Teoria Electromagnetica - Hayt - Solucionario CAPITULO 7 7ma Edicion

CHAPTER CHAPTER 7

7.1. Let V Let V = 2xy 2 z 3 and  =  =  0 . Given point P point P (1 (1,, 2, 1), find: a) V at P : P : Substituting the coordinates into V , V , find V P P =

−

−8 V.

− − − 6xy2z2az , which, when evaluated − c) ρv at P : P : This is ρv = ∇ · D = −0 ∇2 V = −4xz( xz (z 2 + 3y 3 y2 ) C/m3 d) the equation of the equipotential equipotential surface passing passing through P : P : At P , P , we know V = −8 V, 2 3 so the equation will be xy z = −4. b)

at P : P : We use E = V = 2y 2 z 3 ax 4xyz 3 ay at P , P , becomes EP = 8ax + 8 ay 24az V/m

−∇

E

e) the equation of the streamline streamline passing through through P : P : First, E y dy 4xyz 3 2x = = 2 3 = E x dx 2y z y Thus ydy = ydy = 2xdx, and so

1 2 y = x 2 + C 1 2

Evaluating at P at P ,, we find C 1 = 1. Next, E z dz 6xy2 z 2 3x = = = E x dx 2y 2 z 3 z Thus 3xdx = xdx = zdz, zdz, and so

3 2 1 x = z 2 + C 2 2 2

Evaluating at P at P ,, we find C 2 = 1. The streamline is now specified by the equations: y2

− 2x2 = 2

and 3x2

− z2 = 2

f) Does V Does V satisfy satisfy Laplace’s equation? No, since the charge density is not zero. 7.2. Given Given the sphericallyspherically-symmet symmetric ric potential field in free space, V space, V = V 0 e−r/a , find: a) ρv at r = a = a;; Use Poisson’s equation, 2 V = ρv /, /, which in this case becomes

−

ρv 1 d = 2 0 r dr

∇ −  2 dV  −V 0 d  2 r = r e ar2

dr

−r/a

dr



V 0  r − = 2− e ar

−r/a

a

from which which

− 

0 V 0 ρv ( r ) = 2 ar

r −r/a e a

⇒

ρv (a) =

0 V 0 −1 e C/m3 a2

b) the electr electric ic field at r at r = a = a;; this we find through the negative gradient: E(r )

=

V 0 ar = e −∇V = − dV dr a

−r/a

1

ar

⇒

E(a)

=

V 0 −1 e ar V/m a

Titles you can't find anywhere else

Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.











7.2c) the total charge: charge: The easiest way way is to first find the electric electric flux density, density, which from part b is −r/a D = 0 E = (0 V 0 /a) /a)e ar . Then Then the net outward outward flux of D through a sphere of radius r would be Φ(r Φ(r ) = Q encl (r ) = 4πr 2 D = 4π0 V 0 r 2 e−r/a C As r

→ ∞, this result approaches zero, so the total charge is therefore Qnet = 0. 7.3. Let V Let V ((x, y) = 4e2x + f ( f (x) − 3y 2 in a region of free space where ρ where ρ v = 0. It is known that both E x and a nd V V are are zero at the origin. Find f Find f ((x) and V and V ((x, y): Since ρ Since ρ v = 0, we know that ∇2 V = 0, and so

∇ Therefore

2

∂ 2 V ∂ 2 V d2 f 2x V = + = 16 16ee + 2 ∂x 2 ∂y 2 dx

d2 f = dx2

− 6 = 0

16ee2x + 6 ⇒ −16

Now E x =

df = dx

6x + C 1 −8e2x + 6x

∂V df = 8e2x + ∂x dx

and at the origin, this becomes df E x (0) = 8 + dx Thus df/dx Thus df/dx



x=0

= 0(as given)

that C 1 = 0. Integrating again, we find |x=0 = −8, and so it follows that C f ( f (x, y ) = −4e2x + 3x 3x2 + C 2 which at the origin becomes f (0 f (0,, 0) = − 4 + C 2 . Ho Howe weve ver, r, V (0 V (0,, 0) = 0 = 4 + f (0 f (0,, 0). So 2 x 2 2 x f (0 f (0,, 0) = −4 and C 2 = 0. Finally, Finally, f ( f (x, y ) = −4e + 3x 3x , and V ( V (x, y) = 4e − 4e2x + 3x 3 x2 − 3y2 = 3(x 3(x2 − y 2 ).

7.4. Given Given the potential field, field, V ( V (ρ, φ) = (V 0 ρ/d)cos ρ/d)cos φ: a) Show Show that V that V ((ρ, φ) satisfies Laplace’s equation:

∇

2

 

1 ∂ ∂V 1 ∂ 2 V 1 ∂ + 2 = V = ρ ρ ∂ρ ∂ρ ρ ∂φ 2 ρ ∂ρ V 0 ρ V 0 ρ = cos φ sin φ = 0 d d





V 0 ρ cos φ d

−

1 ∂ ρ2 ∂φ





V 0 ρ sin φ d

−

b) Descri Describe be the constant constant-pot -poten ential tial surfaces surfaces:: These These will be surfac surfaces es on which which ρ cos φ is a constan constant. t. At this stage, stage, it is helpful helpful to recall recall that the x coordinate in rectangular coordinates is in fact ρ cos φ, so we identify the surfaces of constant potential as (plane) surfaces of constant x (parallel to the yz plane).











7.5. Given Given the potential field V = (Aρ4 + Bρ −4 )sin4φ )sin4φ: 2 a) Show Show that V = V = 0: In cylindrical coordinates,

∇ ∇

2

   −

1 ∂ ∂V 1 ∂ 2 V V = ρ + 2 ρ ∂ρ ∂ρ ρ ∂φ 2 1 ∂ 1 = ρ(4Aρ (4Aρ3 4Bρ −5 ) sin4φ sin4φ 16(Aρ 16(Aρ4 + Bρ −4 )sin4φ )sin4φ 2 ρ ∂ρ ρ 16 16 1 6 = (Aρ3 + Bρ −5 )sin4φ )sin4φ (Aρ4 + Bρ −4 )sin4φ )sin4φ = 0 2 ρ ρ



−

−

b) Select Select A and B so that V V = 100 V and E = 500 V/m at P ( P (ρ = 1, φ = 22 22..5◦ , z = 2): First, ∂V 1 ∂V E= V = aρ aφ ∂ρ ρ ∂φ = 4 (Aρ3 Bρ −5 )sin4φ )sin4φ aρ + (Aρ ( Aρ3 + Bρ −5 )cos4φ )cos4φ aφ

| |

−∇ − − − and at P , P , EP = −4(A 4(A − B ) aρ .

−

Thus Thus

equations equations are:

4(A − B ). |EP | = ±4(A 4(A 4(A − B ) = ±500



Also, Also, V P A + B B.. Our Our two two P = A +

and A + B = 100 We thus have two pairs of values for A and a nd B B : A = 112. 112.5, B =

12..5 −12

or A =

12..5, B = 112. 112.5 −12

7.6. A parallel-plate parallel-plate capacitor has plates located at z = 0 and z = d. d . The region region between between plates plates 3 is filled with a material containing volume charge of uniform density ρ density ρ 0 C/m , and which has permittivity . Both plates are held at ground potential. a) Determ Determine ine the potential potential field betwe between en plates plates:: We solve solve Poiss Poisson’s on’s equation, equation, under under the assumption assumption that V that V varies varies only with z : 2

∇2V = ddzV 2 = − ρ0 ⇒

V =

−ρ0z2 + C 1z + C 2 2

At z = 0, V = V = 0, and so C so C 2 = 0. Then, at z at z = d, d , V = V = 0 as well, so we find C 1 = ρ 0 d/2 d/2. Finally, V ( V (z ) = (ρ0 z/2 z/ 2)[d )[d z ].

−

b) Determine Determine the electric electric field intensity intensity,, E between plates: Taking the answer to part a part a,, we find E through E

=

d az = − −∇V = − dV dz dz



ρ0 z (d 2



(2z − d) az V/m − z) = ρ20 (2z











7.6c) Repeat a and b and b for the case of the plate at z = d raised to potential V 0 , with the z = 0 plate grounded: Begin with ρ0 z 2 V ( V (z ) = + C 1 z + C 2 2

−

with C 2 = 0 as before, since V since V ((z = 0) = 0. Then V ( V (z = d) d ) = V 0 =

−ρ0d2 + C 1 d ⇒ 2

C 1 =

V 0 ρ0 d + d 2

So that

V 0 ρ0 z z + (d z ) d 2 We recognize this as the simple superposition of the voltage as found in part a and a and the voltage of a capacitor capacitor carrying voltage V voltage V 0 , but without the charged dielectric. The electric field is now V ( V (z ) =

E

=

−

−V 0 az + ρ 0 (2z az = (2z − d) az V/m − dV dz d 2

7.7. Let V Let V = = (cos2φ (cos2φ)/ρ in /ρ in free space. a) Find the volume charge charge density density at point A point A(0 (0..5, 60◦ , 1): Use Poisson’s equation: ρv =

−0

=

−0

    ∇ −  −   2

So at A we find: ρvA =

V =

1 ∂ ρ ∂ρ

∂V 1 ∂ 2 V ρ + 2 ∂ρ ρ ∂φ 2 4 cos2φ cos2φ 30 cos cos 2φ = ρ2 ρ ρ3

1 ∂ 0 ρ ∂ρ cos2φ cos2φ ρ

30 cos(120◦ ) = 0.53

−

120 = −106pC/ 106pC/m3 −12

b) Find the surface charge density on a conductor surface passing passing through B through B(2 (2,, 30◦ , 1): First, we find E: ∂V 1 ∂V E= V = aρ aφ ∂ρ ρ ∂φ cos2φ cos2φ 2 sin 2φ = aρ + aφ 2 ρ ρ2

−∇

−

−

At point B the field becomes EB

=

cos60◦ 4

aρ +

2 sin 60◦ 4

The surface charge density will now be

aφ =

0.125 aρ + 0. 0 .433 aφ











7.8. A uniform volume charge charge has constant constant density ρv = ρ0 C/m3 , and fills the region r < a, in which permittivity  permittivity  as assumed. A conducting spherical shell is located at r = a, a , and is held at ground potential. Find: a) the potential everywhe everywhere: re: Inside the sphere, sphere, we solve Poisson’s Poisson’s equation, assuming assuming radial variation only:

∇

2

1 d V = 2 r dr

  r

2 dV

dr

−ρ0 ⇒ = 

ρ0 r2 C 1 − V ( V (r ) = + + C 2 60

r

We require that V V is finite at the orgin (or as r 0), and so therefore C 1 = 0. Ne Next xt,, 2 V = V = 0 at r = a, a , which gives C 2 = ρ 0 a /6. Outside, Outside, r > a, we know the potential must be zero, since the sphere is grounded. To show this, solve Laplace’s equation:

→

2

∇

1 d V = 2 r dr

  r2

dV = 0 dr

⇒

V ( V (r ) =

C 1 + C 2 r

Requiring V = V = 0 at both r = a = a and at infinity leads to C to C 1 = C 2 = 0. To summarize V ( V (r ) =



ρ 0 (a2 6

0

− r2 )

ra

b) the electric field intensity intensity, E, everywhere: Use E

=

−∇V = −drdV ar = ρ30r ar

Outside (r (r > a), the potential is zero, and so

E

r
= 0 there as well.

7.9. The functions functions V and V 2 (ρ,φ,z) b oth satisfy Laplace’s equation equation in the region a region a < ρ < b, V 1 (ρ,φ,z) ρ,φ,z ) and V ρ,φ,z) both b, 0 φ < 2π , L < z < L; L; each is zero on the surfaces ρ = b for L < z < L; L; z = L for a < ρ < b; b; and z and z = L for a < ρ < b; b ; and each is 100 V on the surface ρ surface ρ = = a a for fo r L < z < L. L. a) In the region specified specified above above,, is Laplace Laplace’s ’s equatio equation n satisfi satisfied ed by the functio functions ns V 1 + V 2 , V 1 V 2 , V 1 + 3, and V 1 V 2 ? Yes for the first three, since Laplace’s Laplace’s equation is linear. linear. No for V 1 V 2 . b) On the boundary surfaces surfaces specified, are the potential potential values given given above obtained from the functions V functions V 1 + V 2 , V 1 V 2 , V 1 +3, and V and V 1 V 2 ? At the 100 V surface (ρ ( ρ = a = a), ), No for all. At the 0 V surfaces, yes, except for V 1 + 3. c) Are the functio functions ns V 1 + V 2 , V 1 V 2 , V 1 + 3, and V 1 V 2 identical with V 1 ? Only V2 is, since it is given as satisfying all the boundary conditions that V that V 1 does. Therefore, by the uniqueness theorem, V 2 = V 1 . The others, others, not satisfyi satisfying ng the boundary boundary conditio conditions, ns, are not the same as V as V 1 .

≤

−

−

−

−

−

−

−











7.10. Consider Consider the parallel-plate parallel-plate capacitor of Problem 7.6, but this time the charged dielectric dielectric exists only between z = 0 and z = b, where b < d. Free space space fills the region region b < z < d. d. Bo Both th plates are at ground potential. potential. No surface charge charge exists at z = b, b , so that both V and D are continuous there. By solving Laplace’s and Poisson’s equations, find: a) V ( V (z ) for 0 < 0 < z < d: d: In Region 1 (z (z < b), we solve Poisson’s equation, assuming z assuming z variation variation only: d2 V 1 ρ0 dV 1 ρ0 z = = + C 1 (z < b) dz 2  dz  In Region 2 (z (z > b), we solve Laplace’s equation, assuming z variation only:

−

−

⇒

d2 V 2 =0 dz 2

⇒

dV 2 = C 1 (z > b) dz

At this stage we apply the first boundary condition, which is continuity of D across the interface at z at z = b. b . Knowing that the electric field magnitude is given by dV/dz by dV/dz,, we write dV 1  dz



z =b

dV 2 =  0 dz





z =b

⇒ −ρ0b + C 1 = 0 C 1 ⇒

C 1 =

−ρ0b + 0

 C 1 0

Substituting the above expression for C 1 , and performing a second integration on the Poisson and Laplace equations, we find V 1 (z ) = and V 2 (z ) =

−

ρ0 z 2 + C 1 z + C 2 (z < b) 2

− ρ20bz0 + 0 C 1z + C 2 

(z > b)

Next, requiring V 1 = 0 at z = 0 leads to C 2 = 0. Then, the requiremen requirementt that V 2 = 0 at z = d leads d leads to 0=

− ρ00bd + 0 C 1 d + C 2 ⇒ 

C 2 =

ρ0 bd 0

− 0 C 1d

ρ0 b (d 0

− z ) − 0 C 1(d − z)

With C 2 and C 2 known, the voltages now become V 1 (z ) =

− ρ20z

2

+ C 1 z and V 2 (z ) =

Finally, to evaluate C 1 , we equate the two voltage expressions at z at z = b: b : V 1

|z=b = V 2|z=b ⇒



ρ0 b b + 2 2 r (d b) C 1 = 2 b + r (d b)

− −



where r = /0 . Subs Substit titut utin ingg C 1 as found above into V 1 and V 2 leads to the final expressions for the voltages:













7.10b) the electric electric field intensity intensity for 0 < 0 < z < d: d: This involves involves taking taking the negative negative gradient of the final voltage expressions of part a. We find E1 =

−

E2

dV 1 dz

az

=

dV 2 dz

−

  − 

ρ0 = z  az

b 2

b + 2 2r (d b) b + r (d b)

ρ0 b2 1 = 20 b + r (d

− −

− b)





az

az

V/m

V/m

(z < b)

(z > b )

7.11. The conducting planes 2x 2x + 3y = 12 and 2x 2x + 3y = 18 are at potentials of 100 V and 0, respectively. respectively. Let  Let  =  =  0 and find: a) V at P (5 P (5,, 2, 6): The planes planes are parallel, parallel, and so we expect expect variati variation on in potential potential in the direct direction ion normal to them. them. Using Using the two two boundar boundary y condtio condtions, ns, our general general potential potential function can be written: V ( V (x, y ) = A(2 A (2x x + 3y 3y and so A so A =

100/ /6. We then write −100 V ( V (x, y ) =

and V and V P P = b) Find

E

= A(2 (2x x + 3y 3y − 18) + 0 − 12) + 100 = A

100 (2x (2x + 3y 3y − 18) = − 50yy + 300 x − 50 − 100 6 3

33.33 V. − 1003 (5) − 100 + 300 = 33.

at P : P : Use E

=

ax + 50 ay V/m −∇V = 100 3

7.12. The derivation derivation of Laplace’s Laplace’s and Poisson’s Poisson’s equations assumed constant constant permittivity permittivity,, but there are cases of spatially-v spatially-varying arying permittivity permittivity in which which the equations will still apply. apply. Consider Consider the vector identity, (ψG) = G ψ + ψ where ψ and G are scalar and vector functions, G, where ψ respectivel respectively y. Determine Determine a general general rule on the allowed allowed directions in which  may vary with respect to the electric field.

∇ ·

·∇

∇·











We seek ρ seek ρ at which V = V = 20 V, and thus we need to solve: 20 = 100

ln(. ln(.012 012/ρ /ρ)) ln(2. ln(2.4)

⇒

ρ =

.012 = 1.01 cm (2. (2.4)0.2

b) E ρmax ρ max : We have E ρ =

dV 100 =− = − ∂V ln(2.4) ∂ρ dρ ρ ln(2.

whose maximum value will occur at the inner cylinder, or at ρ at ρ = = . .5 5 cm: E ρ max max =

100 = 2.28 .005 005 ln(2 ln(2.4)

22.8 kV kV//m × 104 V/m = 22.

c) r if the charge per meter length on the inner cylinder is 20 nC/m: The capacitance per meter length is 2π 0 r Q C = = ln(2. ln(2.4) V 0 We solve for r : r =

(20

× 10

−9

)ln(2. )ln(2.4) = 3.15 2π0 (100)

7.14. Repeat Problem 7.13, but with the dielectric only partially partially filling the volume, within 0 < 0 < φ < π , and with free space in the remaining volume. We note that the dielectric changes with φ, and not with ρ. Also, Also, sin since ce E is radiallydirected directed and varies only with radius, Laplace’s equation equation for this case is valid (see Problem Problem 7.12) and is the same as that which led to the potential and field in Problem 7.13. Theref Therefore ore,, the solutio solutions ns to parts parts a and b are are unchange unchanged d from from Proble Problem m 7.13 7.13.. Part Part c, however, is different. We write the charge per unit length as the sum of the charges along each half of the center conductor (of radius a) =  r 0 E ρ,max Q =  πa) + 0 E ρ,max πa ) =  0 E ρ,max πa)(1 + r ) C/m ρ,max (πa) ρ,max (πa) ρ,max (πa)(1











Subtracting one equation from the other, we find = C 1 (.188 − .179) ⇒ −180 = C

C 1 =

−2.00 × 104

Then

−2.00 × 104 (.188) + C 2 ⇒ Finally, V ( V (φ) = (−2.00 × 104 )φ + 3. 3.78 × 103 V. 20 =

b)

E(ρ):

D(ρ)

× 103

Use E(ρ)

c)

C 2 = 3.78

=  0 E(ρ) = (2. (2.00

=

−∇V = −

1 dV 2.00 104 = ρ dφ ρ

×

aφ

V/m

/ρ) aφ C/m2 . × 104 0 /ρ)

d) ρs on the upper surface of the lower plane: We use ρs = D



· n surface =

2.00

× 104 aφ · aφ = 2.00 × 104 C/m2

ρ

ρ

e) Q on the upper surface of the lower plane: This will be

  .1

Qt =

0

.120

.001

2.00

× 104 0 dρdz = dρdz = 2.00 × 104 0 (.1) ln(120) ln(120) = 8. 8.47 × 10

−8

ρ

C = 84. 84.7 nC

f) Repeat a) to c to c)) for region 2 by letting the location of the upper plane be φ = . = .188 188 2π, and then find ρs and Q on the lower lower surface of the lower plane. plane. Back to the beginning, we use 20 = C 1 (.188 2π ) + C 2 and and 20 2000 = C 1 (.179) + C 2

−

−











7.15ff (cont 7.15 (contin inued ued)) Subtra Subtractin cting g one from from the other, other, we find find 



= C 1 (.009 − 2π ) ⇒ C 1 = 28 28..7 −180 = C Then Then 200 = 28. 28.7(. 7(.179) + C 2 ⇒ C 2 = 194. 194.9. Thus V ( 28.7φ + 194. 194.9 in region 2. Then V (φ) = 28. 

E

=



28..7 aφ V/m − 28 ρ

and and

D

=

28..70 2 aφ C/m − 28 ρ

ρs on the lower surface of the lower plane will now be ρs =

28..70 28 28..70 aφ · (−aφ ) = C/m2 − 28 ρ ρ

The charge on that surface will then be Q b = 28 28..70 (.1) ln(120) ln(120) = 122 122 pC. g) Find the total charge on the lower plane and the capacitance between between the planes: Total charge will be Q be Q net = Q t + Qb = 84 84..7 nC nC + 0.122 nC = 84. 84.8 nC. The capacitance will be C =

Qnet 84 84..8 = = 0.471 nF = 471 pF ∆V 200 20

−

7.16. A parallel-plate parallel-plate capacitor is made using two circular circular plates of radius a radius a,, with the bottom plate on the xy plane, centered at the origin. The top plate is located at z = d, d , with its center on the z axis. Potential Potential V 0 is on the top plate; the bottom plate is grounded. Dielectric Dielectric having having permittivity y fills the region between between plates. The permittivity permittivity is given given by radially-dependent permittivit (ρ) =  0 (1 + ρ/a). ρ/a). Find: a) V ( Sincee  varies in the direction normal to E, Laplace’s equation applies, and we V (z ): Sinc write d2 V 2 V = =0 V ( V (z ) = C 1 z + C 2 dz 2 With the given boundary conditions, C conditions, C 2 = 0, and C and C 1 = V 0 /d. /d. Therefore V Therefore V ((z ) = V 0 z/d V.

∇

b)

E:

This will be

⇒

/d) az V/m. −∇V = −dV/dz az = −(V 0/d) c) Q: First we find the electric flux density: density: D =  E = −0 (1 + ρ/a + ρ/a)( )(V V 0 /d) /d) a E

=

C/m2 . The











where V 0 = 100, a = 5 and b = 20. 20. Sett Settin ingg V ( V (r) = 20, and solving for r produces r = 12 12..5mm. b) Find E r,max r,max : Use E

=

V 0 ar   ar = −∇V = − dV 2 dr r 1 − 1 a

E r,max E (r = a = a)) = r,max = E (

b

100 V 0 = = 26 26..7 V/mm = 26. 26.7 kV kV//m a(1 (a/b)) a/b)) 5(1 (5/ (5/20))

−

−

c) Find r if the surface charge density on the inner sphere is 1. 1.0 µC/m2 : ρs will be equal in magnitude to the electric flux density at r = a. So ρs = (2. (2.67 104 V/m) m)r 0 = −6 2 10 C/m . Thus Thus r = 4.23. Note, Note, in the first first print printing ing,, the given given charge charge density density was 2 100 µC/m , leading to a ridiculous answer of   r = 423.

×

7.18. The hemispher hemispheree 0 < 0 < r < a, a, 0 < θ < π/2, π/2, is composed of homogeneous conducting material of conductivity σ . The flat side side of the hemisph hemisphere ere rests rests on a perfect perfectlyly-con conduc ducting ting plane. plane. No Now, w, the material within the conical region 0 < 0 < θ < α, α, 0 < r < a, a , is drilled out, and replaced with material material that is perfectly-conduct perfectly-conducting. ing. An air gap is maintained maintained between between the r = 0 tip of this new material and the plane. What resistance is measured between the two perfect conductors? Neglect Neglect fringing fringing fields. With no fringing fields, we have θ-variation -variation only in the potential. Laplace’s Laplace’s equation is therefore: 1 d dV 2 V = 2 sin θ = 0 r sin θ dθ dθ



∇



This reduces to

dV C 1 = V ( V (θ) = C 1 lntan lntan (θ/2) θ/ 2) + C 2 dθ sin θ We assume zero potential on the plane (at θ = π /2), which means that C 2 = 0. On the cone (at θ (at θ = α), α), we assume potential V 0 , and so V 0 = C 1 ln tan( tan(α/2) α/2) C 1 = V 0 / ln tan( tan(α/2) α/2) The potential function is now

⇒

⇒

V ( V (θ) = V 0

ln tan( tan(θ/2) θ/2) lntan(α/ lntan(α/2) 2)

α < θ < π/2 π/2













7.19. Two coaxial conducting cones have their vertices at the origin and the z axis as their axis. Cone A has the point A(1, (1, 0, 2) on its surface, while cone B has the point B (0, (0, 3, 2) on its surface. Let V A = 100 V and V B = 20 V. Find: a) α for each cone: Have αA = tan−1 (1/ (1/2) = 26. 26.57◦ and αB = tan−1 (3/ (3/2) = 56. 56.31◦ . b) V at P (1 P (1,, 1, 1): The potential function between cones can be written as V ( V (θ) = C 1 ln tan( tan(θ/2) θ/ 2) + C 2 Then 20 = C 1 ln tan(56 tan(56..31 31/ /2) + C 2 and and 10 1000 = C 1 ln tan(26 tan(26..57 57//2) + C 2 Solving these two equations, we find C 1 = tan−1 ( 2) = 54. 54.7◦ . Thus

√

V P P =

97..7 and C 2 −97

=

41..1. −41

Now at at P , P , θ =

97..7lntan(54. 7lntan(54.7/2) − 41 41..1 = 23. 23.3 V −97

7.20.. A potential 7.20 potential field in free space space is given given as V as V = 100 100 lntan( lntan(θ/2) θ/2) + 50 V. a) Find the maximum maximum value value of Eθ on the surface θ = 40◦ for 0. 0.1 < r < 0. 0 .8 m, 60◦ < φ < 90◦ . First

| | |

E

=

100 100 100 aθ = − aθ = − aθ = − aθ − 1r dV dθ 2r tan(θ/ tan(θ/2)cos 2)cos2 (θ/2) θ/2) 2r sin(θ/ sin(θ/2)cos( 2)cos(θ/ θ/2) 2) r sin θ











7.21. In free space, let ρ let ρ v = 200 2000 /r2.4 . a) Use Poisson’s Poisson’s equation equation to find V find V ((r) if it is assumed that r that r 2 E r that V that V 0 as r as r : With r With r variation only, we have

→

when r → 0, and also → 0 when r

→∞

1 d 2 V = 2 r dr

∇ or

Integrate once:

d dr

  r2

or

dV = dr

dV = dr

333..3r −333

  r2

dV = dr

  r2

dV = dr

200rr − ρv = −200

200rr −200

−2.4

−.4

r.6 + C 1 = −333 333..3r .6 + C 1 − 200 .6

−1.4

+

C 1 = r2

V (in this case) = −E r ∇V (in Our first boundary condition states that r2 E r → 0 when r → 0 Therefore C 1

Integrate again to find:

= 0.

333.3 −.4 333. r + C 2 .4 From our second boundary condition, condition, V V 0 as r as r , we see that C that C 2 = 0. Finally, V ( V (r ) =

→

→∞

V ( V (r) = 833. 833.3r−.4 V











7.22. By appropriate appropriate solution solution of Laplace’s and Poisson’s Poisson’s equations, determine the absolute potential at the center of a sphere of radius a radius a,, containing containing uniform volume charge of density density ρ 0 . Assume permittivity 0 everyw everywher here. e. HIN HINT: T: What must must be true true about the potential potential and the electric electric field at r at r = 0 and at r = a? a ? With radial dependence only, Poisson’s equation (applicable to r

≤ a) becomes 2 ⇒ V 1(r) = − ρ60r0 + C r1 + C 2 (r ≤ a)

 

ρ0 1 = − ∇2V 1 = r12 drd r2 dV dr 0 For region 2 (r (r ≥ a) there is no charge and so Laplace’s equation becomes  2 dV 2  1 d 2 ∇ V 2 = r2 dr r dr = 0 ⇒ V 2(r) = C r3 + C 4 (r ≥ a) Now, as r → ∞, V 2 → 0, so therefore C 4 = 0. Also Also,, as r → 0, V 1 must be finite, so therefore C therefore C 1 = 0. Then, V must V must be continuous across the boundary, r = a: a : 2 ρ0 a C 3 C 3 ρ0 a2 V 1 r=a = V 2 r=a + C 2 = C 2 = + 6 0 a a 60 So now ρ0 2 C 3 C 3 V 1 (r) = (a r2 ) + and V 2 (r ) = 60 a r Finally, since the permittivity is 0 everywhere, the electric field will be continuous at r = a = a.. This is equivalent to the continuity of the voltage derivatives: dV 1 dV 2 ρ0 a C 3 ρ0 a 3 = = C = 3 dr r=a dr r=a 30 a2 30 So the potentials in their final forms are





⇒ −

⇒

−





⇒ −

−

⇒











7.24. The four sides of a square trough are held at potentials of 0, 20, -30, and 60 V; the highest and lowest lowest potential potentialss are on opposite opposite sides. sides. Find Find the potential potential at the center center of the trough: trough: Here we can make good use of symmetry. The solution for a single potential on the right side, for example, with all other sides at 0V is given by Eq. (39): 4V 0 V ( V (x, y) = π

∞



1,odd

 

1 sinh(mπx/b sinh(mπx/b)) mπy sin m sinh(mπd/b sinh(mπd/b)) b

In the current problem, we can account for the three voltages by superposing three solutions of the above form, suitably modified to account for the different locations of the boundary potentials. potentials. Since we want want V at V at the center of a square trough, it no longer matters on what boundary each of the given potentials is, and we can simply write: V (center) V (center) =

4(0 + 20

∞

− 30 + 60)  π

1,odd

1 sinh(mπ/ sinh(mπ/2) 2) sin(mπ/ sin(mπ/2) 2) = 12. 12.5 V m sinh(mπ sinh(mπ))

The series converges to this value in three terms. 7.25. In Fig. 7.7, change change the right side so that the potential potential varies varies linearly linearly from 0 at the bottom of that that side side to 100 V at the top. Solve Solve for the potential potential at the center center of the trough: trough: Since Since the potential reaches zero periodically in y in y and also is zero at x at x = 0, we use the form: ∞

V ( V (x, y) =



V sinh

    mπx

sin

mπy











7.25 (cont (contin inued ued))

Thus Thus V m =

So that finally,

200( 1)m+1 mπb sinh(mπd/b sinh(mπd/b))

−

∞

200 ( 1)m+1 sinh sinh (mπx/b) mπx/b) mπy V ( V (x, y) = sin πb m=1 m sinh sinh (mπd/b) mπd/b) b

 

−

Now, with a square trough, set b = d = 1, and so 0 < x < 1 and 0 < y < 1. The potential potential becomes ∞ 200 ( 1)m+1 sinh sinh (mπx) mπx ) V ( V (x, y) = sin(mπy sin(mπy)) π m=1 m sinh sinh (mπ) mπ )

−

Now at the center of the trough, x = y = y = 0.5, and, using four terms, we have



sinh(π/2) 2) . 200 sinh(π/ V ( V (.5, .5) = π sinh(π sinh(π )

−

1 sinh(3π/ sinh(3π/2) 2) 1 sinh(5π/ sinh(5π/2) 2) + 3 sinh(3π sinh(3π ) 5 sinh(5π sinh(5π )

−



1 sinh(7π/ sinh(7π/2) 2) = 12 12..5 V 7 sinh(7π sinh(7π)

where additional terms do not affect the three-significant-figure answer. 7.26. If X X is a function of x and X  + (x ( x 1)X 1)X 2X = X = 0, assume a solution in the form of an infinite power series and determine numerical values for a2 to a8 if a0 = 1 and a and a 1 = 1: The series solution will be of the form:

−

− −

−

∞

X =



am xm













7.27. It is known that V that V = X Y is Y is a solution of Laplace’s equation, where X where X is is a function of x alone, x alone, and Y is Y is a function of y alone. Determine Determine which of the following potential potential function are also solutions of Laplace’s equation: a) V = V = 100X 100X : We know that 2 X Y = Y = 0, or

∇

∂ 2 ∂ 2 X Y + 2 X Y = 0 ∂x 2 ∂y

⇒





Y X + X Y = 0

= 100X 100X = ∇2X =  0 – No. b) V = 50 50X X Y : Y : Would have ∇2 V = 50∇2 X Y Y = 0 – Yes. c) V = 2X Y + x − 3y : ∇2 V = 2∇2 X Y + 0 − 0 = 0 −

⇒

X  = X

−

Y  = α 2 Y



Therefore,

Yes

d) V = xX Y : Y : 2

2

xXY Y ∂ xX xXY Y + = ∇2V = ∂ xX

∂

[X [ X Y + xX  Y ] Y ] +

∂

[xX [ xXY Y  ]











7.28b. Show that P = Aρ n + Bρ −n satisfies the ρ the ρ equation: From part a, the radial equation is: ρ2

d2 P dP + ρ dρ2 dρ

− n2P = 0

Substituting Aρ Substituting Aρ n , we find ρ2 n(n

−2

1)ρn − 1)ρ

+ ρnρn−1

− n2ρn = n2ρn − nρn + nρn − n2ρn = 0

Substituting B Substituting B ρ−n : ρ2 n(n + 1)ρ 1)ρ−(n+2) So it works.

−(n+1)

− ρnρ

− n2 ρ

−n

= n 2 ρ−n + nρ−n

− nρ n − n2 ρ −

−n

=0











7.29.. Referr 7.29 Referring ing to Chapter Chapter 6, Fig. Fig. 6.14 6.14,, let the inner inner conductor conductor of the transmiss transmission ion line be at a potent potential ial of 100V 100V,, while while the outer outer is at zero potential. potential. Constr Construct uct a grid, grid, 0. 0.5a on a side, and use iteration to find V V at a point that is a units above the upper right corner of the inner conductor. Work to the nearest volt: The drawing is shown below, and we identify the requested voltage as 38 V.











7.30. Use the iteration iteration method to estimate the potentials potentials at points x points x and and y y in the triangular trough of Fig. 7.14. Work only to the nearest volt: The result is shown below. The mirror image of the values shown occur at the points on the other side of the line of symmetry (dashed line). Note that V x = 78V and and V y = 26V.











7.32.. Using 7.32 Using the grid indicate indicated d in Fig. 7.16 7.16,, work work to the nearest nearest volt volt to estimate estimate the potential potential at point A point A:: The voltages at the grid points are shown below, where V where V A is found to be 19 V. Half the figure is drawn since mirror images of all values occur across the line of symmetry (dashed line).













7.33. Conductors having boundaries that are curved or skewed usually do not permit every grid point to coincide with the actual boundary. boundary. Figure 6.16a illustrates illustrates the situation where the potential at V 0 is to be estimated in terms of V 1 , V 2 , V 3 , and V 4 , and the unequal distances h1 , h2 , h3 , and h4 . a) Show Show that











7.34. Consider Consider the configuration configuration of conductors conductors and and potentials potentials shown in Fig. Fig. 7.18. Using the method method described in Problem 7.33, write an expression for V x (not V 0 ): The result result is shown shown below, below, where V where V x = 70V.











7.35c. Use a computer to obtain values values for a 0.25 cm grid. Work to the nearest 0.1 V: Values Values for the left left half of the configura configuration tion are shown in the table below. below. Values alues along the vertica verticall line line of symmetry are included, and the original grid values are underlined.

Teoria Electromagnetica - Hayt - Solucionario CAPITULO 7 7ma Edicion

Recommend Documents