 0 \ \

0

az ¶ ¶z

=

1000 cos ( wt – 50x) wm 0

¶D ¶t

é ¶ ì 1000 ù é ¶ 1000 ¶D ü ù ê ¶y wm cos ( wt – 50x) – 0ú a x – ê ¶x í wm cos ( wt – 50x)ý – 0ú a y + [0 – 0] a z = ¶t þ û 0 û ë ë î 0

[0 – 0] a x

¶D é 1000 –ê {– sin( wt – 50x)}( –50) – 0ùú a y + [0] a z = ¶t wm û ë 0 TM



\

9-8


–50000 ¶D sin ( wt – 50x) a y = wm 0 ¶t

...(5)

Comparing equations (2) and (5) as both are of same form and representing same quantity, we can write, –50000 –20 w e 0 = w m0 \

w m0 =

2500 w e0

...(6)

Putting equation (6) in equation (4), we can represent H in another form as, 1000 H = cos ( wt – 50x) a z = 0.4 w e 0 cos ( wt – 50x ) a z A m æ 2500 ö ç we ÷ è 0ø Now rearranging equation (6) 2500 2500 = w2 = 7 m0 e0 4 ´ p ´ 10 8.854 ´ 10 -12

(

)(

)

...(7)

= 2.24694 ´ 10 20

w = 1.4989 ´ 10 10 rad sec » 1.5 ´ 10 10 rad sec

\

Thus representing values of J d and H by putting values of e 0 and w using equations (2) and (7) as follows

(

) (

)

J d = 20 8.854 ´ 10 –12 cos 1.5 ´ 10 10 t – 50x a y

(

)

J d = 1.7708 ´ 10 10 cos 1.5 ´ 10 10 t – 50x a y A m 2

\ Similarly,

(

H = 0.4 1.5 ´ 10 10

)(8.854 ´ 10 –12 ) cos ( wt – 50x) a z

= 0.053124 cos ( wt – 50x ) a z A/m

Example 9.5.12 Solution : a) For time varying fields, we can write Maxwell's equation as, ¶B ÑÉ = ¶t We can write, ÑÉ =

ax ¶ ¶x Ex

= -

ay ¶ ¶y Ey

az ax ¶ ¶ = ¶z ¶x Ez 0

ay az ¶ ¶ ¶y ¶z [kx - 100 t] 0

¶ ¶ [kx - 100 t] a x + ¶ x [kx - 100 t] a z ¶z

Again E is varying with respect to x and not with z.

TM



9-9


¶B ¶ [kx - 100 t] a z = - ¶ t ¶x

\

ÑÉ =

\

k az = -

¶ ¶ m H = -m [x + 20 t] a z = - 20 m a z ¶t ¶t

( )

... B = m H

Comparing, k = - 20m = - 20 (0.5) = – 5 V/m 2 b) Consider Maxwell's equation derived from Gauss's law for electric fields, Ñ· D = rv \

¶ Dx ¶ Dy ¶ Dz = rv = 0 + + ¶x ¶y ¶z

... Given

From given expressions of D, D x = 5x,

D y = – 2y ,

D z = kx

Putting values of D x , D y and D z , we get, ¶ ¶ ¶ (5x) + ( -2y) + ( kx) = 0 ¶x ¶y ¶z \ \

5-2+k = 0 k = – 3 mC/m 3

Note that in part (a), k is unknown in the expression of E which is expressed in V/m. In the expression k is multiplied with x which is expressed in metres (m). Hence accordingly k is expressed in V/m 2 . While in part (b), k is the part of expression of D which is expressed in m C/m 2 . k is multiplied by z which is expressed in m, in expression of D. Hence k is expressed in mC/m 2 . Example 9.5.13 Solution : Using Maxwell's equation, ¶D Ñ´H = J + ¶t In a free space, conduction current density is zero. So J = 0. \

Ñ´H =

( )

¶E ¶D ¶ eE =e = ¶t ¶t ¶t

As H has no component in a y and a z directions, H y = H z = 0 \

é ¶ Hy ¶ Hx ù é ¶H z ¶H y ù é ¶ Hx ¶ Hz ù a ax + ê ay + ê Ñ´H = ê ú ú ¶ y úû z ¶x û ¶z û ë ¶z ë ¶x ë ¶y

TM



9 - 10


¶ Hx ¶ Hx ay az ¶z ¶y

=

Again H is varying with z only and not with y. ¶ Hx = 0 ¶y

\

¶ Hx ¶E ay = e ¶z ¶t

Ñ´H =

\ \

¶E ¶ [H m e j (w t + b z) ] a y = e ¶z ¶t

\

H m × e j (w t + b z) × ( j b) a y = e

¶E ¶t

Separating variables, æ jb ö ¶ E = ç ÷ H m e j (w t + b z) a y dt èeø Integrating both the sides with respect to corresponding variables, b H m j (w t + b z) æ jb ö e j (w t + b z) e ay = ay E = ç ÷ × Hm × ew e j w è ø Also e = e 0 e r . But for free space e r = 1 \

E =

b × H m j (w t + b z ) e a y V/m e0 w


The displacement current density is given by, JD =

But

¶D ¶t

D = e0 er E = e0 E

… For free space or air e r = 1

¶ ¶ ( e E) = e 0 ( E) ¶t 0 ¶t

\

JD =

\

J D = e0

\

J D = 8.854 ´ 10 12 ´ 80 ´ ( - 6.277 ´ 10 8 ) sin [6.277 ´ 10 8 t - 2.092 y] a z

\

J D = - 0.4446 sin (6.277 ´ 10 8 t - 2.092 y) a z A m 2

¶ [80 cos (6.277 ´ 10 8 t - 2.092 y) a z ] ¶t

Hence the amplitude of displacement current density is |J D| = 0.4446 A m 2 TM



9 - 11


Example 9.5.15 Solution : According to ampere circuital law for free space, Ñ´ H =

Now

Ñ´ H =

¶D = JD ¶t ax ¶ ¶x 0

=

where J D is displacement current density. ay

az

¶ ¶y ¶ ¶z 0 10 6 cos (377 t + 1.2566 ´ 10 - 6 z)

¶ ù é 6 -6 êë 0 - ¶ z {10 cos (377 t + 1.2566 ´ 10 z)} úû a z - [0 - 0] a y + é¶ 6 -6 êë ¶x {10 cos (377 t + 1.2566 ´ 10 z)} -

ù 0ú a z û

= - 10 6 {- 1.2566 ´ 10 - 6 sin (377 t + 1.2566 ´ 10 -6 z)} a x = + 1.2566 sin (377 t + 1.2566 ´ 10 - 6 z) a x But by definition, the displacement current density is given by, ¶D = Ñ´H JD = ¶t \

… For free space

J D = 1.2566 sin (377 t + 1.2566 ´ 10 - 6 z) a x A m 2

Hence the amplitude of the displacement current density is 1.2566 A m 2 . Example 9.5.16 Solution : The magnitude of conduction current density is given by, Magnitude of conduction current I C = JC = Area of cross - section A =

2.5 ´ 10 -6

... (Q conductor has circular cross-section

p(2 ´ 10 -3 ) 2 and hence A = pr 2 )

Thus the magnitude of displacement current density is given by, JC JD

=

s we æ we e ö æ we ö Jc ç ÷ = Jc ç 0 r ÷ ès ø è s ø

\

JD =

\

é 5 ´ 10 8 ´ 8.8542 ´ 10 -12 ´ 1 ù J D = 0.198943 ê ú 35 ´ 10 6 û ë

\

J D = 25.164 pA m 2 TM



9 - 12


Example 9.7.2 Solution : a) From general field relations, E = -Ñ V -

¶A ¶t

¶V ù ¶A é ¶V ¶V = - ê ax + ay + az ú ¶t ¶y ¶z û ë ¶x

[

(

= - 3 ´ 105 sin 3 ´ 10 8 t sin z a y + y cos z a z

)] - 10 -3 y cos z (- 3 ´ 10 8 sin 3 ´ 10 8 t) a z

= - 3 ´ 105 sin 3 ´ 10 8 t sin z a y - 3 ´ 105 y sin 3 ´ 10 8 t cos z a z + 3 ´ 105 y sin 3 ´ 10 8 t cos z a z

(

)

= - 3 ´ 10 5 sin 3 ´ 10 8 t sin z a y V/m

b)

ax ay az B = Ñ Á = ¶ / ¶ x ¶ / ¶ y ¶ / ¶ z Ax Ay Az

From given expression of A A x = 0, \

A y = 0, B =

(

A z = 10 -3

)( y) cos 3 ´ 10 8 t cos z

¶Az ¶Az ax ay ¶y ¶x

As A z is not varying with x,

¶Az =0 ¶x

¶Az ¶ ax = ¶y ¶y

[(10 ) y cos (3 ´ 10 ) t cos z] a -3

8

\

B =

\

B = 10 -3 cos 3 ´ 10 8 t cos z a x Wb/m 2

But

B = m H = mr m0 H

\

10 -3 cos 3 ´ 10 8 t cos z B B = = ax H = mr m0 m0 4 ´ p ´ 10 -7

\

H =

(

x

)

(

) )

( 795.7747 cos (3 ´ 10 8 ) t cos z a x

... m r = 1 given

A/m

qqq TM


10

Electromagnetic Power and Poynting Theorem Solutions of Examples for Practice

Example 10.2.7 Solution : The field vectors can be represented as, E = 150 sin ( wt - bz) a x V/m Now H is mutually perpendicular to E and H m = H =

Em we can write, h0

150 sin ( wt - bz) a y A/m h0

Converting both the sinusoidal functions to cosinusoidal functions, æ pö and E = 150 cosçwt - bz- ÷ a x 2ø è H =

æ 150 pö cosçwt - bz- ÷ a y 2ø h0 è

Writing in phasor form, E = 150 e j ( -bz-p / 2 ) a x H =

and

150 j ( -bz-p / 2 ) e ay h0

Thus the complex conjugate of H is given by, H

*

æ

pö

150 jçè b z+ 2 ÷ø = e ay h0

Hence average power density is given by, 1 Re[ E ´ H * ] Pavg = 2 =

æ150 ö j ( -bz-wt ) 1 ÷ e (150)ç a x ´ e+ j ( wt + bz ) a y 2 è h0 ø

=

1 (150) 2 (a z ) = 29.841 a z watt/m 2 2 120 p

[

]

Now the total power crossing area is given by, (10 - 1) TM



P = Key Point

10 - 2

Electromagnetic Power and Poynting Theorem

(Pavg ) (area ) = (29.84)[(15 ´ 10-3 )´( 30 ´ 10-3 )]

= 13.428 mwatt

The flow of power is normal to the area. The area is in z = 0 plane, so the

direction normal to this plane is a z . Example 10.2.8 Solution : E (z, t) = 100 sin ( wt - bz) a x V m, where E m = 100 a = Side of the square = 25 mm = 25´10- 3 m Now H is mutually perpendicular to E, where H m = \

H (z, t) =

Em h0

Em sin ( wt - bz) a y A m h0

100 sin ( wt - bz) A m 120 p Converting E and H to cosinusoidal functions, æ pö E = 100 cosçwt - bz- ÷ a x and 2ø è =

H =

æ 100 pö cosçwt - bz- ÷a y 120 p 2ø è

In phasor form we can write, E = 100 e j ( - bz-p H = Hence

H* =

2)

100 j ( - bz-p e 120 p 100 j ( bz+p e 120p

2)

a x and 2)

ay

ay

1 R [ E ´ H *] 2 e æ 100 ö j ( -bz-p 2) 1 ÷e = a x ´ e j ( bz+p (100)ç 2 è120p ø

Average power density = Pavg =

[

2)

ay

]

= 13.2629 a z W m 2

... (Q a x á y = a z )

Hence total power passing through area of square of side 25 mm is given by, P = (Pavg ) (Area)= (13.2629) (25´10- 3 ) 2 = 8.289 mW Example 10.2.9 Solution : The magnitude of the electric field intensity is E = 2.2 mV/m The electric field intensity is measured at a distance 3 4 R = 10 km = 10 × 10 m = 10 m TM



10 - 3

Electromagnetic Power and Poynting Theorem

The power is radiated by the station in free space. Hence the magnitude of the magnetic field intensity is given by, 2 . 2´10-3 E –6 = H = = 5.8357 × 10 A/m h0 120p The magnitude of the power density is given by, Power density =|P| = (E) (H) = (2.2 × 10-3 ) (5.8357 × 10– 6) = 12.8385 × 10-9 W/m2 The power is radiated at a distance 10 km over a spherical region symmetrically. The area of the spherical region is given by S = 4pR2 = 4 × p × (104)2 = 1.2566 × 109 m2 Hence the total power radiated over region is given by, 9 Power = (Power density) (Area) = (12.8385 × 10-9 ) (1.2566 × 10 ) = 16.13333 W Example 10.2.10 Solution : |P | = For free space, \

2 1 E0 2 h0

h 0 = 377 8 =

2 1 E0 2 377

\

E 20 = 16 (377)

\

E 0 = 77.666 V/m

qqq

TM


11

Uniform Plane Waves Solutions of Examples for Practice

Example 11.3.5 Solution : i) For air as a medium, the velocity of proagation is 8

v = c = 3 × 10 m/s Then the wavelength is given by, c 3 ´ 10 8 = = 30 m f 10 ´ 10 6 Hence phase constant b is given by, 2p 2 ´ p = = 0.2094 rad/m b = 30 l ii) For air, the intrinsic impedance is given by l =

h = h 0 = 120 p = 377 W The electric field E and the magnetic field H are in phase quadrature. As E is in x-direction, H must be in y-direction so that the wave travels in z-direction. \

H =

6 E cos (w t – bz) a y A/m = 377 h0

iii) The average Poynting vector is given by Pavg =

1 Re [E ´ H * ] 2

We can represent E in phasor form as, E = 6 e j ( wt -bz) a x Similarly we can represent H in phasor form as H =

6 j ( wt -bz ) e ay 377

The complex conjugate of H can be written as H* =

6 - j ( wt -bz ) e ay 377 (11 - 1) TM



11 - 2

Uniform Plane Waves

Hence average Poynting vector is given by

[

]

P avg =

1ì 6 - j( wt -bz ) ùü ay ý 6 e j( wt -bz ) a x ´ é e í úûþ ê 2î ë 377

=

1 æ 36 ö 2 ç ÷ (a ´ ay) = 0.0477 az watt/m 2 è 377 ø x

Example 11.3.6 E = ( 200Ð 30° ) e –j250Z a x

Solution : By comparison,

b = 250 =

2p l

\

l =

2p = 0.0251 m 250

\

f =

c 3 ´ 10 8 = = 11.9522 ´ 10 9 Hz 0.0251 l

But

w = 2pf = 2 ´ p ´ 11.9522 ´ 10 9 = 75.0978 GHz

Intrinsic impedance of free space is 377 W or 120 p. H =

200Ð 30° + 90° –250Z ×e a y = 0.5305Ð120° e –250Z a y A/m 120p

The wave propagates in positive z-direction. Example 11.3.7 Solution :

(

)

E = 800 cos 10 8 t - b y a z V/m

i) For uniform plane wave in free space,

ii)

b =

w 10 8 = 0.3333 rad/m = c 3 ´ 10 8

l =

2p 2p = 18.85 m = 0.3333 b

iii) The magnetic field intensity in the free space is given by, H =

E h0

For free space, h 0 = 120 p = 377 W. Since power flow is in y-direction and E is in z-direction, the direction of H will be + x-direction. \

H =

(

)a

800 cos 10 8 t - b y 377

x

(

)

A/m = 2.122 cos 10 8 t - b y a x A/m

TM



11 - 3

Uniform Plane Waves

iv) H at P (0.1, 1.5, 0.4) at t = 8 ns is given by

[ (

]

)

H = 2.122 cos 10 8 8 ´ 10 -9 - ( 0.3333)(1.5) a x = 2.122 cos (0.3) a x \

H = 2.1219 a x A/m

Example 11.5.6 Solution : The wavelength of an electromagnetic wave may be different in different media, but the frequency remains constant in all media. When an EM wave is in medium 1 i.e. free space, the wavelength is l 1 = 0.2 m, while when it is in medium 2 i.e. perfect dielectric, the wavelength changes to l 2 = 0.09 m. But in both the media, the frequency of an EM wave remains constant i.e. f. When an EM wave is in free space, the velocity of propagation is given by, v 1 = c = 3 ´ 10 8 m/s \ \

f l 1 = c in free space f =

c 3 ´ 10 8 = = 1.5 ´ 10 9 Hz 0.2 l1

Now EM wave enters in perfect dielectric with frequency f = 1.5 ´ 10 9 Hz and wavelength l 2 = 0.09 m. Then the velocity of propagation of an EM wave in perfect dielectric is given by,

(

)

v 2 = f l 2 = 1.5 ´ 10 9 ´ 0.09 = 1.35 ´ 10 8 m/s But velocity of propagation can also be expressed as, 1 1 1 = = v2 = me (m 0 m r ) ( e 0 e r ) (m 0 m r ) e 0 v 2 2 \

er =

1

(

) (

4 ´ p ´ 10 -7 (1) 8.854 ´ 10 -12

)(

1.35 ´ 10 8

)

2

= 4.9315

Example 11.5.7 Solution : Assume lossless medium. \ s = 0. The wavelength is given by, 2p l = b \

b =

2p 2p = 3.59 rad/m = 1.75 l

But for lossless medium, b = w m e = 3.59 \

w =

3.59 = me

3.59

(m 0 m r )( e 0 e r ) TM



11 - 4

Uniform Plane Waves

3.59

\

w =

\

w = 1.522 ´ 10 8 rad/sec

(4 ´ p ´ 10 -7 )(1) (8.854 ´ 10 -12 )(50)

The magnetic field is given by, H =

B = H m e j (w t -b z) m0

The intrinsic impedance is given by, m 0m r = e0 er

m = e

h =

4 ´ p ´ 10 -7 ´ 1 8.854 ´ 10 -12 ´ 50

\ h = 53.278 W But intrinsic impedance can also be expressed interms of the magnitudes of electric and magnetic fields as, Em h = Hm From the given expression of E, the magnitude E m is 20 p . \

Hm =

Em ( 20)( p ) = 1.1793 A/m = h 53.278

Example 11.5.8 Solution : Given : For lossless medium, mr = 2, er = 3, f = 10 MHz = 10 ´ 10 6 Hz = 1 W m2

Pavg

i) The velocity of propagation for lossless medium is given by, n=

1 = me

1

(m 0 mr )( e 0 er )

=

(4 ´ p ´ 10

1

-7

) (

´ 2 ´ 8.854 ´ 10

-12

ii) The wavelength is given by, l =

n 1.2239 ´ 10 8 = = 12.239 m f 10 ´ 10 6

iii) The intrinsic impedance of a lossless medium is given by, h = Note that h 0 =

m0 e0

m 0m r m = = e e0 er

m0 e0

mr = h0 er

mr er

= 377 W = Intrinsic impedance of free space.

TM

)

´3


= 1.2239 ´ 10 8 m sec


\

11 - 5

Uniform Plane Waves

2 = 307.8192 W 3

h = 377

iv) The magnitude of average Poynting vector is given by, Pavg

=

1 Em 2 2 h

1

=

E 2m 1 2 ( 307.8192)

\ \

E 2m = 615.6384

\

Em =

615.6384 = 24.8121 V m

Hence the r.m.s. value of the electric field is given by, E r.m.s. =

Em 2

=

24.8121 = 17.5448 V m 2

Example 11.5.9 Solution : i) For the given medium i.e. fresh water, conductivity s = 0. Assuming medium to be a lossless medium, we can write, attenuation constant a = 0 ii) The phase constant is given by, b = w me = w

(m 0 m r ) ( e 0 e r )

Putting values of w, m 0 , m r , e 0 and e r , b =

(2 ´ p ´ 300 ´ 10 6 ) (4 p ´ 10 -7 )(1) (8.854 ´ 10 -12 )(78) = 55.529

rad/m

iii) The wavelength is given by, l =

2p 2p = 0.1131 m = 55. 529 b

iv) The intrinsic impedance is given by, h =

m = e

m0 mr = e0 er

(4 ´ p ´ 10 -7 )(1) = 42.656 (8.854 ´ 10 -12 )(78)

W

Example 11.6.5 Solution : For lossy dielectric medium, propagation constant is given by, g = \ g =

(

j wm (s + j we)

j 2 ´ p ´ 15.9 ´ 10 6

)(4 ´ p ´ 10 -7 ´ 1) 60 + j (2 ´ p ´ 15.9 ´ 10 6 )(8.854 ´ 10 -12 ´ 50) TM



11 - 6

\

g =

j (125.5413)[ 60 + j 0.0442]

\

g =

[125.5413 Ð90°] [60 Ð0.04°]

\

g = 86.78 Ð45.02°

\

g = a + j b = 61.3413 + j 61.3841

Uniform Plane Waves

Comparing real and imaginary terms, we get, a = 61.3413 Np m and b = 61.3841 rad/m The velocity of propagation is given by, v =

w = b

(2 ´ p ´ 15.9 ´ 10 6 ) = 1.6275 ´ 10 6 61.3841

m/s

The intrinsic impedance is given by,

\

( )(4 ´ p ´ 10 -7 ´ 1) 60 + j (2 ´ p ´ 15.9 ´ 10 6 ) (8.854 ´ 10 -12 ´ 50) j 2 ´ p ´ 15.9 ´ 10 6

h =

j wm = s + j we

h =

j (125.5413) = ( 60 + j 0.0442)

125.5413 Ð90° = 1.4465 Ð 44.98° W 60 Ð0.04°

Example 11.6.6 Solution : The propagation constant in lossy dielectric is given by, g = a + j b = j wm (s + jwe) \

g =

(

[

]

j ( 2 p f) (m 0 m r ) s + j ( 2 p f) ( e 0 e r )

)[2.56 ´ 10 -4 + j (2 ´ p ´ 10 ´ 10 9 )(8.854 ´ 10 -12 ´ 2.3)] j (78.9568 ´ 10 3 )[2.56 ´ 10 -4 + j 1.2795]

\ g = j 2 ´ p ´ 10 ´ 10 9 ´ 4 ´ p ´ 10 -7 \

g =

\

g =

[78.9568 ´ 10

\

g =

101.025 ´ 10 3 Ð 179.98°

\

g = 317.84 Ð 89.99° = 0.0554 + j 317.84

Thus,

a = 0.0554 Np/m

3

]

Ð 90° [1.2795 Ð 89.98°]

and

b = 317.84 rad/m

Example 11.6.7 Solution : For lossy medium, g =

j wm (s + j we) and

TM



11 - 7

Uniform Plane Waves

jwm s + jwe

h =

j wm = j(10 9 ´ p ´ 3 ´ 4 ´ p ´ 10 –7 ) = j(11.8435 ´ 10 3 ) = 11.8435 ´ 10 3 Ð 90º s + j we = 5 ´ 10 –2 + j(10 9 ´ p ´ 2 ´ 8.854 ´ 10 –12 ) = 5 ´ 10 –2 + j 0.05563 =

0.0748Ð 48.05º

Now the propagation constant is given by, jwm(s + jwe) = (11.8435 ´ 10 3 Ð90º )( 0.0748Ð48.05º )

g =

= 29.7639Ð 69.03º \

g = a + jb = 10.6518 + j 27.7925

a)

a = 10.6518 Np/m

b)

b = 27.7925 rad/m

c)

h =

11.8435 ´ 10 3 Ð90º = 397.914 Ð 41.95º W 0.0748Ð48.05º

d) The velocity of propagation is given by, v =

w 10 9 p = = 1.1303 ´ 10 8 m/sec 27.7925 b

Hence electric field intensity can be given as, p E = 20 e + ax cos æç10 9 pt + bx + ö÷ a z V/m è 3ø

K(1)

e) E at x = –1.5 m, t = 3 nsec : Substituting values of x, t, a, b in equation (1), we get p E = 20 e + ( 10.6518 )(–1.5 ) × cosé(10 9 p)( 3 ´ 10 –9 ) + ( 27.7925)( –1.5) + ù a z êë 3 úû K(Calculate cos value in radian mode) = 20( 0.114 ´ 10 –6 )( 0.9802)a z = 2.2348 ´ 10 –6 a z = 2.2348 a z mV/m f)

H

=

E h

=

p 20e ax cosé10 9 pt + + bx – 41.95ù a y A/m úû êë 3 397.914

At x = –1 m, t = 2 nsec : H=

p 41.95 ´ p ö ù 20e( 10.6518 )(–1) é cosê(10 9 p)( 2 ´ 10 –9 ) + + ( 27.7925)(–1) – æç ÷ ay è 3 180 ø úû 397.914 ë = (0.0502) (23.658 ´ 10 –6 ) (– 0.6989)a y = – 0.83 a y mA/m TM



11 - 8

Uniform Plane Waves

g) For 360º phase shift (i.e. for 2 p radian) the distance travelled by wave is l. Then for 20º phase shift, the distance d travelled by the wave is given by, d =

20 ´ l 20 æ 2p ö 40 p = = 0.01256 m = 12.56 mm = ç ÷ 360 360 è b ø 360(27.7925)

h) Now amplitude reduces to 12 V/m. Hence we can write, 12 = 20 × e a( –x ) because amplitude is reducing \

12 = e(10.6518)(–x) 20

Taking ln on both the sides, (10.6518)(–x) = – 0.5108 \

x = + 47.956 ´ 10

Example 11.7.7

–3

m = 47.956 mm

Kept this example for student's practice.

Example 11.7.8 Solution : s = 10 - 3 S/m,

e = 80 e 0 ,

m = m0

Let us first check the type of medium; whether it is dielectric or conducting. s s s 10 - 3 = = 22.47 >> 1 = = we (2pf) ( e) (2pf) (80 e 0 ) 2 ´ p ´ 10 ´ 10 3 ´ 80 ´ 8.854 ´ 10 -12

s is greater than 1, at frequency of 10 kHz, the medium can be we assumed to be a conducting medium.

As value of ratio

For conducting medium different parameters can be obtained as follows. i) Attenuation constant : a = p f ms = \

p fm0 s =

p ´ 10 ´ 10 3 ´ 4 ´ p ´ 10 - 7 ´ 10 - 3

a = 6.2832 ´ 10 - 3 Np m

ii) Phase constant : b = \

p f ms =

p fm0 s =

p ´ 10 ´ 10 3 ´ 4 ´ p ´ 10 - 7 ´ 10 - 3

b = 6.2832 ´ 10 - 3 rad m

iii) Propagation constant : \ g = a + j b = 6.2832 ´ 10 - 3 + j 6.2832 ´ 10 - 3 = 8.8858 ´ 10 - 3 Ð 45º m - 1 iv) Intrinsic impedance : j wm = h = s + j we \

h =

j (2 ´ p ´ f) m 0 s + j (2 ´ p ´ f) (80 e 0 )

j (2 ´ p ´ 10 ´ 10 3 ) (4 ´ p ´ 10 - 7 ) 10 - 3 + j (2 ´ p ´ 10 ´ 10 3 ) (80 ´ 8.854 ´ 10 - 12 ) TM



\

h =

\

h =

11 - 9

Uniform Plane Waves

j 0.07895 0.001 + j 4.4505 ´ 10 - 5 0.07895 Ð 90º 1 ´ 10 - 3 Ð 2.548º

\ h » 8.8853 Ð 43.72º W iv) Wavelength : 2p 2p = = 999.99 m » 1000 m l = b 6.2832 ´ 10 - 3 v) Velocity of propagation : 1 1 v = = = me m 0 (80 e 0 ) \

1 4 ´ p ´10 - 7 ´ 80 ´ 8.854 ´ 10 - 12

v = 0.3352 ´ 10 8 m sec

Example 11.7.9 Solution : The velocity of propagation is given by, v = fl \

f =

v 2.5 ´ 105 = = 1 ´ 10 9 Hz = 1 GHz l 0.25 ´ 10 -3

But the velocity of propagation is also given by, w 2pf = v = b b \

b =

2 p f 2 ´ p ´ 1 ´ 10 9 = 25.1327 ´ 10 3 rad/m = v 2.5 ´ 105

For good conductor, the phase constant is given by, b = p f ms = p f (m 0 m r ) s But for a non-magnetic material, m r = 1

(

)

p ´ 1 ´ 10 9 ´ 4 ´ p ´ 10 -7 ´ s = 25.1327 ´ 10 3

\

b =

\

s = 1.6 ´ 10 5 S/m

Example 11.7.10 Solution : For conducting medium, s = 58 MS/m. So using expressions of h, g and v for good conductor. The intrinsic impedance is given by, j ( 2 p f) (m 0 m r ) j wm = h = s s =

(2 ´ p ´ 100 ´ 10 6 )(4 ´ p ´ 10 -7 ´ 1) Ð90° 58 ´ 10

6

TM

= 3.6896 ´ 10 -3 Ð45° W



11 - 10

Uniform Plane Waves

The propagation constant is given by g = = \

j wms =

( 2 p f) (m 0 m r ) s

Ð90°

(2 ´ p ´ 100 ´ 10 6 )(4 ´ p ´ 10 -7 ´ 1)(58 ´ 10 6 ) Ð90° = 2.1399 ´ 105

Ð45°

g = a + j b = 1.5131 ´ 103 + j 1.5131 ´ 10 5

Comparing real and imaginary terms, we get, a = 1.5131 ´ 105 Np / m , and b = 1.5131 ´ 105 rad/m The velocity of propagation is given by, w 2´p´f 2 ´ p ´ 100 ´ 10 6 = 4.152 ´ 103 m/s v = = = b b 1.5131 ´ 105 Example 11.7.11 Solution : For aluminium with very high conductivity, the propagation constant is given by, g = jwms = j (2pf) (m 0 m r ) s =

j (2 ´ p ´ 2 ´ 10 6 ) (4 ´ p ´ 10 - 7 ´ 1) (40 ´ 10 6 ) =

=

6.3165 ´ 10 8 Ð 90° = 25.1326 ´ 10 3 Ð 45° m - 1

j 6.3165 ´ 10 8

= 17.771 ´ 10 3 + j 17.771 ´ 10 3 m - 1 But

g = a + j b = 17.771 ´ 10 3 + j 17.771 ´ 10 3

By comparing real and imaginary terms, we get, a = 17.771 ´ 10 3 Np m b = 17.771 ´ 10 3 rad m Hence skin depth for aluminium is given by, 1 1 d = = = 56.2714 mm a 17.771 ´ 10 3 Similarly the velocity of propagation is given by, w 2pf 2 ´ p ´ 2 ´ 10 6 v = = = = 707.12 m s b b 17.771 ´ 10 3 Example 11.7.12 Solution : For conducting medium, assume m r = 1. The depth of penetration is given by, 1 1 = d = p f (m 0 m r ) s p f ms d = 0.1 m, f = 1 MHz = 1 ´ 10 6 Hz TM


… Given


\ \

0.1 =

11 - 11

Uniform Plane Waves

1 p ´ 1 ´ 10 6 ´ 4 ´ p ´ 10 -7 ´ 1 ´ s

s = 5.0329 ª/m

Example 11.9.6 Solution : The interface is between perfect dielectric (region 1) and free space (region 2). For region 1, h1 =

=

m1 e1 4 ´ p ´ 10 -7 ´ 1 8.8542 ´ 10 -12 ´ 8.5

Region 1 er1 = 8.5

Region 2

mr1 = 1

Free space

s1 = 0 m1 ––– Ö e1

h1 =

= 129.22 W

h2 = 377 W

For region 2, h 2 = 120 p = 377 W By definitions, 2 h2 2 ( 377 ) Et = 1.4894 = = G = h1 + h2 Ei (129.22 + 377 ) But

E i = 1.5 mV/m

\

E t = t ( E i ) = (1.4894) 1.5 ´ 10 -3 G =

(

) = 2.2341 mV/m

h - h1 Er ( 377 - 129.22) = 0.4894 = 2 = Ei h2 + h1 ( 377 + 129.22)

But

E i = 1.5 mV/m

\

E r = G ( E i ) = ( 0.4894) 1.5 ´ 10 -3

As we know, H t =

Fig. 11.1

Et h2

and

(

Hr =

) = 0.7341 mV/m

Hr we can write, - h1

2.2341 ´ 10 -3 = 5.9259 mA/m and 377 0.7341 ´ 10 -3 = = – 5.681 mA/m - 129.22

Ht = Hr Example 11.9.7

Solution : a) The standard equation for the incident electric field in x-direction is given by, Ei = E 0 e - j b z a x V/m Comparing given Ei with standard equation, we get, b = 1 rad/m

TM



11 - 12

Uniform Plane Waves

Then the wavelength is given by, l =

2p 2p = = 2 p = 6.2831 m 1 b

Now the given Ei is in free space. In free space v = c = 3 ´ 10 8 m/s \

v = c= fl

c 3 ´ 10 8 = 47.75 MHz = 6.2831 l b) For free space, h = h 0 = 120 p = 377 W.

\

f =

A uniform plane wave is propagating in z-direction. E field is in x-direction. Thus to get uniform plane wave in z-direction, H field must be in y-direction. The amplitudes of E and H are relation by, E0 E E or H 0 = 0 = 0 h0 = H0 h0 377 \

H0 =

(2.6525 ´ 10 -3 ) E 0

A/m

Thus the magnetic field of the incident wave is given by, Hi = H 0 e - j z a y = 2.6525 ´ 10 -3 E 0 e - j z a y A/m

(

)

c) The interface is between free space and dielectric (m r = 1, e r = 4) as shown in the Fig. 11.2.

x Dielectric (mr = 1, er = 4)

Free space h2 = h0 = 377 W

For dielectric, h2 = \

h2 =

m = e

m rm 0 er e0

1 ´ 4 ´ p ´ 10 -7

y

4 ´ 8.854 ´ 10 -12

z

z=0

= 188.36 W

Fig. 11.2

The transmission coefficient is given by, t =

2h 2 Et = Ei h1 + h 2

But

Ei = E0

\

Et = E0

2 (188.36) = 0.6663 E 0 ( 377 + 188.36)

As the incident wave has the electric field in x-direction, the electric field in the transmitted wave will be also in x-direction. Then the field in the transmitted wave is given by, Et = E t e - j b z a x V/m But

E t = (0.6663) E 0 V/m and b = 1 rad/m TM



\

11 - 13

Uniform Plane Waves

Et = (0.663 E 0 ) e - j z a x V/m

The transmitted wave is in medium 2. For medium 2, h 2 = 188.36 W. As the incident wave has the magnetic field in y-direction, the transmitted wave will also have the magnetic field in y-direction. The amplitudes of transmitted electric and magnetic fields are related by h 2 as, Et h2 = Ht \

Ht =

0.6663 E 0 Et = = 3.5373 ´ 10 -3 E 0 A/m h2 188.36

Thus equation for magnetic field is given by, Ht = H t e - j b z a y A/m But

H t = 3.5373 ´ 10 -3 E 0 A/m and b = 1 rad/m

\

Ht =

(3.5373 ´ 10 -3 E 0 ) e - j z

a y A/m

Example 11.9.8 Solution : For region 1, e r1 = 8.5, m r1 = 1 and s 1 = 0 means perfect dielectric \

h1 =

m 0m r = e0 er

m = e

4 ´ p ´ 10 -7 ´ 1

8.854 ´ 10 -12 ´ 8.5

\ h1 = 129.22 W For region 2 (free space), h 2 = 120 p = 377 W a) For normal incidence : The reflection coefficient is given by, E r h 2 - h1 377 - 129.22 G = = = E i h 2 + h1 377 + 129.22 \

G =

Er = 0.4894 Ei

The transmission coefficient is given by, 2 ( 377 ) 2h 2 Et t = = = E i h 2 + h1 377 + 129.22 \

t =

Et = 1.4894 Ei

b) For oblique incidence : By Snell's law of refraction, e1 e r1 e 0 sin q t = = = sin q i e2 e r2 e 0

8.5 1 TM



\ \

11 - 14

sin q t = sin (10° ) qt =

sin -1

Uniform Plane Waves

8.5

(0.506) = 30.42°

With perpendicular polarization, at boundary, h cos q i - h 1 cos q t 377 ( cos 10° ) - 129.22 ( cos 30.42° ) Er = 0.5383 = = 2 h 2 cos q i + h 1 cos q t Ei 377 ( cos 10° ) + 129.22 ( cos 30.42° ) The transmission coefficient is given by, 2h 2 cos q i 2 ( 377 )( cos 10° ) Et = = = 1.5383 h 2 cos q i + h1 cos q t Ei 377 ( cos 10° ) + 129.22 ( cos 30.42° ) Example 11.9.9 Solution : For medium 1, i.e. free space, h 1 = 120 p = 377 W For medium 2, i.e. perfect dielectric, h2 =

m = e

m 0m r = e0 er

(4 ´ p ´ 10 -7 )(1) (8.854 ´ 10 -12 )( 3)

\ h 2 = 217.507 W The transmission coefficient is given by, 2 h2 2 ( 217.507 ) Et = 0.7317 t = = = E i h 1 + h 2 ( 377 + 217.507 ) Hence magnitude of transmitted wave is E t = (0.7317) E i V / m The reflection coefficient is given by, h - h1 Er ( 217.507 - 377 ) = – 0.2683 G = = 2 = Ei h2 + h1 ( 217.507 + 377 ) ... Negative sign indicates wave in opposite direction Hence magnitude of reflected wave is E r = (0.2683) E i V / m Interms of magnetic field magnitudes, we can write, Et / h2 h Ht E = = 1 æç t ö÷ Hi Ei / h1 h2 è Ei ø \

Ht Hi

=

377 (0.7317) = 1.2682 217.507

Hence magnitude of transmitted H wave is, H t = (1.2682) H i A/m Ei E = i (2.6525 ´ 10 -3 ) E i 377 h1

But

Hi =

Hence,

H t = (1.2682) (2.6525 ´ 10 -3 E i ) = 3.3639 ´ 10 -3 E i A/m

TM



11 - 15

Uniform Plane Waves

Similarly - Er / h1 E Hr = = - r = - ( - 0.2683) = 0.2683 Ei / h1 Ei Hi Hence magnitude of reflected H wave is, H r = (0.2683) H i = (0.2683) (2.6525 ´ 10 -3 E i ) = (0.7116 ´ 10 -3 ) E i A/m Example 11.9.10 Solution : As medium 1 is free space, h1 = 120 p = 377 W For medium 2, s = 0 indicates it is lossless dielectric. For lossless dielectric, intrinsic impedance is given by, m 0m r m h2 = = e e0 er \

h2 =

4 ´ p ´ 10 -7 ´ 9 8.854 ´ 10 -12 ´ 4

Medium 2 Material with er = 4, mr = 9

Medium 1 Free space

and s = 0

Fig. 11.3

= 565.1 W

The transmission coefficient is given by, t =

2 h2 2 (565.1) = 1.1996 = h1 + h 2 377 + 565.1

The reflection coefficient is given by, h - h1 565.1 - 377 = 0.1996 G = 2 = h 2 + h1 565.1 + 377 Example 11.10.5 Solution : Medium 1 is glass, for which e r1 = 9 ; while medium 2 is air, for which e r2 = 1. The critical angle is given by, e2 q c = sin -1 = sin -1 e1

e r2 e r1

1 9

\

q c = sin -1

\

q c = sin -1 (0.3333) = 19.47º

Example 11.10.6 Solution : For medium 1 i.e. air : e 1 = e 0 e r1 = e 0 For medium 2 i.e. glass : e r = e 0 e r2 = 9 e 0 Hence Brewster angle is given by, e2 9 e0 q B = tan - 1 = tan - 1 = 71.565º e1 e0 TM

... (Assume e r = 9 for glass)



11 - 16

Uniform Plane Waves

Now light is incident at Brewster's angle. Hence angle of incidence is q i = 71.565º. \ By Snell's law, cm e h 1 1 sin q i Now m 1 = m 2 = m 0 . sin q t = 1 sin q i = h2 cm e 2 2

\ \

æ e0 ö 1 sin q t = ç ÷ sin q i = sin (71.565º ) 3 è 9 e0 ø 1 q t = sin - 1 é (sin 71.565º ) ù = 18.434º úû êë 3

qqq

TM


50999882VAKX_Electromagnetic Field Theory_Solution Manual

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