Capacitance

CAPACITANCE

Contents Topic

Page No.

Theory

01 - 02

Exercise - 1

03 - 21

Exercise - 2

22 - 29

Exercise - 3

30 - 33

Exercise - 4

34 - 36

Answer Key

37 - 39

Syllabus Capacitance ; Parallel plate capacitor with and without dielectrics ; Capacitors in series and parallel ; Energy stored in a capacitor.

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1.

CAPACITANCE OF AN ISOLATED SPHERICAL CONDUCTOR : C = 4 ! # 0# r R i n a m e d i u m C = 4! #0 R in air * *

2.

This This sphe sphere re is at at in infini finite te dist dista ance nce fro from m all all the the con condu duct ctor ors s. T he capacitance C = 4! #0 R exists between the surface of the sphere & earth earth .

SPHERICAL CAPACITOR : It consists consists of two concentri concentric c spherical spherical shells shells as shown shown in figure. figure. Here capacitan capacitance ce of region region be twee tw ee n th e two shells is C1 and that outside the shell is C 2. We have C1 =

4! #0 ab

and C2 = 4! #0 b

b$a

Depending on connection, it may have different combinations of C1 and C2.

3.

PARALLEL PLATE CAPACITOR : (i)

UNIFORM DI-ELECTRIC M EDIUM : If two parallel plates each of area A & separated by a distance d are charged with equal & opposite charge Q, then the system is called a parallel plate capacitor & its capacitance is given by b y, C=

#0 #r A

;

MEDIUM PARTLY AIR :

C=

#0 A

with air as medium d d This result is only valid when the electric field between plates of capacitor is constant.

(i i)

in a m e diu m

C=

#0 A % ( d $ ' t $ #t * & r )

When a di-electric slab of thickness t & relative permittivity #r is introduced between the plates plates of an air capacitor, then the distance between

% t ( * irrespective of the position of & #r )

the plates is effectively reduced by ' t $ the di-electric slab .

(i ii )

4.

COMPOSITE M EDIUM :

C=

#0 A t1

t3

t2

#r1 +#r 2 +#r 3

CYLINDRICAL CAPACITOR : It consist of two co-axial cylinders of radii a & b, the outer conductor is earthed . The di-electric constant of the medium filled in the space between the cylinder is

#r . The capacitance per unit unit length is C =

5.

2!#0# r Farad

,-

!n b a

m

.

CONCEPT OF VARIATION OF PARAMETERS: As capacitance of a parallel plate capacitor isC =

#0

kA d

, if either of k, A or d varies in the region r egion between the

plates, we choose a small dc in between the plates and for total capacitance of system. If all dC's are in series

1 CT

/.

dx

#0 k ( x ) A (x )

,

If al all dC's are in parallel CT =

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. dC

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1.

CAPACITANCE OF AN ISOLATED SPHERICAL CONDUCTOR : C = 4 ! # 0# r R i n a m e d i u m C = 4! #0 R in air * *

2.

This This sphe sphere re is at at in infini finite te dist dista ance nce fro from m all all the the con condu duct ctor ors s. T he capacitance C = 4! #0 R exists between the surface of the sphere & earth earth .

SPHERICAL CAPACITOR : It consists consists of two concentri concentric c spherical spherical shells shells as shown shown in figure. figure. Here capacitan capacitance ce of region region be twee tw ee n th e two shells is C1 and that outside the shell is C 2. We have C1 =

4! #0 ab

and C2 = 4! #0 b

b$a

Depending on connection, it may have different combinations of C1 and C2.

3.

PARALLEL PLATE CAPACITOR : (i)

UNIFORM DI-ELECTRIC M EDIUM : If two parallel plates each of area A & separated by a distance d are charged with equal & opposite charge Q, then the system is called a parallel plate capacitor & its capacitance is given by b y, C=

#0 #r A

;

MEDIUM PARTLY AIR :

C=

#0 A

with air as medium d d This result is only valid when the electric field between plates of capacitor is constant.

(i i)

in a m e diu m

C=

#0 A % ( d $ ' t $ #t * & r )

When a di-electric slab of thickness t & relative permittivity #r is introduced between the plates plates of an air capacitor, then the distance between

% t ( * irrespective of the position of & #r )

the plates is effectively reduced by ' t $ the di-electric slab .

(i ii )

4.

COMPOSITE M EDIUM :

C=

#0 A t1

t3

t2

#r1 +#r 2 +#r 3

CYLINDRICAL CAPACITOR : It consist of two co-axial cylinders of radii a & b, the outer conductor is earthed . The di-electric constant of the medium filled in the space between the cylinder is

#r . The capacitance per unit unit length is C =

5.

2!#0# r Farad

,-

!n b a

m

.

CONCEPT OF VARIATION OF PARAMETERS: As capacitance of a parallel plate capacitor isC =

#0

kA d

, if either of k, A or d varies in the region r egion between the

plates, we choose a small dc in between the plates and for total capacitance of system. If all dC's are in series

1 CT

/.

dx

#0 k ( x ) A (x )

,

If al all dC's are in parallel CT =

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. dC


6.

COMBINATION OF CAPACITORS : (i)

CAPACITORS IN S ERIES :

In this arrangement all the capacitors when unchar ged get the same charge Q but the potential difference across each will differ (if the capacitance are unequal).

1 C eq. (i i)

=

1 C1

+

1 C2

+

1 C3

+ ........ +

1

.

Cn

CAPACITORS IN PARALLEL : When one plate of each capacitor is connected to the positive termi termina nall of the the batt batter ery y & the the othe otherr plat plate e of

each each capacitor capacitor is

connected to the negative terminals of the battery, then the capacitors are said to be in parallel connection. The capaci capacitor tors s have have the same pot potent entia iall differe difference nce,,

V but but the the

charge on each one is different (if the capacitors are unequal). Ceq. = C 1 + C 2 + C 3 + ...... + Cn .

7.

ENERGY STORED IN A CHARGED CAPACITOR : Capacitance C, charge Q & potential difference V ; then energy stored is

1 Q2 U= CV = QV = . This energy is stored in the electrostatic field set up in the di-electric 2 2 2 C 1

2

1

medium between the conducting plates of the capacitor .

8.

HEAT EAT PRODUCED IN SWITCHING IN CAPACITIVE CIRCUIT Due to charge flow always some amount of heat is produced when a switch is closed in a circuit which can be obtained by energy conservation as – Heat = Work done by battery – Energy absorbed by capacitor.

9.

SHARING OF CHARGES : When two charged conductors of capacitance C1 & C2 at potential V1 & V2 respectively are connected by a conducting wire, the charge flows from higher potential conductor to lower potential conductor, until the potential of the two condensers becomes equal. The common potential (V) after sharing of charges; V=

net ch arg e net capa apaci tan ce

=

q1

q2

C1 + C 2

=

C1V1 + C 2 V2 C1 + C 2

.

charges after sharing q1 = C1V & q 2 = C2V. In this proces s energy is lost in the connecting wire as heat . This loss loss of energy is Uinitial $ Ureal =

C1 C 2 2 ,C1 + C 2 -

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(V 1 $ V 2)2 .

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PART - I : OBJECTIVE QUESTIONS * Marked Questions are having more than one correct option.

Section (A) : Definition of capacitance, Circuits with capacitor and use of KCL and KVL A-1.

In the figure initial status of capacitor and their connection is shown. Which of the following is incorrect about this circuit :

(A) Final charge on each capacitor will be zero (B) Final total electrical energy of the capacitors will be zero (C) Total charge flown fr om A to D is 30µC (D) Total charge flown fro m A to D is – 30µC A-2.

One plate of a capacitor is connected with a spring as shown s hown in figure. Area of both the plates is A. In steady state separation between the plates is 0.8 d (spring was unstretched and the distance between the plates was d when the capacitor was uncharged). The force constant of the spring is approximately-

(A)

A-3.

4å o AE 2 d

3

(B)

2.50 0 AE d2

(C)

6å o E 2 Ad

3

(D)

å

o AE 3

3

2d

A circuit has a section AB AB shown in the figure. The emf em f of the source s ource equals 0 = 10V, 10V, the capacitor capacitances are equal to C1 = 1.0 1F and C2 = 2.0 1F, the potential difference 2A $ 2B = 5.0V. The voltage across each capacitor are

(A) V1 =

5V 10 V , V2 = 3 3

(B) V1 =

10 V 10 V , V2 = 3 3

(C) V1 =

10 V 5V , V2 = 3 3

(D) V1 =

5V 5V , V2 = 3 3

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A-4.

A 1 µF capacitor is connected in the circuit shown below. The e.m.f. of the cell is 3 volts and internal resistance is 0.5 ohms. The resistors R 1 and R2 have values 4 ohms and 1 ohm respectively. The charge on the capacitor in steady state must be :

(A) 2 1 C A-5.

(C) 1.33 1 C

(D) zero

A capacitor of capacitance C is charged to a potential difference V from a cell and then disconnected from it. A charge +Q is now given to its positive plate. The potential difference across the capacitor is now : (A) V

A-6.

(B) 1 1 C

(B) V +

Q C

(C) V +

Q 2C

(D) V –

Q 2C

In the given arrangement of capacitors 6µC charge is added to point A, find the charge on upper capacitor:

3C

2C

(A) 3 µC A-7.

(B) 1 µC

C

(C) 2 µC

(D) 6 µC

In the circuit shown, switch S2 is closed first and is k ept closed for a long time. Now S1 is closed. Just after that instant the current through S1 is:

(A)

0 R1

towards right

(B)

(C) zero

A-8.

A

(D)

0 R1 2

towards left

0

R1

Initially switch S is connected to position 1 for a long time. The net amount of heat generated in the circuit after it is shifted to position 2 is

(A)

C ,01 + 0 2 -0 2 2

(B) C,01 + 0 2 - 0 2

(C)

C ,01 + 0 2 -2 2

(D) C ,01 + 0 2 -2


A-9.

A parallel plate capacitor having capacitance C is connected to a source of constant emf E. Which of the following statements is correct ? (A) Net charge supplied by the battery to the capacitor is equal to CE. (B) Net charge supplied by the battery to the capacitor is equal to 2 CE. (C) Net charge supplied by the battery to capacitor is equal to zero. (D) None of these.

A-10.* Two capacitors of capacitances 1µF and 3µF are charged to the same voltages 5V. They are connected in parallel with oppositely charged plates connected together. Then : (A) Final common voltage will be 5 V (B) Final common voltage will be 2.5 V (C) Heat produced in the circuit will be zero .

(D) Heat produced in the circuit will be 37.51J.

A-11.

A capacitor of capacitance C is charged to a potential difference V0. The charging battery is disconnected and the capacitor is connected to a capacitor of unknown capacitance Cx. The P.D. across the combination is V. The value of Cx should be : (A)

A-12.

(C)

A-14.

C(V $ V0 ) V

0 0 A1

CV (C) V 0

(B)

d

00 (A1 + A 2 )

(D)

2d

(D)

CV0 V

00 A 2 d

00 A1A 2 d

In the circuit shown, the energy stored in 11F capacitor is (A) 40 1J

(B) 64 1J

(C) 32 1J

(D) none

If charge on left plane of the 51F capacitor in the circuit segment shown in the figure is –201C, the charge on the right plate of 31F capacitor is :

(A) +8.57 1C A-15.

(B)

The capacitance of capacitor of plate areas A1 and A2 (A1 < A2) at a distance d is :

(A)

A-13.

C( V0 $ V ) V

(B) –8.57 1C

(C) +11.42 1C

(D) –11.42 1C

The plates S and T of an uncharged parallel plate capacitor are connected across a battery. The battery is then disconnected and the charged plates are now connected in a system as shown in the figure. The system shown is in equilibrium. All the strings are insulating and m assless. The magnitude of charge on one of the capacitor plates is: [Area of plates = A] (A)

2mgA #0

(C) mgA #0

(B)

(D)

4mgA #0 k 2mgA #0 k


A-16.

A parallel plate capacitor has an electric field of 105V/m between the plates. If the charge on the capacitor plate is 11C, then the force on each capacitor plate is (A) 0.1Nt

A-17.

(B) 0.05Nt

(C) 0.02Nt

(D) 0.01Nt

A capacitor is connected to a battery. The force of attraction between the plates when the separation between them is halved (A) remains the same

(B) becomes eight times (C) becomes four times

(D) becomes two times

Section (B) : Combination of capacitors B-1.

In the figure shown the equivalent capacitance between 'A' and 'B' is :

(A) 3.75 F B-2.

(C) 21 F

(B) (N - 1) V

(D) N2V

(C) N V

10 identical capacitors are connected as shown. The capacitance of each capacitor is 30 1F. Find the equivalent capacitance between A and B.

A (A) 30 1F B-4.

(D) 16 F

N identical capacitors are connected in parallel to a potential difference V. These capacitors are then reconnected in series such that positively charged plate of one capacitor is connected to negatively charged plate of the other, their charges being left undisturbed. The potential difference obtained is :

(A) zero B-3.

(B) 2 F

(B) 60 1F

B (C) 120 1F

(D) 3

The equivalent capacitance between point A and B is:

(A) 1 1F

(B)

21F

(C)

31F

(D) 41F


B-5.

Consider the circuit shown in the figure. Charge stored in capacitor of capacitance

C

2

is

C C 2

C

V (A) CV

B-6.

(B)

C V 4

C (C)

C V 2

(D) 2CV

Fig (a) Shows two capacitors connected in series and joined to a battery. The graph in fig (b) shows the variation in potential as one moves from left to right on the branch containing the capacitors if -

(A) C1 > C2

(B) C1 = C2

(C) C1 < C2

(D) The information is not sufficient to decide the relation between C1 and C2 B-7.

In the circuit shown, a potential difference of 60V is applied across AB. The potential difference between the point M and N is

(A) 10 V B-8.

B-9.

(B) 15 V

(C) 20 V

(D) 30 V

In the circuit shown in figure, the ratio of charges on 51F and 41F capacitor is : (A) 4/5

(B) 3/5

(C) 3/8

(D) 1/2

On each side of a polygon of n sides a capacitor of capacitance C is placed as shown in figure. Equivalent capacitance across A and B is

C

C A

(A)

(n $ 1)C n

(B)

nC n $ 1

C

B

(C) (n – 1)C

(D) nC


B-10.

From a supply of identical capacitors rated 8 1F, 250 V, the minimum number of capacitors required to form a composite 16 1F, 1000 V is : (A) 2

B-11.

B-13.

B-14.

30 0 A 2d

V2

50 0 A 12d

V2

(C)

00A 2d

(B) 15 V

(C) 20 V

(D) 30 V

V2

(D)

00 A d

V2

In the circuit shown in figure, the ratio of charges on 51F and 41F capacitor is : (A) 4/5

(B) 3/5

(C) 3/8

(D) 1/2

The minimum number of capacitors each of 3 1F required to make a circuit with an equivalent capacitance 2.25 1F is (B) 4

(C) 5

(D) 6

From a supply of identical capacitors rated 8 1F, 250 V, the minimum number of capacitors required to form a composite 16 1F, 1000 V is : (B) 4

(C) 16

(D) 32

What is the equivalent capacitance of the system of capacitors between A & B

(A) B-17.

(B)

(A) 10 V

(A) 2 B-16.

(D) 32

In the circuit shown, a potential difference of 60V is applied across AB. The potential difference between the point M and N is

(A) 3 B-15.

(C) 16

A, B, C, D, E, F are c onducting plates each of area A and any two consecutive plates separated by a distance d. The net energy stored in the system after the switch S is closed is:

(A) B-12.

(B) 4

7 6

C

(B) 1.6 C

(C) C

(D) None

Two capacitor having capacitances 8 1F and 16 1F have breaking voltages 20 V and 80 V. They are combined in series. The maximum charge they can store individually in the combination is (A) 160 1C

(B) 200 1C

(C) 1280 1C

(D) none of these


B-18.

Three plates A, B and C each of area 0.1 m2 are separated by 0.885 mm from each other as shown in the figure. A 10 V battery is used to charge the system. The energy stored in the system is

(A) 1 1J B-19.

(C)

3

0 0 AE

4

d

40 0 AE d

(C) 10 2 1J –

(D) 10 3 1J –

(D)

2

0 0 AE

3

d

0 0 AE 2d

(B)

6! #0 ln 2

(C)

! #0 2ln 2

(D) None

Four metallic plates arearranged as shown in the figure. If the distance between each plate then capacitance of the given system between points A and B is (Given d << A) (A)

(C)

00A

(B)

d

30 0 A

(D)

d

20 0 A d

40 0 A d

Find the equivalent capacitance across A & B

(A)

28 3

1f

(C) 15 1F B-23.

(B)

#

0 (A) 3 ln 2

B-22.

1J

Three long concentric conducting cylindrical shells have radii R, 2R and 2 2 R. Inner and outer shells are connected to each other. The capacitance across middle and inner shells per unit length is:

1

B-21.

1

Five conducting parallel plates having area A and separation between them d, are placed as shown in the figure. Plate number 2 and 4 are connected wire and between point A and B, a cell of emf E is connected. The charge flown through the cell is

(A)

B-20.

–

(B) 10

(B)

15 2

1F

(D) none

The diagram shows four capacitors with capacitances and break down voltages as mentioned. What should be the maximum value of the external emf source such that no capacitor breaks down? [Hint: First of all find out the break down voltages of each branch. After that compare them.]

B-24.

(A ) 2.5 kV

(B) 10 / 3kV

(C) 3 kV

(D) 1 kV

Three capacitors 2 1F, 3 1F and 5 1F can withstand voltages to 3V, 2V and 1V respectively. Their series combination can withstand a maximum voltage equal to (A) 5 Volts (B) (31/6) Volts (C) (26/5) Volts (D) None ETOOSINDIA.COM CAPACITANCE (JEE MAINS) # 9 India's No.1 Online Coaching for JEE Main & Advanced 3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Section (C) : Equation of charging and discharging C-1.

In the circuit shown the capacitor is initially uncharged. The charge passed through an imaginary circular loop parallel to the plates (also circular) and having the area equal to half of the area of the plates, in one time constant is:

(A) 0.63

C-2.

0C

(B) 0.37

(C)

0 C 2

(D) zero

An uncharged capacitor is connected in series with a resistor and a battery. The charging of the capacitor starts at t = 0. The rate at which energy in capacitor is stored : (A) first increases then decreases (B) first decreases then increases (C) remains constant

C-3.*

0C

(D) continuously decreases

The figure shows, a graph of the current in a discharging circuit of a capacitor through a resistor of resistance 10 4.

(A) The initial potential difference across the capacitor is 100 volt. 1 (B) The capacitance of the capacitor is 10 !n 2 F.. (C) The total heat p roduced in the circuit will be

500 joules. !n2

(D) The thermal power in the resistor will decrease with a time constant

C-4.

1 second. 2 !n 2

Three identical capacitors are given a charge Q each and they are then allowed to discharge through resistance R1, R 2 and R3 separately. Their charges, as a function of time are shown in the graph below. The smallest of the three resistances is

(A) R3

(B) R2

(C) R1

(D) cannot be predicted


C-5.

A graph between current & time during charging of a capacitor by a battery in series with a resistor is shown. The graphs are drawn for two circuits. R1, R2, C1, C2 and V1V2 are the values of resistance, capacitance and EMF of the cell in the two circuits. If only two parameters (out of resistance, capacitance, EMF) are different in the two circuits. What is /are the correct option(s)

(A) V1 = V2; R1 > R2, C1> C2 (C) V1 < V2, R1< R2, C1 = C2

(B) V1 > V2, R1 > R2 ; C1 = C2 (D) V1 < V2, C1< C2, R1 = R2

C-6.*

Capacitor C1 of capacitance 1 mircofarad and capacitor C2 of capacitance 2 microfarad are separately charged fully by a common battery. The two capacitors are then separately allowed to discharge through equal resistors, at time t = 0 : (A) the current in each of two discharging circuits at t = 0 are equal and non-zero. (B) The current in the two discharging circuits at t = 0 are equal (C) The currents in the two discharging circuits at t = 0 are unequal. (D) Capacitor C1 loses 50% of its initial charge sooner than C2 loses 50% of its initial charge.

C-7.

A capacitor of capacitance C is charged to a potential difference V from a cell and then disconnected from it. A charge +Q is now given to its positive plate. The potential difference across the capacitor is now (A) V

Q

(B) V +

C

(C) V +

Q 2C

(D) V –

Q C

, if V < CV

C-8.

A capacitor of capacitance C is initially charged to a potential difference of V volt. Now it is connected to a battery of 2V Volt with opposite polarity. The ratio of heat generated to the final energy stored in the capacitor will be (A) 1.75 (B) 2.25 (C) 2.5 (D) 1/2

C-9.

A conducting body 1 has some initial charge Q, and its capacitance is C. There are two other conducting bodies, 2 and 3, having capacitances : C2 = 2C and C3 5 3. Bodies 2 and 3 are initially uncharged. "Body 2 is touched with body 1. Then, body 2 is removed from body 1 and touched with body 3, and then removed." This process is repeated N times. Then, the charge on body 1 at the end must be (A) Q/3N

C-10.

C-11.

(B) Q/3N

–

1

(C) Q/N3

(D) None

A charged capacitor is allowed to discharge through a resistance 24 by closing the switch S at the instant t = 0. At time t = l n 2 1s, the reading of the ammeter falls half of its initial value. The resistance of the amm eter equal to (A) 0

(B) 24

(C) 3

(D) 2M4

A capacitor C = 100 1F is connected to three resistor each of resistance 1 k4 and a battery of emf 9V. The switch S has been closed for long time so as to charge the capacitor. When switch S is opened, the capacitor discharges with time constant

(A) 33 ms

(B) 5 ms

(C) 3.3 ms

(D) 50 ms

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C-12.

In the circuit shown in figure C1=2C2. Switch S is closed at time t=0. Let i1 and i2 be the currents flowing through C1 and C2 at any time t, then the ratio i1 / i2 (A) is constant (B) increases with increase in time t (C) decreases with increase in time t (D) first increases then decreases

C-13.

In the circuit shown, when the key k is pressed at time t = 0, which of the following statements about current I in the resistor AB is true (A) I = 2mA at all t (B) I oscillates between 1 mA and 2mA (C) I = 1 mA at all t (D) At t = 0, I = 2mA and with time it goes to 1 mA

C-14.

In the R –C circuit shown in the figure the total energy of 3.6 ×10 3 J is dissipated in the 10 4 resistor when the switch S is closed. The initial charge on the capacitor is –

(A) 60 1C

C-15.

(C) 60

2 1C

(D)

60 2

1C

A charged capacitor is allowed to discharge through a resistor by closing the key at the instant t =0. At the instant t = (ln 4) 1s, the reading of the ammeter falls half the initial value. The resistance of the ammeter is equal to

(A) 1 M4

C-16.

(B) 120 1C

(B) 14

(C ) 24

(D) 2M4

In the circuit shown, the cell is ideal, with emf = 15 V. Each resistance is of 34. The potential difference across the capacitor is

(A) zero

(B) 9 V

(C) 12 V

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(D) 15 V


Section (D) : Capacitor with dielectric D-1.*

A parallel plate air capacitor is conn ected to a battery. The quantities charge, voltage, electric f ield and energy associated with this capacitor are given by Q 0, V 0, E 0 and U 0 respectively. A dielectric slab is now introduced to fill the space between the plates with battery still in connection. The corresponding quantities now given by Q, V, E and U are related to the previous one as (A) Q > Q 0 (B) V > V 0 (C) E > E0 (D) U > U0

D-2.

A parallel plate capacitor (without dielectric) is char ged and disconnected from a battery. Now a dielectric is inserted between the plates. The electric force on a plate of the capacitor will: (A) decrease (B) increase (C) remain same (D) depends on the width of the dielectric.

D-3.

Two parallel plate capacitors of capacitances C and 2C are connected in parallel and charged to a potential differ ence V by a battery. The battery is then disconnect ed and the space between the plates of capacitor C is completely filled with a material of dielectric constant K. The potential difference across the capacitors now becomes. (A)

D-4.

V

(B)

K +1

2V K+2

(C)

3V K+2

(D)

3V K+3

Two conductors of thickness d are inserted inside a parallel capacitor of thickness 3d and capacitance C 0. The capacitance of new arrangement is :

d

d

3d (A) C 0 D-5.*

(C) 3C 0

(D)

C 0

3 A parallel plate capacitor of plate area A and plate seperation d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E and W denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted) and the work done on the system, in question, in the process of inserting the slab, then

(A) Q =

D-6.

(B) 2C 0

0 0 AV d

(B) Q =

0 0 KAV

V (C) E =

d

(D) W = –

Kd

0 0 AV 2 % 1 ( '1 $ * 2 d & K )

A parallel plate capacitor made from two square plates of side a and separation b (<< a ) is charged by a battery of emf V . After disconnecting the battery, a conductor of thickness slightly less than b is inserted as shown in figure. The potential energy of the system is x b a

0 0 a 3 V 2 (A) 2b (a $ x )

0 0 a 2 V 2 (B) 2(a $ x )

(C)

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0 0a 2 2b

(a $ x ) V 2

(D)

0 0a 3 (a $ x ) 2

2b

V 2



D-7.

A parallel plate capacitor without any dielectric has capacitance C0. A dielectric slab is made up of two dielectric slabs of dielectric constants K and 2K and is of same dimensions as that of capacitor plates and both the parts are of equal dimensions arranged serially as shown. If this dielectric slab is introduced (dielectric K enters first) in between the plates at constant speed, then variation of capacitance with time will be best represented by :

(A)

D-8.

D-10.

(C)

(D)

Two parallel plate air filled capacitors each of capacitance C, are joined in series to a battery of emf V. The space between the plates of one of the capacitors is then completely filled up with a uniform dielectric having dielectric constant K. The quantity of charge which flows through the battery is (A)

D-9.

(B)

CV 2

% K – 1 ( '' ** & K + 1 )

(B)

CV 2

% K + 1 ( '' ** & K – 1 )

% K – 1 ( ** & K + 1 )

(C) CV ''

% K + 1 ( ** & K – 1 )

(D) CV ''

Two identical capacitor C 1 and C2 are connected in ser ies with a battery. They are fully charged. Now a dielectric slab is inserted between the plates of C 2. The potential difference across C1 will :

(A) increase

(B) decrease

(C) remain same

(D) depend on internal resistance of the cell

A parallel plate capacitor has two layers of dielectric as shown in figure. This capacitor is connected across a battery. The graph which shows the variation of electric field (E) and distance (x) from left plate.

(A)

(B)

(C)

(D)


D-11.

In the figure shown a parallel plate capacitor has a dielectric of width d/2 and dielectric constant K = 2. The other dimensions of the dielectric are same as that of the plates. The plates P1 and P2 of the capacitor have area 'A' each. The energy of the capacitor is :

(A)

D-12.

#0 AV 2 3d

d

3 #0 AV 2 (C) 2 d

(D)

2 #0 AV 2 3d

In the figure a capacitor of capacitance 2µF is connected to a cell of emf 20 volt. The plates of the capacitor are drawn apart slowly to double the distance between them. The work done by the external agent on the plates is :

(A) – 200 µJ D-13.

(B)

2 #0 AV 2

(B) 200 µJ

(C) 400 µJ

(D) – 400 µJ

A dielectric slab of relative permittivity 0r and thickness t is inserted into the capacitor. Then,

(A) the capacitance of the system increases by

00 A %

t ' 1 –

&

(B)

1(

0r *)

qfree 0 / r qbound 0r – 1

(C) the fraction change in the energy stored is 0r –1

%

(D) the plates are m oved apart by a relative distance t ' 1 –

&

1(

* 0r ) to recover the original energy stored.


D-14.

A parallel plate capacitor has two layers of dielectric as shown in figure. This capacitor is connected across a battery. The graph which shows the variation of electric field (E) and distance (x) from left plate.

(A)

(B)

(C)

(D)

D-15.

A capacitor stores 601C charge when connected across a battery. When the gap between the plates is filled with a dielectric , a charge of 1201C flows through the battery. The dielectric constant of the material inserted is : (A) 1 (B) 2 (C) 3 (D) none

D-16.

In the adjoining figure, capacitor (1) and (2) have a capacitance ‘C’ each. When the dielectric of dielectric consatnt K is inserted between the plates of one of the capacitor, the total charge flowing through battery is (A)

KCE K +1

from B to C

(B)

( K $ 1)CE (C)

D-17.

2(K + 1)

KCE K +1

from C to B

(K $ 1)CE from B to C

(D)

2( K + 1)

from C to B

The distance between plates of a parallel plate capacitor is 5d. Let the positively charged plate is at x=0 and negatively charged plate is at x=5d. Two slabs one of conductor and other of a dielectric of equal thickness d are inserted between the plates as shown in figure. Potential versus distance graph will look like :

(A)

(B)

(C)

(D)


D-18.

The distance between the plates of a charged parallel plate capacitor is 5 cm and electric field inside the –

plates is 200 Vcm 1. An uncharged metal bar of width 2 cm is fully immersed into the capacitor. The length of the metal bar is same as that of plate of capacitor. The voltage across capacitor after the immersion of the bar is : (A) zero

D-19.

(B) 400 V

Condenser A has a capacity of 15 condenser B has a capacity 1

(C) 600 V

(D) 100 V

1F when it is filled with a medium of dielectric constant 15. Another

1F with air between the plates. Both are charged separately by a battery of

100V . After charging, both are connected in parallel without the battery and the dielectric material being removed. The common potential now is (A) 400V D-20.

(B) 800V

(C) 1200V

(D) 1600V

Two identical capacitors 1 and 2 are connected in series to a battery as shown in figure. Capacitor 2 contains a dielectric slab of dielectric constant k as shown. Q1 and Q2 are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are Q’1 and Q’2. Then

(A)

(C)

D-21.

Q16 Q1 Q62 Q2

k + 1

/

(B)

k

/

k + 1 (D)

2k

Q62 Q2 Q16 Q1

/

/

k + 1 2

k 2

Four identical plates 1, 2, 3 and 4 are placed parallel to each other at equal distance as shown in the figure. Plates 1 and 4 are joined together and the space between 2 and 3 is filled with a dielectric of dielectric constant k = 2. The capacitance of the system between 1 and 3 & 2 and 4 are C1 and C2 respectively. The

C1

ratio

(A)

C2

5 3

is :

(B) 1

(C)

3 5

(D)

5 7


PART - II : MISLLANEOUS QUESTIONS

1. COMPREHENSION : COMPREHENSION # 1 Capacitor C3 in the circuit is a variable capacitor (its capacitance can be varied). Graph is plotted between potential difference V1 (across capacitor C1) versus C 3. Electric potential V 1 approaches on asymptote of 10 V as C3 5 3 .

C1 V C2

1.

C1 The ratio of the capacitance C will be 2 (A)

2.

2 3

(B)

4 3

(C)

3 4

(D)

3 2

The value of C3 for which potential across C1 will become 8 V (A) 1.5 C 1

3.

C3

(B) 2.5 C 1

(C) 3.5 C 1

(D) 4.5 C 1

The ratio of energy stored in capacitor C1 to that of total energy when C3 5 3 (A) Zero

(B)

1 3

(C) 1

(D) Data insufficient

COMPREHENSION # 2 The potential energy of a charged conductor or a capacitor is stored in electric field. The energy per unit volume is called the energy density (u). Energy density in a dielectric media is given by u =

1 0 0KE 2 . This 2

2

relation shows that the energy stored per unit volume depends on E . If E is the electric field in a space of volume dV, then total stored energy in an electrostatic field is given by U =

throughout the volume, then total energy stored can be given by 4.

.

1 K0 0E 2 V . 2

The energy density in the electric field created by a point charge falls off with distance from the point charge as. (A)

5.

1 0 0K E 2 dV and if E is uniform 2

1 r

(B)

1 r

(C)

2

1 r

3

(D)

1 r4

A charges q1 is placed at the centre of a spherical conducting shell of radius R. Conducting shell has a total charge q2. Electrostatic potential energy of the system is q12 + 2q1q2 (A) 8 !00R

q22 + 2q1q2 (B) 8!0 0R

q12 + q1q2 (C) 4!00R

q22 + q1q2 (D) 4!00R


6.

Let ua and ud represent the energy density in air and in a dielectric respectively, for the same field in both. Let K = dielectric constant. Then (A) ua = ud (B) ua = Kud (C) ud = Kua (D) ua = (K –1)ud

7.

A parallel plate capacitor is connected to a battery. The plates are pulled apart with a uniform speed. If x is the separation between the plates, then the rate of change of electrostatic energy of the capacitor is proportional to(B) x2

(A) x

(C)

1 x

(D)

1 x2

COMPREHENSION # 3 The charge across the capacitor in two different RC circuits 1 and 2 are plotted as shown in figure.

8.*

Choose the correct statement(s) related to the two circuits. (A) Both the capacitors are charged to the same charge. (B) The emf's of cells in both the circuit are equal. (C) The emf's of the cells may be different. (D) The emf E1 is more than E2

9.

Identify the correct statement(s) related to the R1, R2, C1 and C2 of the two RC circuits. (A) R1 > R2 if E1 = E2

(B) C1 < C2 if E1 = E2

(C) R1C1 > R2C2

(D)

R1 R2

<

C2 C1

COMPREHENSION # 4 In the circuit as shown in figure the switch is closed at t = 0.

10.

At the instant of closing the switch (A) the battery delivers maximum current. (B) no current flows through C (C) Voltage drop across R2 is zero. (D) the current through the battery decreases with time finally becomes zero.

11.*

A long time after closing the switch (A) voltage drop across the capacitor is E. E

(B) current through the battery is R + R 1 2

% R 2 E ( * (C) energy stores in the capacitor is C '' 2 & R 1 + R 2 )* 1

2

(D) current through the capacitor becomes zero.

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2. ASSERTION AND REASON 12.

Statement-1 : The electrostatic force between the plates of a charged isolated capacitor decreases when dielectric fills whole space between plates. Statement-2 : The electric field between the plates of a charged isolated capacitance decreases when dielectric fills whole space between plates. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true.

13.

Statement-1

If temperature is increased, the dielectric constant of a polar dielectric decreases whereas that of a non-polar dielectric does not change significantly. Statement-2 The magnitude of dipole moment of individual polar molecule decreases significantly with increase in temperature. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true.

14.

Statement-1 : The heat produced by a resistor in any time t during the charging of a capacitor in a series circuit is half the energy stored in the capacitor by that time. Statement-2 : Current in the circuit is equal to the rate of increase in charge on the capacitor. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true.

3. MATCH THE COLUMN 15.

Consider the situation shown. The switch S is open for a long tim e and then closed. Then:

C

C E

S

Column I

Column II 2

CE

(A) Charge flown through battery when S is closed

(p)

(B) Work done by battery.

(q)

2 CE

2 2

CE

(C) Change in energy stored in capacitor.

(r)

4

(D) Heat developed in the system. CE 2 (t) 8 (s)

CE 4


16.

In the figure shown, area of each plate is A. Match the following : 1

2

3 2d

4

5

6

d

V

Column-I

Column-II

(A) (B)

Charge on plate 3 Charge on plate 5

(p) (q)

(C)

Potential difference between plates 2 and 3

(r)

(D)

Potential difference between plates 2 and 5

(s)

zero V

00 A 2d

00 A d

(t) 17.

V V

none of these

Arrangements of Parallel Plates

Capacitance

(A)

(p)

300 A 2d

(B)

(q)

300 A d

(C)

(r)

200 A 3d

(D)

(s)

200 A d

(t) none of these

4. TRUE & FALSE : 18.

If the charge on capacitor is constant, on increasing the separation (still keeping it very less change) between its plates the force between the plates does not change.

19.

If the potential difference between two plates of a capacitor is constant, on increasing the plate's separation the electric field remains constant.

5. FILL IN THE BLANKS : 20.

21.

Two parallel plate capacitors of capacitances C and 2C are connected in parallel and charged to a potential difference V. The battery is then disconnected and the region between the plates of capacitor C is completely filled with a material of dielectric constant K. T he potential difference across the capacitors now becomes .......... The capacity of a conductor ........................ when an earth connected uncharged conductor is brought near it.

22.

Capacity of parallel plate capacitor ........................ by decreasing the separation between two plates.

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MIXED OBJECTIVE * Marked Questions are having more than one correct option. 1.

The plates of a parallel plate capacitor are charged with surface densities 71 and 72 respectively. The electric field at points : (A) inside the region between the plates will be zero (B) above the upper plate & below the lower plate will be zero (C) every where in the space will be zero (D) inside the region between the plates will be uniform & non-zero

2.

Two large conducting plates A and B have charges Q1 and Q2 on them. The charges on the sides 1, 2, 3, and 4 respectively are :

3.

(A) q1 = q4 =

Q1 + Q 2 Q1 $ Q 2 and q2 = – q3 = 2 2

(B) q1 = q3 =

(C) q2 = q3 =

Q1 + Q2 Q1 $ Q 2 and q1 = q4 = 2 2

(D) q1 = q2 = q3 = q4 =

(B) X

(C) 1/X

(D) 1/X2

Two metal spheres of radii a and b are connected by a thin wire. Their separation is large compared with their dimensions. The capacitance of this system is : (A) 4!#0ab

5.*

Q1 + Q 2 2

A parallel plate capacitor is connected to a battery. The plates are pulled apart with a uniform speed. If X is the separation between the plates, then the rate of change of the electrostatic energy of the capacitor is proportional to : (A) X2

4.

Q1 + Q 2 Q1 $ Q2 and q2 = q4 = 2 2

(B) 2!#0(a + b)

(C) 4!#0(a + b)

(D) 4!#0(a2 + b2)/2

An uncharged capacitor having capacitance C is connected across a battery of emf V. Now the capacitor is disconnected and then reconnected across the same battery but with reversed polarity. Then : (A) after reconnection, thermal energy produced in the circuit will be equal to 2CV2. (B) after reconnection, thermal energy produced in the circuit will be equal to two-third of the total energy supplied by the battery. (C) after reconnection, no energy is supplied by the battery. (D) after reconnection, whole of the energy supplied by the battery is converted into heat.


6.

A parallel plate capacitor is filled with a uniform dielectric. Maximum charge that can be given to it, does not depend upon : (A) dielectric constant of the dielectric. (B) dielectric strength of the dielectric. (C) separation between the plates. (D) area of the plates.

7.

In the given figure, a capacitor of non-parallel plates is shown. The plates of capacitor are connected by a cell of emf V0. If 7 denotes surface charge density and E denotes electric field. Then :

B A

V0

(A)

8.

7A

>

(B) EF > ED

F

(C) EF = ED

(D)

7A

=

7B

In the circuit, capacitor is initially uncharged. The equivalent resistance will be (in steady – state) :

(A) 1 4 9.

7B

D

(B) 3 4

(C) 4 4

(D) 5 4

In the circuit shown the cells are ideal & of equal e.m.f. , the capacitance of the capacitor is C & the resistance of the resistor is R . X is first joined to Y and then to Z . After a long time the total heat produced in the resistor will be :

(A) equal to the energy finally stored in the capacitor (B) half of the energy finally stored in the capacitor (C) twice the energy finally stored in the capacitor (D) 4 times the energy finally stored in the capacitor . 10.*

In the circuit shown, all the capacitors are initially uncharged. When s witch S is closed, a total charge of 121C passes through point A and a charge of 81C passes through point B.

A S

C1

C2=31 F

9V

(A) Value of capacitance of C1 is 2 1 F (C) Value of capacitance of C3 is 21 F

B

C3

C4=41 F

(B) Value of capacitance of C1 is 4 1 F (D) Value of capacitance of C3 is 61 F


11.

An ideal cell is connected across a capacitor as shown in figure. T he initial separation between the plates of a parallel plate ca pacitor is d. The lower plate is p ulled down with a uniform velocity v. Neglect the resistance of the circuit. Then the variation of charge on capacitor with time is given by

C

E v

q

q

(A)

q

(B)

(C)

t 12.*

q (D)

t

t

t

A parallel plate capacitor of capacitance 'C' has charges on its plates initially as shown in the figure. Now at t = 0, the switch 'S' is closed. Select the correct alternative(s) for this circuit diagram.

A B

S

t=0 -20c 0c

0

(A) In steady state the charges on the outer sur faces of plates 'A' and ' B' will be same in magnitude and sign. (B) In steady state the charges on the outer sur faces of plates 'A' and ' B' will be same in magnitude and opposite in sign. (C) In steady state the charges on the inner s urfaces of the plate s 'A' and 'B' will be same in magnitude and opposite in sign. 5 0 2C (D) The work done by the cell by the time steady state is reached is . 2 13.

An isolated parallel plate capacitor of capacitance C has four surfaces with charges Q1, Q2, Q3 and Q4 as shown in figure. T he potential difference between the plates is

Q1

Q3

Q2

Q4

(A)

Q1 + Q 2 + Q 3 + Q 4 2C

(B)

Q 2 + Q 3 2C

(C)

Q 2 $ Q3 2C

(D)

Q1 + Q 4 2C


14.

15.

Two metallic spheres of radii a and b are separated by a distance d as shown in figure. The capacity of the system is :

(A) 4!#0 /(1/a + 1/b – 2/d)

(B) 2!#0 /(1/a – 1/b + 1/d)

(C) 4!#0 /(1/a + 1/b – 1/d)

(D) 4!#0(a + b)

A capacitor of capacity C is charged to a steady potential difference V and connected in series with an open key and a pure resistor 'R'. At time t = 0, the key is closed. If I = current at time t, a plot of log I against 't' is as shown in (1) in the graph. Later one of the parameters i.e. V, R or C is changed keeping the other two constant, and the graph (2) is recorded. Then

(A) C is reduced 16.

(C) R is reduced

(D) R is increased

The distance between plates of a parallel plate capacitor is 5d. Let the positively charged plate is at x=0 and negatively charged plate is at x=5d. Two slabs one of conductor and other of a dielectric of equal thickness d are inserted between the plates as shown in figure. Potential versus distance graph will look like :

(A)

17.

(B) C is increased

(B)

(C)

(D)

A capacitor is charged fully using a cell. With the cell connected, the capacitor plates are slowly pulled apart so that new capacitance becomes half of the original capacitance. Let the work done by pulling agent be w (A) Energy absorbed by the cell will be less than w (B) Energy absorbed by the cell will be more than w (C) Energy stored in the capacitor will increase by w (D) There will be heat loss in this process.

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18.*

A parallel plate capacitor of area A and separation d is charged to potential difference V and removed from the charging source. A dielectric slab of constant K = 2, thickness d and area

A is inserted, as 2

shown in the figure. Let 71 be free charge density at the conductor-dielectric surface and 72 be the charge density at the conductor-vacuum surface.

A 71

d

72

K –

–

71

72

(A) The electric field have the sam e value inside the dielectric as in the free space between the plates. (B) The ratio

71 2 is equal to . 72 1

(C) The new capacitance is

3 #0 A 2d

(D) The new potential difference is 19.

2 V 3

A and C are concentric conducting spherical shells of radius a and c respectively. A is surrounded by a concentric dielectric medium of inner radius a, outer radius b and dielectric constant k. If sphere A is given a charges Q, the potential at the outer surface of the dielectric is.

(A)

Q 4!00kb

(B)

Q % 1 1 ( '' + ** (C) 4!00 & a k(b $ a) ) 4!00 b Q

(D) None of these

20.

If n drops, each of capacitanc e C and charged to a potential V, coalesce to form a big drop, th e ratio of the energy stored in the big drop to that in each small drop will be (B) n4/3 : 1 (A) n : 1 (C) n5/3 : 1 (D) n2 : 1

21.

Figure shows a part of network of a capacitor and resistors. The potential indicated at A, B and C are with respect to the ground. The charge on the capacitor in steady state is

A +4V B +6V

24

44

84

11F

10V

44 C +8V (A) 4 1C

(B) 6 1C

(C) 10 1C

(D) 16 1C


22.

A student charges a capacitor in such a manner that it stores energy of 1 J. Now he wants to increase the potential energy to 4 J. He should : (A) quadruple the potential difference across the capacitor without changing the carge (B) double the potential difference across the capacitor without changing the charge (C) double both the potential difference and charge (D) double the charge without changing the potential difference

23.

The capacitance (C) for an isolated conducting sphere of radius (a) is given by 4 !00a. If the sphere is enclosed with an earthed concentric sphere. The ratio of the radii of the spheres being

n ( n $ 1)

then the

capacitance of such a sphere will be increased by a factor

n (A) n

(B)

(n $ 1)

(C)

( n $ 1)

(D) a . n

n

24.

A parallel plate capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with the capacitor are given by Q 0, V 0, E 0 and U0 respectively. A dielectric slab is introduced between plates of capacitor but battery is still in connection. The corresponding quantities now given by Q, V, E and U related to previous ones are (A) Q > Q0 (B) V > V0 (C) E > E0 (D) U < U0

25.

In the transient circuit shown the time constant of the circuit is : (A)

(C)

26.

5 3 7 4

RC

(B)

RC

(D)

5 2 7 3

RC

RC

Find heat produced on closing the switch S (A) 0.0002 J

(B) 0.0005 J

(C) 0.00075

(D) zero

Multiple Choice Questions : 27.

Two capacitors of 2 1F and 3 1F are charged to 150 volt and 120 volt respectively. The plates of capacitor are connected as shown in the figure. A discharged capacitor of capacity 1.5 1F falls to the free ends of the wire. Then (A) charge on the 1.5 1F capacitors is 180 1C (B) charge on the 21F capacitor is 120 1C (C) positive charge flows through A from right to left. (D) positive charge flows through A from left to right.

28.

In the circuit shown, each capacitor has a capacitance C. The emf of the cell is E. If the switch S is closed (A) positive charge will flow out of the positive terminal of the cell (B) positive charge will enter the positive terminal of the cell (C) the amount of charge flowing through the cell will be CE. (D) the amount of charge flowing through the cell will be 4/3 CE. ETOOSINDIA.COM



29.

In the circuit shown initially C1, C2 are uncharged. After closing the switch (A) The charge on C2 is greater that on C1 (B) The charge on C1 and C2 are the same (C) The potential drops across C1 and C2 are the same (D) The potential drops across C2 is greater than that across C1

30.

A circuit shown in the figure consists of a battery of emf 10 V and two capacitance C1 and C2 of capacitances 1.0 1F and 2.0 1F respectively. The potential difference VA – VB is 5V (A) charge on capacitor C1 is equal to charge on capacitor C2 (B) Voltage across capacitor C1 is 5V. (C) Voltage across capacitor C2 is 10 V (D) Energy stored in capacitor C1 is two times the energy stored in capacitor C2.

31.

If Q is the charge on the plates of a capacitor of capacitance C, V the potential difference between the plates, A the area of each plate and d the distance between the plates, the force of attraction between the plates is (A)

1 % Q 2 (

' * 2 '& 00 A )*

(B)

1 % CV 2 (

' 2 '&

d

* * )

(C)

1 % CV

2 ( ' * 2 '& A0 0 )*

(D)

1 % ' Q

2

4 '& !0 0 d

2

( * * )

32.

A capacitor C is charged to a potential difference V and battery is disconnected. Now if the capacitor plates are brought close slowly by some distance : (A) some +ve work is done by external agent (B) energy of capacitor will decrease (C) energy of capacitor will increase (D) none of the above

33.

Four capacitors and a battery are connected as shown. The potential drop across the 7 1F capacitor is 6 V. Then the: (A) potential difference across the 3 1F capacitor is 10 V (B) charge on the 3 1F capacitor is 42 1C (C) e.m.f. of the battery is 30 V (D) potential difference across the 12 1F capacitor is 10 V.

34.

The capacitance of a parallel plate capacitor is C when the region between the plate has air. This region is now filled with a dielectric slab of dielectric constant k. The capacitor is connected to a cell of emf E, and the slab is taken out (A) charge CE(k – 1) flows through the cell (B) energy E2C(k – 1) is absorbed by the cell. (C) the energy stored in the capacitor is reduced by E2C(k – 1) (D) the external agent has to do

1 2

E2C(k – 1) amount of work to take the s lab out.

35.

A parallel plate air-core capacitor is connected across a source of constant potential difference. When a dielectric plate is introduced between the two plates then : (A) some charge from the capacitor will flow back into the source. (B) some extra charge from the source will flow back into the capacitor. (C) the electric field intensity between the two plate does not change. (D) the electric field intensity between the two plates will decrease.

36.

A parallel plate capacitor of plate area A and plate seperation d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E and W denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted) and the work done on the system, in question, in the process of inserting the slab, then (A) Q =

37.

0 0 AV

V

0 0 AV 2 % 1 ( '1 $ * 2 d & K )

(C) E = K d (D) W = – d A parallel plate capacitor has a parallel slab of copper inserted between and parallel to the two plates, without touching the plates. The capacity of the capacitor after the introduction of the copper sheet is : (A) minimum when the copper slab touches one of the plates. (B) maximum when the copper slab touches one of the plates. (C) invariant for all positions of the slab between the plates. (D) greater than that before introducing the slab.

d

(B) Q =

0 0 KAV


38.

Two thin conducting shells of radii R and 3R are shown in the figure. The outer shell carries a charge +Q and the inner shell is neutral. The inner shell is earthed with the help of a switch S. (A) With the switch S open, the potential of the inner sphere is equal to that of the outer. (B) When the switch S is closed, the potential of the inner sphere becomes zero. (C) With the switch S closed, the charge attained by the inner sphere is – Q/3. (D) By closing the switch the capacitance of the system increases.

39.

The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now a dielectric of dielectric constant K is inserted to fill the whole space between the plates with voltage source remaining connected to the capacitor. (A) the energy stored in the capacitor will become K$times (B) the electric field inside the capacitor will decrease to K$times (C) the force of attraction between the plates will increase to K2 –times (D) the charge on the capacitor will increase to K$times

40.

A parallel-plate capacitor is connected to a cell. Its positive plate A and its negative plate B have charges +Q and –Q respectively. A third plate C, identical to A and B, with charge +Q, is now introduced midway between A and B, parallel to them. Which of the following are correct? (A) The charge on the inner face of B is now

$

3Q 2

(B) There is no change in the potential difference between A and B. (C) The potential difference between A and C is one-third of the potential difference between B and C. (D) The charge on the inner face of A is now Q 2 . 41.

In the circuit shown in the figure, the switch S is initially open and the capacitor is initially uncharged. I1, I2 and I3 represent the current in the resistance 24, 44 and 84 respectively. (A) Just after the switch S is closed, I1 = 3A, I2 = 3A and I3 = 0 (B) Just after the switch S is closed, I1 = 3A, I2 = 0 and I3 = 0 (C) long time after the switch S is closed, I1 = 0.6 A, I2 = 0 and I3 = 0 (D) long after the switch S is closed, I1 = I2 = I3 = 0.6 A.

42.

The circuit shown in the figure consists of a battery of emf 0 = 10 V ; a capacitor of capacitance C = 1.0 1F and three resistor of values R1 = 24, R2 = 24 and R3 = 14. Initially the capacitor is completely uncharged and the switch S is open. The switch S is closed at t = 0. (A) The current through resistor R3 at the moment the switch closed is zero. (B) The current through resistor R3 a long time after the switch closed is 5A. (C) The ratio of current through R1 and R2 is always constant. (D) The maximum charge on the capacitor during the operation is 51C.

43.

In the circuit shown in figure C1 = C2 = 21F. Then charge stored in

44.

(A) capacitor C1 is zero

(B) capacitor C2 is zero

(C) both capacitor is zero

(D) capacitor C1 is 40 1C

A capacitor of capacity C is charged to a steady potential difference V and connected in series with an open key and a pure resistor 'R'. At time t = 0, the key is closed. If I = current at time t, a plot of log I against 't' is as shown in (1) in the graph. Later one of the parameters i.e. V, R or C is changed keeping the other two constant, and the graph (2) is recorded. Then

(A) C is reduced

(B) C is increased

(C) R is reduced

(D) R is increased

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PART-I IIT-JEE (PREVIOUS YEARS PROBLEMS) * Marked Questions are having more than one correct option. 1.

A parallel combination of 0.1 M 4 resistor and a 10 1 F capacitor is connected across a 1.5 volt source of negligible resistance. The tim e required for the capacitor to set charged upto 0.75 volt is approximately (in seconds) : [JEE - 97' 2/100]

(A) 3

(B) loge 2

(C) log10 2

(D) zero

2.

A leaky parallel plate capacitor is filled completely with a dielectric having dielectric constant k = 5 and electrical conductivity 7 = 7.4 x 10 -12 4 -1 m -1. If the charge on the plate of the capacitor at t = 0 is Q = 8.8 1C, then calculate the leakage current at the instant t = 12 s. [JEE - 97' 5/100]

3.

An electron enters the region between the plates of a parallel plate capac itor at a point equidistant from either plate. The capacitor plates are 2 x 10$2 m apart and 10 $1 m long. A potential difference of 300 volt is kept across the plates . Assuming that the initial velocity of the electron is parallel to the capacitor plates, calculate the largest value of the velocity of the electron so that they do not fly out of the capacitor at the other end. (take mass of electron = 9 × 10 31 kg) [JEE - 97' 5/100] –

4.

Two capacitors A and B with capacito rs 3µF and 2µF are charged to a potential differe nce of 100 V and 180 V respectively. The plates of the capacitors are connected as shown in fig. with one wire fro m each capacitor free. The upper plate of A is positive and that of B is negative. An uncharged 2µF capacitor C with lead wires falls on the free ends to complete the circuit. Calculate. [JEE - 97' 5/100]

(i) The final charge on the three capacitors and (ii) The amount of electrostatic energy stored in the system before and after the completion of the circuit. 5*.

A dielectric slab of thickness d is inserted in a parallel plate capacitor whose negative plate is at x = 0 and positive plate is at x = 3d. The slab is equidistant from th e plates. The capacitor is given some charge. As x goes from 0 to 3d. [JEE - 98' 2/200] (A) The magnitude of the electric field remains the same (B) The direction of the electric field rem ains the same (C) The electric potential increases continuously (D) The electric potential increases at first, then decreases and again increases.

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6.

In the circuit shown in the figure, the battery is an ideal one, with e.m.f. V. The capacitor is initially uncharged. The switch S is closed at time t = 0. [JEE - 98' 8/200] S

A

B

(a) Find the charge Q on the capacitor at time t. (b) Find the current in AB at time t. What is its limiting value as t 7.

For the circuit shown, which of the following statements is true?

(A) With (B) With (C) With (D) With 8.

53 ?

S1 closed, V1 = 15 V, V 2 = 20 V S3 closed, V1 = V2 = 25 V S1 and S2 closed, V1 = V 2 = 0 S1 and S2 closed, V1 = 30 V, V2 = 20 V

[JEE - 99' 2/200]

In the given circuit with steady current the potential drop across the capacitor must be :

[JEE(Scr)- 2001' 3/105]

(A) V 9.

(C) V/3

(D) 2V/3

Consider the situation shown in the figure. The capacitor A has a charge q on it whereas B is uncharged. The charge appearing on the capacitor B a long time after the switch is closed is : [JEE(Scr) - 2001' 3/105]

(A) zero 10.

(B) V/2

(B) q/2

(C) q

(D) 2 q

Two identical capacitors have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is: [ JEE(Scr) 2002' 3/105]

(A)

1 C (V 12 - V22) 4

(B)

1 C (V 12 + V 22) 4

(C)

1 C (V 1 - V2)2 4

(D)

1 C (V 1 + V2) 2 4


11.

Dotted line represents the charging of a capacitor with resistance X. If resistance is made 2X then which will be the graph of charging [JEE Scr. 2004' 3/84]

(A) P

(B) Q

(C) R

(D) S

12.

An uncharged capacitor of capacitance 4µF, a battery of emf 12 volt and a resistor of 2.5 M4 are connected in series. The time after which VC = 3VR is (take !n2 = 0.693) (A) 6.93 seconds (B) 13.86 seconds (C) 7 seconds (D) 14 seconds [JEE Scr. 2005' 3/84]

13.

In the given circuit the capacitor C is uncharged initially and switch ‘S’ is closed at t = 0. If charge on capacitor at time ‘t’ is given by equation Q = Q 0 (1 – e 8t ). Find value of Q0 and 8 ? –

[JEE Mains 2005' 4/60]

14.

A circuit is connected as shown in the figure with the switch S open. When the switch is closed, the total amount of charge that flows from Y to X is [JEE 2007' 3/81]

3 1F

X

34

6 1F

64 Y 9V

(B) 54 1C

(A) 0 15.

(C) 27 1C

(D) 81 1C

A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant d initially. Suppose the liquid level decreases at a constant speed V, the time 3 constant as a function of time t is [JEE' 2008 ; 3/163 ] Figure :

K = 2. The level of liquid is

C d

d 3

6 00 R (A) 5d 3 V t +

(B)

6 00 R 5d – 3 V t

(D)

(C)

R

(15d + 9 V t ) 0 0 R 2d2 – 3d V t – 9 V 2 t 2 (15 d – 9 V t ) 0 0 R 2d 2 + 3d V t – 9 V 2 t 2

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16.

At time t = 0, a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have no charge initially, at what time (in seconds) does the voltage across them become 4 V? [Take : !n 5 = 1.6, !n 3 = 1.1]

17.

A 2 1F capacitor is charged as shown in figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is [JEE' 2011 ]

(A) 0%

18.*

[JEE' 2010 ; 3/163 ]

(B) 20%

(C) 75%

(D) 80%

In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switch S1 is pressed first to fully charge the capacitor C1 and then released. The Switch S2 is then pressed to charged the capacitor C2 After some time, S2 is released and then S3 is pressed. After some time, [JEE Advanced 2013]

S

S

1

2V

S

2

C

1

3

C

2

V

0

0

(A) the charge on the upper plate of C1 is 2CV0. (B) the charge on the upper plate of C1 is CV0. (C) the charge on the upper plate of C2 is 0. (D) the charge on the upper plate of C2 is –CV0.

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PART-II AIEEE (PREVIOUS YEARS PROBLEMS) * Marked Questions are having more than one correct option. 1.

If there are n capacitors of capacitance C in parallel connected to V volt source, then the energy stored is equal to : [AIEEE-2002, 4/300]

(2)

(1) CV 2.

1 2 nCV 2 –

–

1 2 CV 2n [AIEEE-2002, 4/300] –

(4) 10

3

–

The work done in placing a charge of 8 × 10 18 coulomb on a condenser of capacity 100 micro-farad is : [ AIEEE-2003, 4/300] 32 26 10 (1) 16 × 10 joule (2) 3.1 × 10 joule (3) 4 × 10 joule (4) 32 × 10 32 joule –

4.

(4)

Capacitance (in F) of a spherical conductor having radius 1m, is : (1) 1.1 × 10 10 (2) 10 6 (3) 9 × 10 9 –

3.

2

(3) CV

–

–

–

A fully charged capacitor has a capacitance ‘C’. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity ‘s’ and mass ‘m’. If the temperature of the block is raised by ‘9T’, the potential difference ‘V’ across the capacitance is : [AIEEE-2005, 4/300] 2mC 9T s

(1)

(2)

mC9T s

(3)

ms 9T C

(4)

2ms 9T C

5.

A parallel plate capacitor is made by stacking n equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is ‘C’, then the resultant capacitance is : [AIEEE-2005, 4/300] (1) (n – 1)C (2) (n + 1) C (3) C (4) nC

6.

A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be [AIEEE-2007, 3/120] (1) 1 (2) 2 (3) 1/4 (4) 1/2

7.

A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capacity C and is charged to a potential V volts. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is : (1)

1 (K –1)CV2 2

(2) CV2(K – 1)/K

(3) (K – 1)CV2

[AIEEE-2007, 3/120] (4) zero

8.

A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separation between its plates is ‘d’. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant k1 = 3 and thickness d/3 while the other one has dielectric constant k2 = 6 and thickness 2d/3. Capacitance of the capacitor is now : [AIEEE-2008, 3/105] (1) 45 pF (2) 40.5 pF (3) 20.25 pF (4) 1.8 pF

9.

Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t1 is the time taken for the energy stored in the capacitor to reduce to half its initial value and t2 is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio t1 /t2 will be [AIEEE-2010, 8/144] (1) 1

10.

(2)

1 2

(3)

1 4

(4) 2

Two capacitors C1 and C2 are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential on each one can be made zero. Then : [JEE Mains 2013] (1) 5C1 = 3C2 (2) 3C1 = 5C2 (3) 3C1 + 5C2 = 0 (4) 9C1 = 4C2


NCERT QUESTIONS 1.

A parallel plate capacitor with air between the plates has a capacitance of 8 pF(1 pF= 1F). What will be the capacitance if the distance between the plates is reduced by half , and the space between then is filled with a substance to dielectric constant 6 ?

2.

Three capacitors of capacitances 9 pF are connected in series. (a) What is the total capacitance of the combination ? (b) Determine the charge on each capacitor of t he combination is connected to a 120 V supply?

3.

Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. (a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected is connected to a 100 V supply ?

4.

In a parallel plate capacitor with air between the plates, each plate has an area of 6 x10| 3 m 2 and the distance between the plates is 3 mm . Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor ?

5.

Explain what would happen if in the capacitor given in question 4, a 3 mm thick m ica sheet (of dielectric constant = 6 ) were inserted between the plates. (a) while the voltage supply remained connected. (b) after the supply was disconnected.

6.

A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor ?

7.

A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor . How much electrostatic energy is lost in the process ?

8.

An electrical technician requires a capacitance of 2 1 F in a circuit across a potential differe nce of 1 kV.. A large number of 1 1 F capacitors are available to him each of which can withstand a potential diff erence of not more than 400 V. Suggest a possible arrangement that requires the minimum num ber of capacitors.

9.

What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [you will realise from your answer why ordinary capacitors are in the range of 1 F or less. However, electrolytic capacitors do have a much larger capacitance ( 0.1 F ) because of very minute separation between the conductors,]

10.

Obtain the equivalent capacitance of the network in Fig. For a 300 V supply, determine the charge and voltage across each capacitor.

–

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11.

The plates of a parallel plate capacitor h ave an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply. (a) How much electrostatic energy is stored by the capacitor? (b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u and the magnitude of electric field E between the relation.

12.

A 4 1 F capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 1 F capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ?

13.

Show that the force on each plate of a par allel plate capacitor has a magnitud e equal to (½ ) QE, where Q is the charge on the capacitor ,and E is the magnitude of electric field between the plates. Explain the origin of the factor ½.

14.

A spherical capacitor consists of two concentric spherical conductors, held in position by suitable 4 ! 0 : r1 r2 insulating supports . Show that the supports. Show that the capacitor is given by C / r1 $ r2 where r1 and r2 are the radii of outer and inner spheres respectively.

15.

A spherical capacitor has an inner sphere of radius 12cm and an sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 1 C. The space between the concentric spheres is filled with a liquid of dielectric constant 32. (a) Determine the capacitance of the capacitor. (b) What is the potential of the inner sphere? (c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.

16.

A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner c ylinder is given a charge of 3.5 1 C. Determine the capacitance of the system and the potential; of the inner cylinder. Neglect end effects ( i,e., bending of field lines at the ends).

17.

A parallel plate capacitor is to be designed with a voltage rating 1kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm 1 ( Dielectric strength is maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation ).For safety, we should like the f ield never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have capacitance of 50 pF. –

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Exercise # 1 PART-I A-1.

(C)

A-2.

(B)

A-3.

A-8.

(C)

A-9.

(C)

A-15.

(A)

A-16.

B-5.

(B)

B-12.

(C)

A-4.

(A)

A-5.

(C)

A-6.

(A)

A-7.

(B)

A-10.* (BD)

A-11.

(B)

A-12.

(A)

A-13.

(C)

A-14.

(A)

(B)

A-17.

(C)

B-1.

(B)

B-2.

(C)

B-3.

(B)

B-4.

(D)

B-6.

(C)

B-7.

(D)

B-8.

(C)

B-9.

(B)

B-10.

(D)

B-11.

(C)

(D)

B-13.

(C)

B-14.

(B)

B-15.

(D)

B-16.

(B)

B-17.

(A)

B-18.

(B)

B-19.

(B)

B-20.

(B)

B-21.

(B)

B-22.

(B)

B-23.

(A )

B-24.

(B)

C-1.

(D)

C-2.

(A)

C-3.*

(ABCD) C-4.

(C)

C-5.

(C)

C-6.*

(BD)

C-7.

(C)

C-8.

(B)

C-9.

(A)

C-10.

(A)

C-11.

(D)

C-12.

(B)

C-13.

(D)

C-14.

(B)

C-15.

(C )

C-16.

(C)

D-1.*

(AD)

D-2.

(C)

D-3.

(C)

D-4.

(C)

D-5.*

(ACD)

D-6.

(A)

D-7.

(B)

D-8.

(A)

D-9.

(A)

D-10.

(A)

D-11.

(D)

D-12.

(B)

D-13.* (BD)

D-14.

(A)

D-15.

(C)

D-16.

(D)

D-17.

(B)

D-18.

(C)

D-19.

(B)

D-20.

(C)

D-21.

(B)

PART-II 1.

(A)

2.

(B)

3.

(D)

4.

(D)

5.

(B)

6.

(C)

7.

(D)

8.*

(AC)

9.

(D)

10.

(AC)

11.*

(BCD)

12.

(D)

13.

(C)

14.

(D)

15.

(A) – q ; (B) – p ; (C) – r ; (D) – r

16.

(A) – t ; (B) – s ; (C) – q ; (D) – p

17.

(A) – q ; (B) – p ; (C) – s ; (D) – r

18.

True

21.

increases

22.

4

2.

n=3

3.

VAB =

6.

30 V

7.

C

8.

0

A #0 V d

, –

2A #0 V

12.

d

2

14.

UI =

17.

C ''

%

3kq1 10 r

where q1

( ** R 3 + R R & 1 3 ) E

% 3 ( ' *V & K + 2 )

False

20.

4.

13 #0 A 10 d

increases

1.

11.

19.

/$

2q

18.

5

Exercise # 2 0

,1 + 3; + ; -

= 10V

2

9. 60 1c , A to B

; UII = 2 K ( q + q1 ) 2 35 r

–

(a) q = 0.05(1 – e

13.

15.

32 23 25

1F

0

0

5. 10.

3Q/2C

8 3

1F

A

24 d 0.8

16.

9J

) 1C; (b) 0.125 1J

t/2

ETOOSINDIA.COM


V0

–

2t/Rc

CV02

2

#0 A

21.

(i)

22.

(a)

23.

150 mJ

26.

W=

28.

(i) 0.2 × 10-8 F, 1.2 × 10-5 J ; (ii) 4.84 × 10-5 J ; (iii) 1.1 × 10-5 J

3

4

20.

q 2d

(a) I =

5

; (b)

1

19.

R

e

1

4 % #0 AVa ( % #0 A ( ' * , Q5 = ; (ii) Q = ' * 3 d 3 & ) d & )

100 7

1 2

3

24.

% &

C0 V02 '1$

1 K

25. (i) 1.5 × 104 V/m, 4.5 × 104 V/m, (ii) 75 V, 225 V, (iii) 8 × 10 7 C/m2 –

12mC

*

27.

31.

q=

% #0 AVa ( ' * & d )

volts; (b) 28.56 mC, 42.84 mC, 71.4 mC, 22.88 mC

30.

32.

2

CV % 1 $ t / RC '1 $ e * 2 & 2

12 volt 29.

4.425 × 10-9 Ampere

A% V0 (1 / n > C = C0 ?' * $1< = 0.01078 1F, n = 20, No @?& V ) =<

33. QA = 90 1C, QB = 150 1C, QC = 210 1C, Ui = 47.4 mJ, Uf = 18 mJ

Exercise # 3 PART-I 2.

Q7 e t7 / #0 k B 0.2 1A i= k #0 –

1.

(D)

4.

(i) QA = 90 µC, QB = 150 µC, QC = 210 µC

5*.

(BC)

6.

(a) q =

CV V [1 $ e$2t/3RC] (b) i = 2 2R

7.

(D)

8.

(C)

9.

13.

Q0 =

17.

(D)

R 2 VC R1 + R 2 18.*

& 8 =

(A)

(R1 + R 2 ) CR1R 2

3.

2 30

× 108 m/s

(ii) 18 mJ

A ?@1 $

1 $2 t / 3RC > V e ; i = as t 5 3 <= 2R 3

10.

(C)

11.

(B)

12.

(B)

14.

(C)

15.

(A)

16.

t = 2 sec

5.

(1)

6.

(4)

(BD)

PART-II 1.

(2)

2.

(1)

3.

(4)

8.

(2)

9.

(3)

10.

(2)

4.

(4)

7.

(4)

Exercise # 4 2.

96 pF

4.

18 pF, 1.8 x 10-9 C.

5.

(a) V = 100 V, C = 108 pF , Q = 1.08 x 10-8 C, (b) Q = 1.8 x 10-9 C, C = 108 pF, V = 16.6 V

6.

1.5 x 10-8 J

7.

(a) 3 pF, (b) 40 V

3. (a) 9 pF, (b) 2 x 10-10 C, 3 x 10-10 C, 4 x 10-10 C

1.

6 x 10-6 J.


Capacitance

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