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CAPACITOR

CAPACITOR Y L L A with BUS Capacitance, Parallel Plate S capacitor and without dielectric, capacitors in series and parallel, Energy stored in a capacitor.

POSITION VECTOR : Capacitor is a device for storing electric charge and energy. It consists of a pair of conductors carrying equal and opposite charges (generally). Magnitude of this charge is known as the charge on the capacitor. Potential difference (V) between the two conductors is proportional to the charge on the capacitor (Q). Q V; Q = CV Here the proportionality constant C is known as the capacitance of the capacitor. Capacitance depends on the size and shape of the plates and the material between them. The SI unit of capacitance is farad (F).

PARALLEL PLATE CAPACITORS : A parallel plate capacitor consists of two equal flat parallel metal plates facing each other and separated by a dielectric of electric permittivity .The plates may be square, rectangular or circular in shape. For calculating the capacity of a capacitor, we first calculate electric field at a point

dV dr

between the plates and then using relation E

compute the potential difference

between the plates. Finally dividing the magnitude of charge (given to one plate) by the potential difference between the plates, we get the capacity. In case of parallel plate capacitor as shown in figure. The field at P

E

2 2

or,

or,

V

[ as E d

dV

dx

dV ] dx

i.e., V

0

d

q A A V d / d

So,

C

or

KA C 0 d

V x0

dV dx 0

d

[as

P

K] 0

We can derive the capacity of a spherical capacitor in a similar

[Type text]

V0

xd

SPHERICAL CAPACITOR : q 1 1 ; way V= 4 0 a b

b q a

E

CAPACITOR

4 0

q C= = 1 1 V

a

=

b

4 0 ab ba

If the radius of the outer sphere tends to infinity, b , the capacitance reduces to C = 4 0 a which is the capacitance of the isolated sphere.

CYLINDRICAL CAPACITOR : E=

a r b 2 0 r

Q

a

V = Va Vb = - Edr b

ln

or

V=

a

2 0

b l

b a

2 0 q b C= = = n V V a ILLUSTRATION : 01 A parallel plate capacitor has plates of area 200cm2 and separation between the plates 1.00mm. What potential difference will be developed if a charge of 1.00nC (i.e., 1.00 x 10-9C) is given to it. Now if separation between the plates is increased to 2.00mm, what will be the new potential difference? SOLUTION : The capacitance of the capacitor is C

0A d

F 200 x 10 -4 m 2 x m 1 x 10 -3 m

= 8. 85 x 10-12 = 0.177 x 10-9F = 0.177nF. The potential difference between the plates is

V

Q 1nC 5.65 volts. C 0.177 nF

If the separation is increased from 1.00mm to 2.00mm the capacitance is decreases by a factor of 2. If the charge remains the same, the potential difference will increase by a factor of 2. Thus, the new potential difference will be 5.65volts x 2 = 11.3 volts.

COMBINATION OF CAPACITORS SERIES COMBINATION : Capacitors connected as shown in the figure are said to be connected in series. In series combination the charges on individual

Q

Q C1 V1

[Type text]

Q

Q C2

V2

Q

Q

C3 V3

CAPACITOR

condensers are equal and the total p.d. across the combination is to shared by the capacitors. Q = C 1V1 C 2V2 C 3V3 and V = V1 V2 V3 Effective capacitance of the combination C can be found from the relation.

1 1 1 1 = + + C C C C 1 2 3

PARALLEL COMBINATION : In this combination p.d. across each of capacitors is same but the charge supplied at points A and B is shared by capacitors. V=

Q1

Q2 Q3 Q1 = = and total C1 C2 C3

charge

B

Q3 - Q3

C = C1 + C 2 + C 3 ILLUSTRATION : 02 Find the equivalent capacitance between points A and B of the circuit shown, each capacitance = C

Q2 - Q2

A

Q = Q1 + Q 2 + Q 3

- Q1

A

B

SOLUTION : Equivalent circuit is Each branch equivalent capacitance is There are four branches in parallel C eq =

C C C C + + + = 2C 2 2 2 2

C . 2

B

A

ILLUSTRATION : 03 The figure shown is a system of parallel conductors. Each plate is of equal area A and equally separated by d. Find the equivalent capacitance of the system between a and b SOLUTION : By joining the points of same potential, the arrangement of conductors may be reduced as shown in figure. If the capacitance between two successive plates is

A given by C 0 then, the equivalent capacitance of the d system is given by C eq

DIELECTRICS :

[Type text]

3C 3 0 A 2 2 d

a

2 1 4 3

2 3 b

CAPACITOR

When a dielectric is introduced between conductors of a capacitor, its capacitance increases. A dielectric is characterized by a constant 'K' called dielectric constant.

DIELECTRIC CONSTANT : When a dielectric is placed in an external electric field, polarization occurs and it develops an electric field in opposition to the external one. As a result total field inside it decreases. If E is the total field inside the dielectric when it is placed in an external field

E 0 , then its dielectric constant 'K' is given as K =

E0 ( k 1) E

If a dielectric completely occupies the space between the conductors of a capacitor its capacitance increases 'K' times. Hence in presence of a dielectric with dielectric constant ' K ', the capacitance of a parallel plate capacitor =

K 0 A d

ENERGY STORED IN A CAPACITOR : The energy stored in a capacitor is equal to the work done to charge it. Let q be the instantaneous charge on either plate of the capacitor and the potential difference between the plate is V=

q . The work done to transfer an infinitesimal charge dq from the negative C q dq C

plate to the positive plate is dW = Vdq =

[The charge moves through the wires, not across the gap between the plates] Q

W = total work done to transfer charge Q =

q

C dq = 0

QV 1 Q2 = = C V2 2 2 2C 1 2 3 4

This work done is stored as electrostatic energy U=

1 1 0A C V2 = 2 2 d

E 2 d 2 = 12

0 E 2 (Ad)

a

ie.,

b

Energy density (u) = energy per unit Volume =

1 0 E2 2

If dielectric is introduced then U =

1 K 0 E2 2

This energy is stored in a capacitor in the electric field between its plates.

FORCE ON A DIELECTRIC IN A CAPACITOR : Let us consider a small displacement dx of the dielectric as shown in figure, keeping the net force on it always zero.

Welectrostatic + [Where

WF

WF = 0

denotes

the

work

done

in

F x [Type text]

dx

CAPACITOR

displacement dx]

WF = - Welec. = U

- F dx =

F=

Q2 2

1 Q2 Q2 d = dc C 2 2C 2 dc dx

=

1 dc V2 2 dx

2C [Considering capacitor has battery connected to it, i.e., V = Q/C ]

ILLUSTRATION : 04 Two capacitors of capacitances 20pf and 50pf are connected in series with a 6-volt battery, find (A) The potential difference across each capacitor (B) The energy stored in each capacitor SOLUTION : (A)

Equivalent capacitance

C=

C1C 2 50 x 20 = C1 C 2 50 20

Charge on C1 = charge on C 2 =

=

100 pF 7

100 600 x6= 7 7

20pF

50pF

pC Potential difference across C1 50pC =

600 = 7 x 50

6V

1.71 V and across C 2 20pC = (B)

600 = 4.28 V 7 x 20

1 x 50x 1.71 2 = 73.5 pJ 2 1 Energy in C 2 = E 2 = x 20 x 4.28 2 = 184 pJ 2 Energy in C1 = E1 =

ILLUSTRATION : 05 A 5F capacitor is charged to 12 volt. The positive plate of this capacitor is now connected to the negative terminal of a 12 V battery and vice versa. Calculate the heat developed in the connecting wires. SOLUTION : When capacitor is connected with battery the charge appears on one plate be Q = CV and - Q on the other plate. If the capacitor is now disconnected and connected to the same battery again with opposite polarity then - Q appear on first plate and + Q on second plate.

Total charge flown from battery is 2Q W= charge x potential = 2QV

Q = CV

W = 2C V 2

W = 2x5x 10 6 x 12 2

[Type text]

CAPACITOR

= 1.44 mJ ILLUSTRATION : 06 A capacitor stores 50C charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of 100C flows through the battery. Find the dielectric constant of the material inserted. SOLUTION : Initial charge = 50C = Q1 Amount of charge flows = 100C

With dielectric total charge = (50+100) =150C Initial capacity C =

Q 50C = V V

Final capacity C ' = K =

150C Q' = V V

150 / V C' = =3 50 / V C

ILLUSTRATION : 07 In the above circuit, find the potential difference across AB. SOLUTION : Let us mark the capacitors as 1, 2, 3 and 4 for identification. As is clear, 3 and 4 are in series, and they are in parallel with 2. The 2,3, 4 combination is in series with 1.

C 34

C 3 .C 4 4f , C3 C 4

C eq

8 12 4.8f , 8 12

C 2,3,4 8 4 12f

The 'q' on 1 is 48C, thus V1

q C eq V 4.8 10 48C

q 6V c

48c V1 8F 6V.

VPQ 10 6 4V

By symmetry of 3 and 4, we say, VAB 2V. ILLUSTRATION : 08 What is VA VB in the arrangement shown? What is the condition such that VA VB 0 SOLUTION : Let charge be as shown (Capacitors in series have the same charge)

[Type text]

q

q'

CAPACITOR

Take loop containing C 1 , C 2 and

E

C1C 2 q q E 0 q E C1 C 2 C1 C 2 From loop containing C 3 , C 4 and E Similarly,

q' q' E 0 C3 C 4

C3C 4 C3 C 4

q' E

Now, VA VB

q q' C2 C4

C3 C1 C1 C 2 C 3 C 4

= E.

C1.C 4 C 3 .C 2 VA VB E. C1 C 2 . C 3 C 4 For VA VB 0 C1C 4 C 2 C 3 0 or

C1 C 3 C2 C4

S

ILLUSTRATION : 09 A 8F capacitor C1 is charged to V0 = 120volt. The charging battery is then removed and the capacitor is connected in parallel to an uncharged 4F capacitor C2. (A) What is the potential difference V across the combination? (B) What is the stored energy before and after the switch S is thrown ?

V0

C1

C2

SOLUTION : (A) Let q 0 be the charge on C1 initially Then q 0 C1V0 when C1 is connected to

C 2 in parallel, the charge q 0 is distributed between C1 and C 2 . Let q1 and q 2 be the charges on C1 and C 2 respectively. Now let V be the potential difference across each condenser. Now q 0 q1 q 2 or C1V0 C1V C 2 V

V

C1 8F 120V V0 C1 C 2 8F 4F = 80volt.

(B) Initial energy stored

U0

[Type text]

1 1 C1V02 (8 x 10-6) (120)2 2 2

CAPACITOR

= 5 .76 x 10-2joule Final energy stored U= =

1 1 C1V 2 C2V 2 2 2 1 1 (8 x 10-6) (80)2 + (4 x 10-6) (80)2 2 2

= 3.84 x -2

10 joule. Final energy is less than the initial energy. The loss of energy appears as heat in connecting wires. ILLUSTRATION : 10 From the given figure find the value of the capacitance C if the equivalent capacitance between points A and B is to be 1F. All the capacitances are in F.

C1

A C

SOLUTION : The capacitors C3 and C4 are in parallel, therefore their resultant capacity C8 is 4. The capacitors C5 and C6 are in series, therefore, their resultant capacity C 9 is 4. These are shown in figure (A)

C2 C8

A

Now the capacitor C2 and C8 are in series. Their resultant capacity C10 is

8 . Capacitors C 7 and C 9 are 3

in parallel. Their resultant capacity C11is 8. These are shown in figure. (B) C1 and C11 are in series. The equivalent capacitance is 8/9. The parallel combination of 8/3 and 8/9 gives a resultant capacitance 32/9 as shown in figure. (C)

1 1 9 1 C 32

C

or

1 23 C 32

32 F 23

ILLUSTRATION : 11 Five identical conducting plates 1, 2, 3, 4 and 5 are fixed parallel to and equidistant from each other as shown in figure. Plates 2 and 5 are connected by a conductor while 1 and 3 are joined by another conductor. The junction of 1 and 3 and the plate 4 are connected to a source of constant e.m.f.

V0 . Find

(i) The effective capacity of the system between the

[Type text]

8 1 4 4

C9 C7

4

B

CAPACITOR

terminals of the source (ii) The charge on plates 3 and 5 Given d = distance between A = area of either face of each plate.

any

two

successive

plates

and

SOLUTION : (i) The equivalent circuit is shown in figure (B). The system consists of four

capacitors i.e., C12 , C 32 C 34 and C 54 . The capacity of each capacitor is K 0

A C 0 The d

effective capacity across the source can be calculated as follow: The capacitors C12 and C32 are in parallel and hence their capacity is C 0 + C0 = 2C0. The capacitor C54 is in series with effective capacitor of capacity 2C 0. Hence the resultant capacity will be

C 34 is again in parallel. Hence the effective capacity

Further = C0 (ii)

C0 x 2 C0 C 0 2C 0

C 0 x 2C 0 5 5 A C 0 K 0 C 0 2C 0 3 3 d

Charge on the plate 5 = charge on the upper half of parallel combination

2 2 K 0 AV0 C0 d 3 3

Q 5 V0

Charge on plate 3 on the surface facing 4

V0 C 0

K 0 AV0 d

Charge on plate 3 on the surface facing 2 = [Potential difference across (3 - 2)] C 0 =

Q3

V0

C0 AV0 C 0 K 0 C 0 2C 0 3d

K 0 AV0 K 0 AV0 AV0 = K 0 d d 3d

1 4 A 1 3 3 K 0 d V0

ILLUSTRATION : 12 (A) Find the effective capacitance between points X and Y in the

given figure.

Assume

that

C 2 10 F and other

capacitors are 4F each. (B) Find the capacitance of a system of identical capacitors between points A and B shown in figure. SOLUTION : (A) The circuit is redrawn in figure as the two arms are

[Type text]

CAPACITOR

balanced, no current flows through C 2 , C 3 and C 4 are in series, hence their equivalent capacitance

=

2F

Similarly

the

equivalent

capacitance

of

C1 and

C 5 = 2F.

Corresponding to points X and Y these two are in a parallel combination. Hence the effective capacitance between X and Y is 2 + 2 = 4F. (B) The arrangement of capacitors shown in figure is equivalent to the arrangement shown in figure. The arrangement is connected in parallel. Hence equivalent capacitance C is given by C = C1 + C2 + C3

WORKED OUT OBJECTIVE PROBLEMS : EXAMPLE : 01 A parallel plate capacitor is connected across a 2V battery and charged. The battery is then disconnected and glass slab is introduced between the plates. Which of the following pairs of quantities decrease? (A) Charge and potential difference (B) potential difference and energy stored (C) energy stored and capacitance (D) capacitance and charge Ans: (B) SOLUTION : The introduction of a dielectric slab increases the capacitance. The charge remains unchanged. Potential difference and energy stored decreases. EXAMPLE : 02 Three capacitors of capacitances 3F, 9F and 18F are connected once in series and then in parallel. The ratio of equivalent capacitances in the two cases(C S/CP) will be (A) 1 : 15 (B) 15 : 1 (C) 1 : 1 (D) 1 : 3 Ans: (A) SOLUTION :

C P 3 9 18 30F 1 1 1 1 1 C S 3 9 18 2

CS 2F

Now

CS 2F 1 C P 30F 15

EXAMPLE : 03 A number of capacitors each of capacitance 1F and each one of which get punctured if a potential difference just exceeding 500volt is applied, are provided. Then an arrangement suitable for giving a capacitor of capacitance 2F across which 3000 volt may be applied requires at least

[Type text]

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CAPACITOR

(A) 18 component capacitors (C) 72 component capacitors Ans: (C) SOLUTION :

(B) 36 component capacitors (D) 144 component capacitors

3000 6 500 1 The capacitance of series combination. = F 6 Number of capacitors required in series =

To obtain a capacitor of 2F, we should use 12 such combinations. Total number of capacitors required = 12 x 6 = 72 EXAMPLE : 04 A capacitor of capacitance 1F withstands a maximum voltage of 6kV, while another capacitor of capacitance 2F, the maximum voltage 4kV. If they are connected in series, the combination can withstand a maximum of (A) 6kV (B) 4kV (C) 10kV (D) 9kV Ans: (D) SOLUTION : When the two condensers are connected in series.

Q

C

2 x1 2 F 2 1 3

and

2 E 3 The potential of condenser C1 is given by

V1

Q 2 E 6kV E 9kv C1 3

v2

Q E E 12KV c2 3

To avoid break down E 9KV EXAMPLE : 05 Seven capacitors each of capacitance 2F are to be connected to obtain a

10 F. Which of the following combination is possible. 11

capacitance of

(A) 5 in parallel 2 in series (B) 4 in parallel 3 in series (C) 3 in parallel 4 in series (D) 2 in parallel 5 in series Ans: (A) SOLUTION : 5 capacitors in parallel gives 5 x 2 F = 10F capacity. Further, two capacitors in series gives a capacity 1F. When the two combinations are connected in series, they give

10 x 1 10 = F. 10 1 11

a resultant capacitance

EXAMPLE : 06 Condenser A has a capacity of 15F when it is filled with a medium of dielectric

[Type text]

Page

CAPACITOR

constant 15. Another condenser B has a capacity 1F with air between the plates. Both are charged separately by a battery of 100V. After charging, both are connected in parallel without battery and the dielectric material being removed. The common potential now is (A) 400V (B) 800V (C) 1200V (D) 1600V Ans: (B) SOLUTION : Charge on capacitor A is given by q1 C1 x V = (15 x 10-6) (100) = 15 x 10-4C Charge on capacitor B is given by q2 = C2 x V = (1 x 10-6) (100) = 10-4C Capacity of condensers A after removing dielectric 10 -6 C1 15 x 15 K

C'

1 F

Now when both condenser are connected in parallel their capacity will be 1F + 1F = 2F Common potential V = =

q C

15 x 10 -4 (1 x 10 -4 ) 800V . 2 x 10 -6

EXAMPLE : 07 Two capacitors 2F and 4F are connected in parallel. A third capacitor of 6 F capacity is connected in series. The combination is then connected across a 12V battery. The voltage across 2F capacity is (A)2V (B) 6V (C) 8V (D) 1V Ans: (B) SOLUTION : Resultant capacitance of condensers of capacity 2F and 4F when connected in parallel. C' 2 4 6 F This is connected in series with a capacitor of capacity 6F in series. The resultant capacity C is given by

1 1 1 1 or C 3 F C 6 6 3 Charge on combination

q = (3 x 10-6) x (12) = 36 x 10-6C

Let the charge on 2F capacitor be q1 , then

q1 q q1 q or q1 3 2 4 Now potential across 2F condenser

[Type text]

q1 12 x 10-6C =

q1 2 x 10 -6

12 x 10 -6 2 x 10 -6

6V

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CAPACITOR

EXAMPLE : 08 The capacitance of the system of parallel plate capacitor shown in the figure isA

2 0 A 1 A 2 (A) A1 A 2 d

2 0 A 1 A 2 (B) A 2 A1 d

A (C) 0 1 d

A (D) 0 2 d

1

d A2

Ans : (C) SOLUTION : Since the electric field between the parallel charge plates is uniform and independent of the distance, neglecting the fringe effect, the effective area of the plate of area A 2 is A1. Thus the

A capacitance between the plates is C 0 1 d

(C)

A1 0

d

E

A2

EXAMPLE : 9 The charge flowing across the circuit on closing the key K is equal to

C V 2

(A) CV

(B)

(C)2CV

(D) Zero

SOLUTION : When the key K is kept open the charge drawn from the source is Q C 'V Where C' is the equivalent capacitance given by C'

C 2

C V 2

Therefore Q =

Whey the key K is closed, the capacitor 2 gets short circuited and the charge in the circuit

Q1 CV

Charge flowing is Q Q1 Q

(B)

C V 2

EXAMPLE : 10 The figure shows a spherical capacitor with inner sphere earthed. The capacitance of the system is (A)

4 0 ab ba

(C) 4 0 b a

[Type text]

(B)

4 0 b 2 ba

a

b

(D) None of these

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CAPACITOR

SOLUTION : Let V be potential of the outer sphere. Thus we can consider two capacitors, one between the outer sphere and inner sphere and the other between outer sphere and infinity. Thus,

C1 4 0

ab ba

C 2 4 0 b C 4 0

C

(B)

a

b

ab 4 0 b ba

C1

C2

4 0 b 2 ba

EXAMPLE : 11 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 (B) V

(A) V

Q C

(C) V

Q 2C

(D) V

Q , if V < C

CV SOLUTION : In the figure given below, left X and Y be positive and negative plates. After charging from the cell, the inner faces of X and Y have charges CV , as shown in (A). The outer surfaces have no charge. When charge Q is given to X, let the inner faces of X and Y have charges q Then, by the principle of charge conservation, the outer

faces have charges Q CV q for X and q CV for Y, as shown in (B). Now, the outer faces must have equal charges. Q CV q q CV or 2q 2CV Q or q CV

Q 2

Potential difference

q Q V C 2C

(C) EXAMPLE : 12 In an isolated parallel - plate capacitor of capacitance

[Type text]

Q1

Q3

Q2

Q4

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CAPACITOR

C, the four surfaces have charges Q1 , Q2 , Q3 and Q4 as shown. The potential difference between the plates is (A)

Q Q3 Q1 Q 2 Q 3 Q 4 (B) 2 2C 2C

(C)

Q 2 Q3 2C

(D)

Q1 Q 4 2C

SOLUTION : Plane conducting surfaces facing each other must have equal and opposite charge densities. Here, as the plate areas are equal, Q2 = - Q3 The charge on a capacitor means the charge on the inner surface of the positive plate-(in this case Q2) Potential difference between the plates = charge on the capacitor capacitance. Potential difference =

Q 2 2Q 2 Q 2 Q 2 Q 2 Q 3 C 2C 2C 2C

(C) ***

[Type text]

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CAPACITOR

SINGLE ANSWER OBJECTIVE TYPE QUESTIONS : LEVEL - 1 : 1.

2.

3.

A capacitor of capacitance C is charged to potential difference V 0 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) V0 B) V0 + (Q/C) C) V0 + (Q/2C) D) V0 - (Q/C) A dielectric of dielectric constant 3 fills up three fourths of the space between the plates of a parallel plate capacitor. The percentage of energy stored in the dielectric is A) 25% B) 50% C) 75% D) 100% A parallel plate capacitor having capacitance C 0 is connected to a battery of emf E. It is then disconnected from the battery and a dielectric slab of dielectric constant k completely filling the air gap of the capacitor is inserted in it. If U indicates the change in energy, then A) U = 0

B) U =

1 0E2 (k - 1) 2

C) U =

1 1 0E2 1 k 2

D)U

1 1 1 0E2 k 2 4.

5.

The work done in increasing the voltage across the plates of the capacitor from 5V to 10V is W. The work done in increasing the voltage from 10V to 15V will be A) W B) 4/3 W C) 5/3 W D) 2W A parallel plate capacitor of plate area A and plate separation 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 work done on the system, in questions, in the process of inserting the slab, then A) Q W

6.

7.

8.

0 AV d

0 AV 2 2d

B) Q

0 KAV d

C) E =

V Kd

D)

1 1 K

When the capacitance in an oscillator circuit of frequency f is increased nine times, the frequency of the oscillator is reduced to: A) f/9 B) f/6 C) f/4 D) f/3 64 small drops of water having the same charge & same radius are combined to form one big drop. The ratio of capacitance of big drop to small drop is: A) 4 : 1 B) 1 : 4 C) 2 : 1 D) 1 : 2 A parallel plate condenser is connected to a battery. The plates are pulled apart with a uniform speed. If x is the separation between the plates, then the time rate of charge of the electrostatic energy of the condenser is proportional to

[Type text]

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CAPACITOR

9.

A) x2 B) x C) 1/x D) 1/x2 A spherical condenser has inner and outer spheres of radii a and b respectively. The space between the two is filled with air. The difference between the capacities of two condensers formed when outer sphere is earthed and when inner sphere is earthed will be B) 4 0 a

A) zero a)]

10.

ab ba

D) 40

12.

13.

14.

15.

D) 4 0 a [b/(b -

The sphere shown in the figure are connected by a conductor. The capacitance of the system is: A) 4 0

11.

C) 4 0 b

B) 4 0 a

C) 4 0 b

a2 ba

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 the plates. The capacitor is given some charge. As X goes from 0 to 3d. A) the electric potential increases at first, then decreases and again increases B) the electric potential increases continuously C) the direction of the electric field remains the same D) the magnitude of the electric field remains the same Two similar conducting balls are placed near each other in air. The radius of each ball is r and the separation between the centers is d (d >> r). The capacitance of two balls system when they are connected by a wire is: A) 80r B) 40r C) 40r loge (r/d) D) 4 loge 0 (r/d) A hollow sphere of radius 2R is charged to V volt and another smaller sphere of radius R is charged to V/2 volt. Then the smaller sphere is placed inside the bigger sphere without changing the net charge on each sphere. The potential difference between the two spheres would be A) 3V/2 B) V/4 C) V/2 D) V A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates which results in: A) reduction of charges on the plates and increase of potential difference across the plates B) increase in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates C) decrease in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates D) none of the above Force acting on a charged particle kept between the plates of a charged condenser is F. If one of the plates of the condenser is removed, force acting on the same particle will become:

[Type text]

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CAPACITOR

16.

17.

A) zero B) F/2 C) F D) 2F A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with this capacitor are given by Q 0, V0, E0 and U0 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 in the previous ones as: A) Q > Q0 B) V > V0 C) E > E0 D) U > U0 Figure shows two capacitors connected in series and7joined to a battery. The graph shows the variation in potential as one moves from left to right on the branch containing the capacitors: A) C1 > C2 B) C1 = C2 C) C1 < C2 D) the information is not sufficient to decide the relation between C 1 and C2

LEVEL - II : 1.

2.

3.

4.

Identical charged 103 oil drops each of radius 0.9nm are combined to form a single drop. The electrical capacity of the drop is 1) 0.1 PF 2) 1 PF 3) 10 PF 4) 100 PF A parallel plate condenser is charged end isolated. The distance between the plate is increased by 2mm and a dielectric slab of thickness 3 mm is introduced between the plates. If the potential difference between the plates remains same, the dielectric constant of the dielectric slab is 1) 2 2) 3 3) 4 4) 5 Two identical parallel plate condensers are connected in series. A cell of e.m.f of 20V is connected between their ends. A dielectric slab of constant 4 is placed between the plates of one of the condensers. The potential difference across condenser with dielectric slab is 1) 4V 2) 10V 3) 16V 4) 18V Four identical parallel metal plates each of area A are placed with separation ‘d’ between adjacent plates as shown. The capacitance between the plates P and Q is

3 0 A d 2 0 A 3) d 1)

5.

6.

7.

2 0 A 3d 0 A 4) 2d 2)

Two spherical conductors of radii 3cm and 6cm are in contact. A charge 10 -9 is given to them. The potential of the smaller sphere is 1) 67V 2) 33V 3) 50V 4) 100V A parallel plate condenser is charged and isolated. The energy stored by the condenser is E. The separation of the plate is doubled and then the space is completely filled with a dielectric of constant 5. The energy stored by the condenser now is 1) E 2) 0.8E 3) 0.4E 4) 2.5E Two condensers charged to potentials 50V and 80V are connected in parallel, the common potential is 60V. The capacities of the condensers are in the ratio of 1) 2 : 1 2) 1 : 2 3) 3 : 4 4) 4 : 3

[Type text]

Page

CAPACITOR

8.

The capacitor of 4F charged to 50V is connected to another capacitor of 2F charged to 100V. The total energy of the combination is 1)

9.

4 x102 J 3

2)

3 x102 J 2

3) 3 x102 J

4)

8 x102 J 3

The radii of two charged metal spheres are 5cm and 10cm both having same charge 75c. If they are connected by a wire, the quantity of charge transferred through the wire is 1) 75c 2) 50c 3) 25c 4) 15c

10.

11.

Two identical capacitors have equivalent capacity of 2F when they are connected in series. If they are connected in parallel and charged to a potential of 200V, the energy stored in the system is 1) 18 x 10-4 J 2) 18 x 10-4 J 3) 0.16 J 4) 0.36 The capacity of a parallel plate condenser is C. When half the space between the plates is filled with a slab of dielectric constant K as shown in the figure. If the slab is removed from the condenser, then the capacity of the condenser becomes 1) 3)

12.

13.

14.

15.

16.

2KC K 1

K 1 C 2K

2) 4)

K 1 C 2K KC 2K 1

Two parallel plate capacitors C and 2C are connected in parallel and charged to P.D ‘V’. The battery is then disconnected and the region between the plates of the capacitor ‘C’ is completely filled with a material of dielectric constant ‘3’. The P.D. across the capacitors now becomes 1) V 2) 3V 3) 3V/5 4) 4V/5 A condenser of capacity 10F is connected between the terminals of a battery of potential difference 100V and is charged. The amount of work done by the battery to charge the condenser is 1) 0.05 J 2) 0.025 J 3) 0.1 J 4) 0.0125 J Two condensers of capacities C1 = 5 F – 100V and C2 = 10F – 100V are connected in series. The maximum potential difference that may be applied between them without damaging the condensers is 1) 200 V 2) 175 V 3) 150 V 4) 300 V ‘n’ identical charged spheres combine together to form a large sphere. The ratio of the potential of small sphere to potential of large sphere is 1 : 9. The ratio of the energy stored in small sphere to the energy stored in the large sphere is 1) 1 : 3 2) 1 : 27 3) 1 : 81 4) 1 : 243 The capacity of parallel plate condenser with air between the plates is 10F. If the space between the plates is completely filled with two dielectric slabs each of thickness is equal to half of the separation between the plates, whose dielectric constants are 3 and 2. What will be the percentage change in capacity of the condenser? 1) 133.3 % 2) 140 % 3) 166.6 % 4) 280 %

[Type text]

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CAPACITOR

17.

18.

19.

20.

21.

The work done in charging a capacitor from 5V to 10V is W. The work done in charging the capacitor from 10V to 15V is 1) 4 W/3 2) 5 W/3 3) 2 W/5 4) 5 W/2 The voltages across C1 and C2 are in the ratio 2 : 3. When C2 is completely filled with paraffin, the voltage ratio became 3 : 2. The dielectric constant of paraffin is 1) 2.25 2) 13/6 3) 27/8 4) 6 A parallel plate capacitor of capacity 5F and plate separation 6cm is connected to a 1V battery and is charged. A dielectric of dielectric constant 4 and thickness 4cm is introduced into the capacitor. The additional charge that flows into the capacitor from the battery is 1) 2C 2) 3C 3) 5C 4) 10C A metal sphere ‘A’ of radius a is charged to a potential ‘V’. What will be its potential if it is enclosed by a spherical conducting shell ‘B’ of radius b and the two are connected by a conducting wire? 1) V 2) (a/b) V 3) (b/a) V 4) zero The capacity of a parallel plate capacitor is 5F. When a glass plate of same area as the plates but thickness half of the distance between the plates is placed between the plates of the capacitor, its potential difference reduces to 2/5 of the original value. The dielectric constant of the glass is 1) 1.5 2) 2.5 3) 5 4) 2s

V0 E 0 d Hint:

V 22.

Qd A 0

Qd A 0

t 1 1 d 1 k

2 V t 1 3 1 1 1 1 ; 1 5 V0 d k 5 2 k

Two spheres A and B of radii 4cm and 6cm are given charges of 80C and 40C. If they are connected by a wire, the amount of charge flowing from one to other is 1) 20C from A to B 2) 16C from A to B 3) 24C from B to A 4) 32C from A to B Hint: Q = (C1 – C2)V = i.e. Q =

23.

C2 C1 C1V1 C2 V2 C2 C1 Q1 Q 2

r2 r1 Q1 Q2 r1 r2

C1 C2

C1 C2

2 x 120 24C from B to A 10

Three capacitor of capacitance 10F, 15F, 20F are in series with a cell.

The

charge drawn from the cell is 60C. If they are connected in parallel with the same cell, then the charge drawn from the cell is 1) 385 C 2) 485 C 3) 585 C 4) 685 C Hint: 24.

qp qs

Cp E Cs E

Cp Cs

qp 6 x

45 x13 585C 60

The plates of a parallel plate capacitor are horizontal and parallel. A thin conducting sheet P is initially placed parallel to both the plates and nearer to the lower plate. From t = 0 onwards, the sheet P is moved at constant speed vertically

[Type text]

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CAPACITOR

25.

upwards so that it is always parallel to the capacitor plates. At t = 20 milli seconds, it is nearer to the upper plates. Then during the time interval from t = 0 to t = 20 milli seconds, the capacity of the capacitor will 1) increase gradually 2) decrease gradually 3) remains constant 4) first increases and then decreases Hint: Since the thickness of conducting plate is constant through out, capacity remains constant A capacitor is charged to 200V. A dielectric slab of thickness 4mm is inserted. The distance between the plates is increased by 3.2mm to maintain the same potential difference. Find the dielectric constant of the slab 1) 3 2) 4 3) 5 4) 6 Hint: V1 = V2;

26.

27.

C0

C 1

t 1 1 d k

30.

1

t 1 d

C 1 6 0 k C 7

Between the plates of a parallel plate capacitor of capacity C, two parallel plates of the same material and area same as the plate of the original capacitor are placed. If the thickness of these plates is equal to I/5 th of the distance between the plates of the original capacitor, then the capacity of the new capacitor is 1) 5/3 C 2) 3/5 C 3) 3C/10 4) 10C/3 Hint: d1 = d, d2 =

29.

d 2 d1 1 1 d 2 t 1 k k 1 t

A capacitor is charged with a dielectric to V volts. If the dielectric of constant K is removed then ___ is true a) capacity decreases by k times b) electric field intensity decreased by k times c) potential increases by k times d) charge increases by k times 1) a, b, c 2) a, b, c, d 3) a, c 4) b, d The capacitance of a capacitor becomes 7/6 times its original value if a dielectric slab of thickness t = 2/3d is introduced in between the plates ‘d’ is the distance between the plates. The dielectric constant of dielectric field is 1) 14/11 2) 11/14 3) 7/11 4) 11/7 Hint:

28.

Qd1 Q A 0 A 0

C2 d1 5 5 3 2 C2 C d, t = d & C1 d 2 3 3 5 5

A parallel plate capacitor of capacity Co is charge to a potential V0. A) The energy stored in the capacitor when the battery is disconnected and the plate separation is double is E1 B) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is E2, the E1/E2 value is 1) 4 2) 3/2 3) 2 4) ½ A capacitor of capacity of 10F is charged to 40V and a second capacitor of capacity 15F is charged to 30V. If they are connected in a parallel the amount of charge that flows from the smaller capacitor to higher capacitor in C is 1) 30 2) 60 3) 200 4) 250

[Type text]

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CAPACITOR

C1V1 C2 V2 & Q = (V1 – V)C1 = 60C C1 C2

Hint: V = 31.

A parallel plate capacitor of capacity 100F is charged by a battery of 50 volts. The battery remains connected and if the plates of the capacitor are separated so that the distance between them becomes half the original distance, the additional energy given by the battery to the capacitor in joules is 1) 125 x 10-3 2) 12.5 x 10-3 3) 1.25 x 10-3 4) 0.125 x 10-3 2 Hint: V = ½(C2 – C1)V & C2 = 2C

32.

A parallel plate capacitor of capacity 5F and plate separation 6cm is connected to a I volt battery and is charged. A dielectric of dielectric constant 4 and thickness 4cm introduced into the capacitor. The additional charge that flows into the capacitor from the battery is 1) 2C 2) 3C 3) 5C 4) 10C

C0 V t 1 = 10C; Q = Q2 – Q1 Hint: Q1 C0 V = 5C; Q2 = C2V = 1 1 d k 33.

A 20F capacitor is charged to 5V end isolated. It is then connected in parallel with an uncharged 30F capacitor. The decrease in the energy of the system will be 1) 25 J 2) 100 J 3) 125 J 4) 150 J Hint: V

34.

C1V1 C2 1 & U1 C1V12 U U1 C1 C2 2 C1 C2

A dielectric of thickness 5cm and dielectric constant 10 is introduced in between the plates of a parallel plate capacitor having plate area 500 sq cm and separation between plates 10cm. The capacitance of the capacitor is (0=8.8 x 10-12 SI units) 1) 8 PF 2) 6PF 3) 4PF 4) 20PF

A 0 1 d t 1 k

Hint: C = 35.

A 4F capacitor is charged by 200V battery. It is then disconnected from the supply and is connected to another uncharged capacitor of 2F capacity. The loss of energy during this process is _____ 1) 0 2) 5.33 x 10-2 3) 4 x 10-2 4) 2.67 x 10-2 Hint: U =

36.

CV 1 C1C2 V1 V2 2 ; V2 1 1 2 C1 C2 C1 C2

Energy E is stored in a parallel plate capacitor C 1. An identical uncharged capacitor C2 is connected to it kept in contact with it fro a while and then disconnected. The energy stored in C2 is ___ 1) E/2 2) E/3 3) E/4 4) 0 Hint: U1 =

1 C1V 2 , Potential on second capacitor in contact with the first one V 2 = 2 2

V/2; U2 =

[Type text]

1 V E C 2 2 4

Page

CAPACITOR

37. 38. is

If the capacity of a spherical conductor is 1PF, then its diameter is 1) 9 x 10-15m 2) 9 x 10-3m 3) 9 x 10-5m 4) 18 x 10-7m A 700PF capacitor is charged by a 50V battery. The electrostatic energy stored by it 1) 17 x 10-5 J

2) 13.6 x 10-9 J

3) 9.5 x 10-9 J

4) 8.75 x 10-

7

J 39.

40.

Two equal capacitors are first connected in parallel and then in series. The ratio of the total capacities in the two cases will be 1) 2 : 1 2) 1 : 2 3) 4 : 1 4) 1 : 4 Two condensers of capacity 2C and C are joined in parallel and charged up to potential V. The battery is removed and the condensor of capacity ‘C’ is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be 1)

41.

3V K

C1 V1 C2 V2 C1 C2

V1 V2 2

3)

V K2

4)

V K

2)

C1V2 C 2 V1 C1 C2

4)

C1V2 C 2 V1 C1 C2

3)

A number of capacitors are connected as shown in figure. The equivalent capacity is given by 1) nC 2) n(n + 1)C 3) 2n(n + 1)C

43.

3V K2

Two condensers C1 and C2 in a circuit are joined as shown in figure. The potential at A is V 1 and that of B is V2. The potential of point D will be 1)

42.

2)

4)

n n 1 C 2

Three capacitors of capacitances 3F, 10F and 15F are connected in series to a voltage source of 100V. The charge on 15F is

44.

45.

46.

47.

1) 50C 2) 100C 3) 200C 4) 280C In a parallel plate capacitor of capacitance ‘C’, a metal sheet is inserted between the plates parallel to them. If the thickness of the sheet is half of the separation between the plates, the capacitance will be 1) C/2 2) 3C/4 3) 4C 4) 2C A 10 micro farad capacitor is charged to 500 V and then its plates are joined together through a resistance of 10 ohm. The heat produced in the resistance is 1) 500 J 2) 250 J 3) 125 J 4) 1.25 J A capacitor is charged to 200 volt. It has a charge of 0.1C. When it is discharged, energy liberated will be 1) 1 J 2) 10 J 3) 14 J 4) 20 J Half of the separation between two parallel plates of a capacitor is filled with a dielectric medium. The capacitance of the capacitor becomes 5/3 times its original value with full space dielectric. The dielectric constant of the medium K is

[Type text]

Page

CAPACITOR

48.

49.

1) K = 2 2) K = 3 3) K = 4 4) K = 5 A 10F capacitor is charged to a potential difference of 50 V and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes 20 V. The capacitance of second capacitor is 1) 10 F 2) 20 F 3) 30 F 4) 15 F Capacity of a spherical capacitor having two spheres of radii a and b (a > b) separated by a medium of dielectric constant K is given by

Kab in SI system ab 4 0 Kab 3) in SI system a b 1)

2) 4)

4 0 Kab in CGS system ab K a b ab

in CGS system

50.

A capacitor of capacity ‘C’ has charge Q. The stored energy is W. If the charge is increased to 2Q, the stored energy will be 1) 2W 2) W/2 3) 4W 4) W/4

51.

A D.C. potential of 100 volt is connected to the combination as shown in figure. The equivalent capacity between A and B will be equal to 1) 40 F 2) 20 F

52.

3) 30 F 4) 10 F The capacitance of four plates, each of area A arranged as shown in figure as

2 0 A d 4 0 A 3) d 1)

3 0 A d 5 0 A 4) d 2)

53.

Identify the wrong statement of the following a) the resultant capacity ‘C’ is less than the capacitance of smallest capacitor in series combination b) the resultant capacity ‘C’ is greater than greatest capacitance in parallel combination c) In series combination, charge on capacitor plates is inversely proportional to capacitance of capacitor 1) a is wrong 2) c is wrong 3) b is wrong 4) all are wrong 54. A parallel plate capacitor of area A, plate separation d with the electric capacity C0 is filled with three different dielectric materials with constants K1, K2 and K3 as in the figure. If these three are replaced by a single dielectric, its dielectric constant K is ___ 1)

1 1 1 1 K K1 K 2 2K 3

2)

3)

1 K1 K 2 2K 3 K K1 K 2

4) K

[Type text]

1 1 1 K K1 K 2 2K 3 K1K 3 K 2 K3 K1 K 3 K 2 K 3

Page

CAPACITOR

55.

56.

57.

58.

59.

60.

61.

62.

63.

Four metallic plates, each with a surface area of one side A, are placed at a distance d apart. The outer plates are connected to terminal X and the inner plates to terminal Y. The capacitance of system between X and Y is 1) 0A/d 2) 20A/d 3) 30A/d 4) 40A/d Four metallic plates, each with a surface area of one side A are placed at a distance d apart, these plats are connected as shown in figure. The capacitance of the system between X and Y is 1) 0A/d 2) 20A/d 3) 30A/d 4) 40A/d An infinite ladder is made as shown in figure using capacitors C1 = 1 F and C2 = 2F. The equivalent capacitance of the ladder, in F is 1) 1 2) 2 3) 0.75 4) 0.5 In the circuit segment shown VA - VB = 19V. The p.d. across 3 F capacitor is 1) 7V 2) 8V B A 3) 23V 4) 4V A parallel plate capacitor with a dielectric constant K = 3 filling the space between the plates is charged to a potential difference V. The battery is then disconnected and the dielectric slab is withdrawn and replaced by another dielectric slab having K = 2. The ratio of energy stored in the capacitor before and after replacing the dielectric slab by new one is 1) 3 : 2 2) 9 : 4 3) 4 : 9 4) 2 : 3 Two parallel capacitors of capacitance C and 2C are connected in parallel and charged to a potential difference V. The battery is then disconnected and the capacitor C is completely filled with a material of dielectric constant k. The potential difference across the capacitors is now 1) 2V/k 2) 3V/k 3) 3V/(k+2) 4) 2V/(k+3) Initially the capacitors C1 and C2 shown in figure have equal capacitances. If a dielectric plate (k = 2) is introduced in capacitor C2, then potential difference across its plates and charge 1) both will decrease 2) both will increase 3) p.d. will increase but charge will decrease 4) p.d. will decrease but charge will increase Five identical capacitor plates, each of area A, are arranged such that adjacent plates are at a distance d apart. The plates are connected to a source of emf V as shown in figure. The charge on plate 1 is q and that on 4 is q', where 1) q' = q 2) q' = 2q 3) q' = -2q 4) q' = 3q For the circuit shown in figure which of the following statements is true 1) With S1 closed, V1 = 15V, V2 = 20V 2) With S3 closed, V1 = V2 = 25V 3) With S1 and S2 closed, V1 = V2 = 0 4) With S1 and S3 closed, V1 = 30V, V2 = 20V

[Type text]

Page

CAPACITOR

64.

Two identical capacitors, having the same capacitance C. One of them is charged to a potential V1 and the other to V2. The negative ends of the capacitor are connected together. When the positive ends are also connected, the decrease in energy of the combined system is 1)

65.

66.

67.

1 C ( V12 V2 2 ) 4

69.

70.

72.

1 C ( V1 V2 ) 2 4

4)

1 C ( V1 V2 ) 2 4

2) (3/2)

CV 2

3) (25/6) CV 2

4) (9/2) CV 2

a composite capacitor having capacitance 8 F? 1) 4 Condensers in series 8 such groups in parallel 2) 2 Condensers in series and 16 such groups in parallel 3) 8 Condensers in series and 4 such groups in parallel 4) All of them in series. Find out the effective capacitance between points P and Q. It will be equal to 1) 9 F 2) 4.5 F 3) 1 F 4) 6 F An uncharged parallel plate capacitor having a dielectric of constant K is connected to a similar air filled capacitor charged to a potential V. The two share the charge and the common potential is V'. The dielectric constant K is

V V V V

2)

V V V

3)

V V V

4)

V V V

Two condensers each having capacitance C and breakdown voltage V are joined in series. The capacitance and the breakdown voltage of the combination will be 1) 2C and 2V 2) C/2 and V/2 3) 2C and V/2 4) C/2 and 2V Two identical metal plates are given positive charges Q 1 and Q2 (< Q1) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C, the potential difference between them is 1)

73.

3)

You are given thirty two capacitors each having capacity 4 F. How do you connect all of them to prepare

1) 71.

1 C (V13 V2 2 ) 4

Each edge of a cube (figure) made of wire contains a capacitor of capacitance C. Find the effective capacitance of this bank of capacitors between opposite corners A and G. 1) 5C/6 2) 4C/3 3) 3C/4 4) 6C/5 Consider the situation shown in figure. The capacitor A has charge q on it whereas B is uncharged. The charge appearing on the capacitor B a long time after switch is closed is 1) zero 2) q/2 3) q 4) 2q A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is similarly charged to potential differences 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the negative terminal of one is connected to the negative terminal of the other. The final energy of the configuration is 1) zero

68.

2)

(Q 1 Q 2 ) 2C

2)

(Q 1 Q 2 ) C

3)

(Q1 Q 2 ) C

4)

(Q1 Q 2 ) 2C

Three very large plates are given charges as shown in the figure. If the cross sectional area of each plate is the same, the final charge distribution on plate C

[Type text]

Page

CAPACITOR

74.

is: 1) +5 Q on the inner surface, +5 Q on the outer surface 2) +6 Q on the inner surface, +4 Q on the outer surface 3) +7 Q on the inner surface, +3 Q on the outer surface 4) +8 Q on the inner surface, +2 Q on the outer surface Two identical sheets of metallic foil are separated by d and capacitance of the system is C and is charged to a potential difference E. Keeping the charge constant, the separation is increased by l. Then the new capacitance and potential difference will be: 1)

75.

0 A ,E d

2)

0 A ,E (d l )

3)

0 A A l l , 1 E 4) 0 , 1 E (d l ) d d d

n conducting plates are placed face to face as shown in figure. Distance between any two plates is d. Area of the plates is A,

A A A A , , ...... (n 1) . The equivalent 2 4 8 2

d

capacitance of the system is 1)

0A

2)

2n d

0A

3)

n

(2 1)d

0A

4)

n

( 2 2) d

0A n

(2 1)d 76.

77.

The adjoining figure shows two identical parallel plate capacitors connected to a battery with switch S close(4). The switch is now opened and the plates are filled with a dielectric of dielectric constant 3. The ratio of the total electrostatic energy stored in both the capacitors before and after the introduction of the dielectric is 1) 2 : 5 2) 3 : 5 3) 5 : 2 4) 5 : 3 A parallel plate capacitor has two layers of dielectric as shown in figure. This capacitor is connected across a battery, then 2 1 the ratio of potential difference across the dielectric layers 1) 4/3 2) 1/2 3) 1/3 4) 3/2 Five conducting plates are placed parallel to each other. Separation between them is d and area of each plate is A. Plate number 1, 2 and 3 are connected with each other and at the same time through a cell of emf E. The charge on plate number 1 is 1) E0A/d 2) E0A/2d 3) 2 E0A/d 4) zero A capacitor of capacity C1, is charged by connecting it across a battery of emf V 0. The battery is then removed and the capacitor is connected in parallel with an uncharged capacitor of capacity C 2. The potential difference across this combination is:

k 2k 6

78.

79.

1)

C2 V0 C1 C 2

[Type text]

2)

C1 V0 C1 C 2

3)

C1 C 2 V0 C2

4)

C1 C 2 V0 C1

Page

CAPACITOR

80.

81. 82.

The capacitance of a parallel plate capacitor is 16F. When a glass slab is placed between the plates, the potential difference reduces to 1/8 th of the original value. What is dielectric constant of glass ? 1) 4 2) 8 3) 16 4) 32 The equivalent capacity across M and N in the given figure is: 1) 5C/3 2) 2/3C 3) C 4) 3/2C Three plates of common surface area A are connected as shown. The effective capacitance between points P and Q will be:

0A d 3 0A 3) 2 d 1)

2) 4)

3 0 A d

2 0 A d

83.

A dielectric slab of dielectric constant K = 5 is covered from all sides with a metallic foil. This system is introduced into the space of a parallel plate capacitor of capacitance 10 F. The slab fills almost the entire space between the plates, but does not touch the plates. The capacitance will become nearly: 1) 2) zero 3) 2 pF 4) 50 Pf

84.

A capacitor of capacitance 1 F can withstand the maximum voltage 6 kV while a capacitor of

85.

capacitance 2.0 F can withstand the maximum voltage 4 kV. If the two capacitors are connected in series, then the two capacitors combined can take up a maximum voltage of: 1) 2.4 kV 2) 5 kV 3) 9 kV 4) 10 kV A spherical conductor of radius 2m is charged to a potential of 120 V. It is now placed inside another hollow spherical conductor of radius 6 m. Calculate the potential of bigger sphere if the smaller sphere is made to touch the bigger sphere: 1) 20 V 2) 60 V 3) 80 V 4) 40 V

86.

A parallel plate capacitor has two layers of dielectrics as shown in figure. Then the ratio of potential difference across the dielectric layers when connected to the battery is: 1)

K1 K2

2)

K1a K 2b

3)

K 2b K1a

4)

K 2a K 1b

LEVEL - III : 1.

A parallel plate capacitor C is connected to a battery and it is charged to a potential difference of V. Another capacitor of capacitance 2C is similarly charged to a potential difference of 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to positive of the other. The final energy of the configuration is A) zero

2.

B) 3/2 CV2

C) 25/6 CV2

D) 9/2 CV2

The amount of heat liberated when a capacitor of C farads charged to a potential difference of V volts is discharged through a resistor of R ohms is H joules. The same capacitor is now charged to a potential difference of 2V and discharged through a

[Type text]

Page

CAPACITOR

resistor of 2 R ohms, then heat liberated is A) 4H 3.

B) 2H

C) H

D) H/2

A capacitor of capacity C1 is charged to a potential V0. The electrostatic energy stored in it is U0. It is connected to another uncharged capacitor of capacitance C2 in parallel. The energy dissipated in the process is A)

C2 U0 C1 C 2

B)

C1 U0 C1 C 2

C C

2 C) 1 C C 2 1

2

U0

D)

C1C 2 U0 2(C1 C 2 ) 4.

5.

In the circuit shown in figure, if the key is turned so that instead of 1, 2 terminals 1, 3 are connected, then the heat liberated in resistor R is A)

1 C(1 ~ 2 ) 2 2

B)

1 C(1 2 ) 2 2

C)

1 1 C12 C 2 2 2 2

D)

1 1 C12 C 2 2 2 2

Three capacitors C1, C2, C3 are connected as shown in figure to one another and to points P1, P2, P3 at potentials V1, V2 and V3. If the total charge on the capacitors is zero, the potential of the point O is A)

C1V1 C 2 V2 C 3 V3 C1 C 2 C 3

B) V1 + V2 + V3

D) V1

C) 0 6.

A parallel plate capacitor having capacitance C has two plates of same area A and thickness t. Figure shows the charges available on the four surfaces of the plates. The potential difference V between the two plates is given by A)

7.

1 q 2 q3 2 C

q2 q3 C

B)

C)

1 q2 q4 2 C

q2 q4 C

D)

The plates of a parallel plate capacitor are separated by d cm. A plate of thickness t cm with dielectric constant k 1 is inserted and the remaining space is filled with a plate of dielectric constant k2. If Q is the charge on the capacitor and area of plates is A cm2 each, then potential difference between the plates is A)

8.

C C2 C V2 3 V3 1 C3 C1 C2

Q t dt 0 A k1 k 2

B)

4Q t dt A k1 k 2

C)

k 4Q k 1 2 A t dt

D)

Q k1 d t 0 A t k 2

Find the capacitance of a system of three parallel plates each of area A separated

[Type text]

Page

CAPACITOR

by distances d1 and d2. The space between them is filled with dielectrics of relative dielectric constants 1 and 2. The dielectric constant of free space is 0

9.

A)

1 2 0 A 1 d 2 2 d1

C)

1 2 A A D) 1 2 0 ( 1 d 1 2 d 2 ) 0 (1 2 ) d1 d 2

1 C V12 V22 4

B)

1 C V12 V22 4

C)

1 C V1 V2 2 4

D)

1 C V1 V2 2 4

One plate of a capacitor is connected to a spring as shown in the 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)

11.

1 2 0 A 1 d1 2 d 2

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 A)

10.

B)

6 0 E 2 Ad

3

B)

4 0 AE 2 d

3

C)

4 0 AE 3 2d

3

D)

2 0 AE d2

A slab of copper of thickness b is inserted in between the plates of parallel plate capacitor as shown in figure. The separation of the plates is d. If b = d/2, then the ratio of capacities of the capacitor after and before inserting the slab will be: A)

2:1

B) 2 : 1

C) 1 : 1

D) 1 :

2 12.

13.

Consider a parallel plate capacitor of capacity 10 F with air filled in the gap between the plates. Now one-half of the space between the plates is filled with a dielectric of dielectric constant 4, as shown in the figure. The capacity of the capacitor changes to: A) 25 F

B) 20 F

C) 40 F

D) 5 F

k=4

Consider the arrangement of three plates X, Y and Z each of area A and separation d. The energy stored when the plates are fully charged is: A)

0 AV 2 2d

[Type text]

B)

0 AV 2 d

Page

CAPACITOR

C)

2 0 AV 2 d

D)

3 0 AV 2 2d

MULTIPLE ANSWER OBJECTIVE TYPE QUESTIONS : 1.

2.

In the arrangement shown in Fig, the charge on capacitor C 2 is 1 C and on capacitorC3 is 2 C. If the capacitance of capacitor C2 is 1 F, then A) p.d across C1 is 2 V

B) Charge on C1 is 3 C

C) Energy stored in system is 4.5 J battery is 4.5 J

D) Energy supplied by

In the arrangement shown in Fig, all capacitors have equal capacities equal to 1 F. If the emf of the battery is 2 V A) The charge on capacitor C1 is 1 C B) The potential difference across C2 is 1 V C) The energy stored in C3 is 2 J

3.

In a parallel plate capacitor, the area of each plate is A and the plate separation is d. The capacitor carries a charge q and the force of attraction between the two plates is F. Then A) F q2

4.

D) Energy supplied by battery is 4 J

B) F d

C) F 1/A

D) F 1/d

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 the plates. The capacitor if given some charge. As x goes from o to 3d A) the magnitude of electric field remains the same B) the direction of electric field remains the same C) the electric potential increases continuously D) the electric potential increases at first, then decreases and again increases

5.

In the circuit shown, some potential difference is applied between A and B. If C is joined to D, A) no charge will flow between C and D B) some charge will flow between C and D C) the equivalent capacitance between C and D will not change D) the equivalent capacitance between C and D will change

6.

The two plates X and Y of a parallel-plate capacitor of capacitance C are given a charge of amount Q each. X is now joined to the positive terminal and Y to the negative terminal of a cell of emf = Q/C A) Charge of amount Q will flow from the positive terminal to the negative terminal of the cell through the capacitor B) The total charge on the plate X will be 2Q C) The total charge on the plate Y will be zero D) The cell will supply C2 amount of energy

[Type text]

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CAPACITOR

7.

A parallel-plate capacitor is charged from a cell and then disconnected from the cell. The separation between the plates is now doubled A) The potential difference between the plates will become double B) The field between the plates will not change C) The energy of the capacitor doubles D) Some work will have to be done by an external agent on the plates

8.

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

9.

A parallel-plate air capacitor of capacitance C0 is connected to a cell of emf and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it A) The potential difference between the plates decreases K times B) The energy stored in the capacitor decreases K times C) The change in energy is

1 10.

1 C02 (K - 1) 2

D) The change in energy is

1 C 02 2

1 K

When a charged capacitor is connected with an uncharged capacitor, then flows A) the change always flows from the changed to the unchanged capacitor B) a steady state is obtained after which no further charge flow occurs C) the total potential energy stored in the capacitors remains conserved D) the charge conservation always holds true

MATCHING TYPE QUESTIONS : A parallel-plate air capacitor of capacitance C0 is connected to a cell of emf and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it A) The potential difference between the plates decreases K times B) The energy stored in the capacitor decreases K times C) The change in energy is

1 C02 (K - 1) 2

1. List – I

[Type text]

List – II

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CAPACITOR

a) energy stored in capacitor by external source b) field inside the dielectric when is placed in between plates of capacitor battery still connected c) charge on capacitor by placing dielectric and battery connection is removed

a) remain same

d) force between two parallel plates when separation is increased

d) decreases

b) half of energy supplied by battery

1Q 2C

c)

2. List – I a) Energy stored in capacitor

List – II e) 40R

b) Capacity of spherical capacitor c) Force between plates of capacitor d) Energy density of electric field

f)

q2 2 0 AK

g)

1 0E2 2

h) q2/2C

PREVIOUS YEAR IIT QUESTIONS : 1.

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 the plates. The capacitor is given some charge. As one goes from 0 to 3d [IIT-JEE 1998] (a) The magnitude of the electric field remains the same (b) The direction of the electric field remains the same (c) The electric potential increases continuously (d) The electric potential increases at first, then decreases and again increases

2.

A parallel plate capacitor is charged to a potential difference of 50 V. It is discharged through a resistance. After 1 second, the potential difference between plates becomes 40 V. Then [Roorkee 1999] (a) Fraction of stored energy after 1 second is 16/25 (b) Potential difference between the plates after 2 seconds will be 32 V (c) Potential difference between the plates after 2 seconds will be 20 V (d) Fraction of stored energy after 1 second is 4/5

3.

Five identical plates each of area A are joined as shown in the figure. The distance between the plates is d. The plates are connected to a potential difference of V volts. The charge on plates 1 – 1 2 3 4 5 and 4 will be [IIT 1984] V

AV 2 0 AV . (a) 0 d d (c)

0 AV 2 0 AV . d d

+

AV 2 0 AV . (b) 0 d d (d)

0 AV 2 0 AV . d d

3F

B

3F

1F 1F

[Type text]

20 A

1F

100V

10

Page C

CAPACITOR

4.

In the figure below, what is the potential difference between the point A and B and between B and C respectively in steady state [IIT 1979] (a) VAB VBC 100V

(b) VAB 75V, VBC 25 V

(c) VAB 25 V, VBC 75 V (d) VAB VBC 50 V 5.

6.

Figure given below shows two identical parallel plate capacitors connected to a battery with switch S closed. V A B The switch is now opened and the free space between the plate of capacitors is filled with a dielectric of dielectric constant 3. What will be the ratio of total electrostatic energy stored in both capacitors before and after the introduction of the dielectric [IIT 1983] (a) 3 : 1

(b) 5 : 1

(c) 3 : 5

(d) 5 : 3

A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is connected to another battery and is charged to potential difference 2V. The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is [IIT 1995] (a) Zero

7.

(b)

25CV 2 6

(c)

3CV 2 2

(d)

In an isolated parallel plate capacitor of capacitance C, the four surface have charges Q1 , Q2 , Q3 and Q4 as shown. The

Q3

Q1 Q2

potential difference between the plates is [IIT-JEE 1999]

8.

(a)

Q1 Q2 Q3 Q4 Q Q3 (b) 2 2C 2C

(c)

Q 2 Q3 2C

(d)

9CV 2 2

Q4

Q1 Q4 2C

For the circuit shown, which of the following statements is true [IIT-JEE 1999] (a) With S1 closed, V1 15V, V2 20 V

S1

(b) With S3 closed V1 V2 25V

V1=30V + –

V2=20V S3 + –

C1=2pF

C2=3pF

S2

(c) With S1 and S2 closed V1 V2 0 (d) With S1 and S3 closed, V1 30V, V2 20V 9.

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

q +

–

+

–

+

–

+

–

+

(a) Zero

[Type text]

(b) q / 2

–

A

[IIT-JEE (Screening) 2001] (c) q

s

B

(d)

Page

CAPACITOR

2q

10.

Point charge q moves from point P to point S along the path PQRS (figure shown) in a uniform electric field

E pointing co-parallel to the positive direction of the X axis. The coordinates of the points P , Q, R and S are (a, b, 0), (2a, 0, 0), (a, b, 0) and (0, 0, 0) respectively. The work done by the field in the above process is given by the expression [IIT 1989]

P S

Q

X

R

(b) qEa

(a) qEa (c)

E

(d) qE [(2a)2 b 2 ]

qEa 2

ASSERTION & REASON : Read the assertion and reason carefully to mark the correct option out of the options given below : (a) If both assertion and reason are true and the reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not the correct explanation of the assertion. (c) If assertion is true but reason is false. (d) If assertion is false but reason is true. 1.

Assertion : If three capacitors of capacitance C1 < C2 < C3 are connected in parallel then their equivalent capacitance Cp > Cs Reason

2.

:

1 1 1 1 C p C1 C2 C3

Assertion : If the distance between parallel plates of a capacitor is halved and dielectric constant is made three times, then the capacitor becomes 6 times. Reason material.

3. Assertion beam.

: Capacity of the capacitor does not depend upon the nature of the : Dielectric breakdown occurs under the influence of an intense light

Reason : Electromagnetic radiations exert pressure. 4. Assertion varied on it.

: The capacity of a given conductor remains same even if charge is

Reason : Capacitance depends upon nearly medium as well as size and shape of conductor. 5. Assertion : The whole charge of a conductor cannot be transferred to another isolated conductor.

[Type text]

Page

CAPACITOR

Reason 6.

: The total transfer of charge from one to another is not possible.

Assertion : Conductors having equal positive charge and volume, must also have same potential. Reason : Potential depends only on charge and volume of conductor.

7. Assertion : The lightening conductor at the top of high building has sharp pointed ends. Reason : The surface density of charge at sharp points is very high resulting in setting up of electric wind. 8.

Assertion : Circuit containing capacitors should be handled cautiously even when there is no current. Reason

9.

: The capacitors are very delicate and so quickly break down.

Assertion : The tyres of aircraft's are slightly conducting. Reason : If a conductor is connected to ground, the extra charge induced on conductor will flow to

ground. 10.

Assertion : A bird perches on a high power line and nothing happens to the bird. Reason

11.

: The level of bird is very high from the ground.

Assertion : Two capacitors are connected in series to a battery. If the dielectric medium is inserted between the plates of one of the capacitors, the energy stored in the system will increase

Reason : On inserting the dielectric medium, the capacity of the capacitor increases *****

K

E

Y

S

SINGLE ANSWER TYPE QUESTIONS LEVEL – I

LEVEL – II 1 2 2 1 3 4 1 3 6 1 4

2 2 2 2 3 4 2 4 6 2 3

3 1 2 3 3 4 3 3 6 3 4

[Type text]

4 2 2 4 3 4 4 4 6 4 3

5 4 2 5 3 4 5 4 6 5 4

6 3 2 6 3 4 6 2 6 6 1

7 1 2 7 1 4 7 3 6 7 2

8 1 2 8 1 4 8 3 6 8 1

9

1

1

1

1

1

1

1

1

1

1

2

3 2 9 3 4 9 3 6 9 1

0 3 3 0 2 5 0 3 7 0 4

1 2 3 1 1 5 1 4 7 1 4

2 3 3 2 3 5 2 2 7 2 4

3 3 3 3 3 5 3 2 7 3 3

4 3 3 4 1 5 4 4 7 4 3

5 4 3 5 4 5 5 2 7 5 3

6 2 3 6 3 5 6 3 7 6 2

7 2 3 7 2 5 7 2 7 7 4

8 1 3 8 4 5 8 2 7 8 4

9 3 3 9 3 5 9 4 7 9 2

0 2 4 0 4 6 0 3 8 0 2

Page

CAPACITOR

8 1 1

8 2 4

8 3 1

8 4 3

8 5 1

8 6 4

LEVEL – III 1

2

3

4

5

6

7

8

9

10

11

12

13

B

A

A

B

A

A

B

A

C

B

B

B

B

MULTIPLE ANSWER OBJECTIVE TYPE QUESTIONS 1

2

3

4

5

6

7

8

9

10

ABC

ABD

AC

BC

AC

ABCD

ABCD

AD

ABD

ABD

MATCH THE COLUMN : 01. a-bc, b-a, c-a, d-a

02. a – h, b –

e, c – f, d – g PRIVIOUS YEAR IIT QUESTIONS 1

2

3

4

5

6

7

8

9

10

BC

AB

C

C

C

C

C

D

A

B

ASSERTION & REASON 1

2

3

4

5

6

7

8

9

10

11

C

B

B

A

D

D

A

C

B

C

C

***

[Type text]

Page

View more...
CAPACITOR Y L L A with BUS Capacitance, Parallel Plate S capacitor and without dielectric, capacitors in series and parallel, Energy stored in a capacitor.

POSITION VECTOR : Capacitor is a device for storing electric charge and energy. It consists of a pair of conductors carrying equal and opposite charges (generally). Magnitude of this charge is known as the charge on the capacitor. Potential difference (V) between the two conductors is proportional to the charge on the capacitor (Q). Q V; Q = CV Here the proportionality constant C is known as the capacitance of the capacitor. Capacitance depends on the size and shape of the plates and the material between them. The SI unit of capacitance is farad (F).

PARALLEL PLATE CAPACITORS : A parallel plate capacitor consists of two equal flat parallel metal plates facing each other and separated by a dielectric of electric permittivity .The plates may be square, rectangular or circular in shape. For calculating the capacity of a capacitor, we first calculate electric field at a point

dV dr

between the plates and then using relation E

compute the potential difference

between the plates. Finally dividing the magnitude of charge (given to one plate) by the potential difference between the plates, we get the capacity. In case of parallel plate capacitor as shown in figure. The field at P

E

2 2

or,

or,

V

[ as E d

dV

dx

dV ] dx

i.e., V

0

d

q A A V d / d

So,

C

or

KA C 0 d

V x0

dV dx 0

d

[as

P

K] 0

We can derive the capacity of a spherical capacitor in a similar

[Type text]

V0

xd

SPHERICAL CAPACITOR : q 1 1 ; way V= 4 0 a b

b q a

E

CAPACITOR

4 0

q C= = 1 1 V

a

=

b

4 0 ab ba

If the radius of the outer sphere tends to infinity, b , the capacitance reduces to C = 4 0 a which is the capacitance of the isolated sphere.

CYLINDRICAL CAPACITOR : E=

a r b 2 0 r

Q

a

V = Va Vb = - Edr b

ln

or

V=

a

2 0

b l

b a

2 0 q b C= = = n V V a ILLUSTRATION : 01 A parallel plate capacitor has plates of area 200cm2 and separation between the plates 1.00mm. What potential difference will be developed if a charge of 1.00nC (i.e., 1.00 x 10-9C) is given to it. Now if separation between the plates is increased to 2.00mm, what will be the new potential difference? SOLUTION : The capacitance of the capacitor is C

0A d

F 200 x 10 -4 m 2 x m 1 x 10 -3 m

= 8. 85 x 10-12 = 0.177 x 10-9F = 0.177nF. The potential difference between the plates is

V

Q 1nC 5.65 volts. C 0.177 nF

If the separation is increased from 1.00mm to 2.00mm the capacitance is decreases by a factor of 2. If the charge remains the same, the potential difference will increase by a factor of 2. Thus, the new potential difference will be 5.65volts x 2 = 11.3 volts.

COMBINATION OF CAPACITORS SERIES COMBINATION : Capacitors connected as shown in the figure are said to be connected in series. In series combination the charges on individual

Q

Q C1 V1

[Type text]

Q

Q C2

V2

Q

Q

C3 V3

CAPACITOR

condensers are equal and the total p.d. across the combination is to shared by the capacitors. Q = C 1V1 C 2V2 C 3V3 and V = V1 V2 V3 Effective capacitance of the combination C can be found from the relation.

1 1 1 1 = + + C C C C 1 2 3

PARALLEL COMBINATION : In this combination p.d. across each of capacitors is same but the charge supplied at points A and B is shared by capacitors. V=

Q1

Q2 Q3 Q1 = = and total C1 C2 C3

charge

B

Q3 - Q3

C = C1 + C 2 + C 3 ILLUSTRATION : 02 Find the equivalent capacitance between points A and B of the circuit shown, each capacitance = C

Q2 - Q2

A

Q = Q1 + Q 2 + Q 3

- Q1

A

B

SOLUTION : Equivalent circuit is Each branch equivalent capacitance is There are four branches in parallel C eq =

C C C C + + + = 2C 2 2 2 2

C . 2

B

A

ILLUSTRATION : 03 The figure shown is a system of parallel conductors. Each plate is of equal area A and equally separated by d. Find the equivalent capacitance of the system between a and b SOLUTION : By joining the points of same potential, the arrangement of conductors may be reduced as shown in figure. If the capacitance between two successive plates is

A given by C 0 then, the equivalent capacitance of the d system is given by C eq

DIELECTRICS :

[Type text]

3C 3 0 A 2 2 d

a

2 1 4 3

2 3 b

CAPACITOR

When a dielectric is introduced between conductors of a capacitor, its capacitance increases. A dielectric is characterized by a constant 'K' called dielectric constant.

DIELECTRIC CONSTANT : When a dielectric is placed in an external electric field, polarization occurs and it develops an electric field in opposition to the external one. As a result total field inside it decreases. If E is the total field inside the dielectric when it is placed in an external field

E 0 , then its dielectric constant 'K' is given as K =

E0 ( k 1) E

If a dielectric completely occupies the space between the conductors of a capacitor its capacitance increases 'K' times. Hence in presence of a dielectric with dielectric constant ' K ', the capacitance of a parallel plate capacitor =

K 0 A d

ENERGY STORED IN A CAPACITOR : The energy stored in a capacitor is equal to the work done to charge it. Let q be the instantaneous charge on either plate of the capacitor and the potential difference between the plate is V=

q . The work done to transfer an infinitesimal charge dq from the negative C q dq C

plate to the positive plate is dW = Vdq =

[The charge moves through the wires, not across the gap between the plates] Q

W = total work done to transfer charge Q =

q

C dq = 0

QV 1 Q2 = = C V2 2 2 2C 1 2 3 4

This work done is stored as electrostatic energy U=

1 1 0A C V2 = 2 2 d

E 2 d 2 = 12

0 E 2 (Ad)

a

ie.,

b

Energy density (u) = energy per unit Volume =

1 0 E2 2

If dielectric is introduced then U =

1 K 0 E2 2

This energy is stored in a capacitor in the electric field between its plates.

FORCE ON A DIELECTRIC IN A CAPACITOR : Let us consider a small displacement dx of the dielectric as shown in figure, keeping the net force on it always zero.

Welectrostatic + [Where

WF

WF = 0

denotes

the

work

done

in

F x [Type text]

dx

CAPACITOR

displacement dx]

WF = - Welec. = U

- F dx =

F=

Q2 2

1 Q2 Q2 d = dc C 2 2C 2 dc dx

=

1 dc V2 2 dx

2C [Considering capacitor has battery connected to it, i.e., V = Q/C ]

ILLUSTRATION : 04 Two capacitors of capacitances 20pf and 50pf are connected in series with a 6-volt battery, find (A) The potential difference across each capacitor (B) The energy stored in each capacitor SOLUTION : (A)

Equivalent capacitance

C=

C1C 2 50 x 20 = C1 C 2 50 20

Charge on C1 = charge on C 2 =

=

100 pF 7

100 600 x6= 7 7

20pF

50pF

pC Potential difference across C1 50pC =

600 = 7 x 50

6V

1.71 V and across C 2 20pC = (B)

600 = 4.28 V 7 x 20

1 x 50x 1.71 2 = 73.5 pJ 2 1 Energy in C 2 = E 2 = x 20 x 4.28 2 = 184 pJ 2 Energy in C1 = E1 =

ILLUSTRATION : 05 A 5F capacitor is charged to 12 volt. The positive plate of this capacitor is now connected to the negative terminal of a 12 V battery and vice versa. Calculate the heat developed in the connecting wires. SOLUTION : When capacitor is connected with battery the charge appears on one plate be Q = CV and - Q on the other plate. If the capacitor is now disconnected and connected to the same battery again with opposite polarity then - Q appear on first plate and + Q on second plate.

Total charge flown from battery is 2Q W= charge x potential = 2QV

Q = CV

W = 2C V 2

W = 2x5x 10 6 x 12 2

[Type text]

CAPACITOR

= 1.44 mJ ILLUSTRATION : 06 A capacitor stores 50C charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of 100C flows through the battery. Find the dielectric constant of the material inserted. SOLUTION : Initial charge = 50C = Q1 Amount of charge flows = 100C

With dielectric total charge = (50+100) =150C Initial capacity C =

Q 50C = V V

Final capacity C ' = K =

150C Q' = V V

150 / V C' = =3 50 / V C

ILLUSTRATION : 07 In the above circuit, find the potential difference across AB. SOLUTION : Let us mark the capacitors as 1, 2, 3 and 4 for identification. As is clear, 3 and 4 are in series, and they are in parallel with 2. The 2,3, 4 combination is in series with 1.

C 34

C 3 .C 4 4f , C3 C 4

C eq

8 12 4.8f , 8 12

C 2,3,4 8 4 12f

The 'q' on 1 is 48C, thus V1

q C eq V 4.8 10 48C

q 6V c

48c V1 8F 6V.

VPQ 10 6 4V

By symmetry of 3 and 4, we say, VAB 2V. ILLUSTRATION : 08 What is VA VB in the arrangement shown? What is the condition such that VA VB 0 SOLUTION : Let charge be as shown (Capacitors in series have the same charge)

[Type text]

q

q'

CAPACITOR

Take loop containing C 1 , C 2 and

E

C1C 2 q q E 0 q E C1 C 2 C1 C 2 From loop containing C 3 , C 4 and E Similarly,

q' q' E 0 C3 C 4

C3C 4 C3 C 4

q' E

Now, VA VB

q q' C2 C4

C3 C1 C1 C 2 C 3 C 4

= E.

C1.C 4 C 3 .C 2 VA VB E. C1 C 2 . C 3 C 4 For VA VB 0 C1C 4 C 2 C 3 0 or

C1 C 3 C2 C4

S

ILLUSTRATION : 09 A 8F capacitor C1 is charged to V0 = 120volt. The charging battery is then removed and the capacitor is connected in parallel to an uncharged 4F capacitor C2. (A) What is the potential difference V across the combination? (B) What is the stored energy before and after the switch S is thrown ?

V0

C1

C2

SOLUTION : (A) Let q 0 be the charge on C1 initially Then q 0 C1V0 when C1 is connected to

C 2 in parallel, the charge q 0 is distributed between C1 and C 2 . Let q1 and q 2 be the charges on C1 and C 2 respectively. Now let V be the potential difference across each condenser. Now q 0 q1 q 2 or C1V0 C1V C 2 V

V

C1 8F 120V V0 C1 C 2 8F 4F = 80volt.

(B) Initial energy stored

U0

[Type text]

1 1 C1V02 (8 x 10-6) (120)2 2 2

CAPACITOR

= 5 .76 x 10-2joule Final energy stored U= =

1 1 C1V 2 C2V 2 2 2 1 1 (8 x 10-6) (80)2 + (4 x 10-6) (80)2 2 2

= 3.84 x -2

10 joule. Final energy is less than the initial energy. The loss of energy appears as heat in connecting wires. ILLUSTRATION : 10 From the given figure find the value of the capacitance C if the equivalent capacitance between points A and B is to be 1F. All the capacitances are in F.

C1

A C

SOLUTION : The capacitors C3 and C4 are in parallel, therefore their resultant capacity C8 is 4. The capacitors C5 and C6 are in series, therefore, their resultant capacity C 9 is 4. These are shown in figure (A)

C2 C8

A

Now the capacitor C2 and C8 are in series. Their resultant capacity C10 is

8 . Capacitors C 7 and C 9 are 3

in parallel. Their resultant capacity C11is 8. These are shown in figure. (B) C1 and C11 are in series. The equivalent capacitance is 8/9. The parallel combination of 8/3 and 8/9 gives a resultant capacitance 32/9 as shown in figure. (C)

1 1 9 1 C 32

C

or

1 23 C 32

32 F 23

ILLUSTRATION : 11 Five identical conducting plates 1, 2, 3, 4 and 5 are fixed parallel to and equidistant from each other as shown in figure. Plates 2 and 5 are connected by a conductor while 1 and 3 are joined by another conductor. The junction of 1 and 3 and the plate 4 are connected to a source of constant e.m.f.

V0 . Find

(i) The effective capacity of the system between the

[Type text]

8 1 4 4

C9 C7

4

B

CAPACITOR

terminals of the source (ii) The charge on plates 3 and 5 Given d = distance between A = area of either face of each plate.

any

two

successive

plates

and

SOLUTION : (i) The equivalent circuit is shown in figure (B). The system consists of four

capacitors i.e., C12 , C 32 C 34 and C 54 . The capacity of each capacitor is K 0

A C 0 The d

effective capacity across the source can be calculated as follow: The capacitors C12 and C32 are in parallel and hence their capacity is C 0 + C0 = 2C0. The capacitor C54 is in series with effective capacitor of capacity 2C 0. Hence the resultant capacity will be

C 34 is again in parallel. Hence the effective capacity

Further = C0 (ii)

C0 x 2 C0 C 0 2C 0

C 0 x 2C 0 5 5 A C 0 K 0 C 0 2C 0 3 3 d

Charge on the plate 5 = charge on the upper half of parallel combination

2 2 K 0 AV0 C0 d 3 3

Q 5 V0

Charge on plate 3 on the surface facing 4

V0 C 0

K 0 AV0 d

Charge on plate 3 on the surface facing 2 = [Potential difference across (3 - 2)] C 0 =

Q3

V0

C0 AV0 C 0 K 0 C 0 2C 0 3d

K 0 AV0 K 0 AV0 AV0 = K 0 d d 3d

1 4 A 1 3 3 K 0 d V0

ILLUSTRATION : 12 (A) Find the effective capacitance between points X and Y in the

given figure.

Assume

that

C 2 10 F and other

capacitors are 4F each. (B) Find the capacitance of a system of identical capacitors between points A and B shown in figure. SOLUTION : (A) The circuit is redrawn in figure as the two arms are

[Type text]

CAPACITOR

balanced, no current flows through C 2 , C 3 and C 4 are in series, hence their equivalent capacitance

=

2F

Similarly

the

equivalent

capacitance

of

C1 and

C 5 = 2F.

Corresponding to points X and Y these two are in a parallel combination. Hence the effective capacitance between X and Y is 2 + 2 = 4F. (B) The arrangement of capacitors shown in figure is equivalent to the arrangement shown in figure. The arrangement is connected in parallel. Hence equivalent capacitance C is given by C = C1 + C2 + C3

WORKED OUT OBJECTIVE PROBLEMS : EXAMPLE : 01 A parallel plate capacitor is connected across a 2V battery and charged. The battery is then disconnected and glass slab is introduced between the plates. Which of the following pairs of quantities decrease? (A) Charge and potential difference (B) potential difference and energy stored (C) energy stored and capacitance (D) capacitance and charge Ans: (B) SOLUTION : The introduction of a dielectric slab increases the capacitance. The charge remains unchanged. Potential difference and energy stored decreases. EXAMPLE : 02 Three capacitors of capacitances 3F, 9F and 18F are connected once in series and then in parallel. The ratio of equivalent capacitances in the two cases(C S/CP) will be (A) 1 : 15 (B) 15 : 1 (C) 1 : 1 (D) 1 : 3 Ans: (A) SOLUTION :

C P 3 9 18 30F 1 1 1 1 1 C S 3 9 18 2

CS 2F

Now

CS 2F 1 C P 30F 15

EXAMPLE : 03 A number of capacitors each of capacitance 1F and each one of which get punctured if a potential difference just exceeding 500volt is applied, are provided. Then an arrangement suitable for giving a capacitor of capacitance 2F across which 3000 volt may be applied requires at least

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CAPACITOR

(A) 18 component capacitors (C) 72 component capacitors Ans: (C) SOLUTION :

(B) 36 component capacitors (D) 144 component capacitors

3000 6 500 1 The capacitance of series combination. = F 6 Number of capacitors required in series =

To obtain a capacitor of 2F, we should use 12 such combinations. Total number of capacitors required = 12 x 6 = 72 EXAMPLE : 04 A capacitor of capacitance 1F withstands a maximum voltage of 6kV, while another capacitor of capacitance 2F, the maximum voltage 4kV. If they are connected in series, the combination can withstand a maximum of (A) 6kV (B) 4kV (C) 10kV (D) 9kV Ans: (D) SOLUTION : When the two condensers are connected in series.

Q

C

2 x1 2 F 2 1 3

and

2 E 3 The potential of condenser C1 is given by

V1

Q 2 E 6kV E 9kv C1 3

v2

Q E E 12KV c2 3

To avoid break down E 9KV EXAMPLE : 05 Seven capacitors each of capacitance 2F are to be connected to obtain a

10 F. Which of the following combination is possible. 11

capacitance of

(A) 5 in parallel 2 in series (B) 4 in parallel 3 in series (C) 3 in parallel 4 in series (D) 2 in parallel 5 in series Ans: (A) SOLUTION : 5 capacitors in parallel gives 5 x 2 F = 10F capacity. Further, two capacitors in series gives a capacity 1F. When the two combinations are connected in series, they give

10 x 1 10 = F. 10 1 11

a resultant capacitance

EXAMPLE : 06 Condenser A has a capacity of 15F when it is filled with a medium of dielectric

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CAPACITOR

constant 15. Another condenser B has a capacity 1F with air between the plates. Both are charged separately by a battery of 100V. After charging, both are connected in parallel without battery and the dielectric material being removed. The common potential now is (A) 400V (B) 800V (C) 1200V (D) 1600V Ans: (B) SOLUTION : Charge on capacitor A is given by q1 C1 x V = (15 x 10-6) (100) = 15 x 10-4C Charge on capacitor B is given by q2 = C2 x V = (1 x 10-6) (100) = 10-4C Capacity of condensers A after removing dielectric 10 -6 C1 15 x 15 K

C'

1 F

Now when both condenser are connected in parallel their capacity will be 1F + 1F = 2F Common potential V = =

q C

15 x 10 -4 (1 x 10 -4 ) 800V . 2 x 10 -6

EXAMPLE : 07 Two capacitors 2F and 4F are connected in parallel. A third capacitor of 6 F capacity is connected in series. The combination is then connected across a 12V battery. The voltage across 2F capacity is (A)2V (B) 6V (C) 8V (D) 1V Ans: (B) SOLUTION : Resultant capacitance of condensers of capacity 2F and 4F when connected in parallel. C' 2 4 6 F This is connected in series with a capacitor of capacity 6F in series. The resultant capacity C is given by

1 1 1 1 or C 3 F C 6 6 3 Charge on combination

q = (3 x 10-6) x (12) = 36 x 10-6C

Let the charge on 2F capacitor be q1 , then

q1 q q1 q or q1 3 2 4 Now potential across 2F condenser

[Type text]

q1 12 x 10-6C =

q1 2 x 10 -6

12 x 10 -6 2 x 10 -6

6V

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EXAMPLE : 08 The capacitance of the system of parallel plate capacitor shown in the figure isA

2 0 A 1 A 2 (A) A1 A 2 d

2 0 A 1 A 2 (B) A 2 A1 d

A (C) 0 1 d

A (D) 0 2 d

1

d A2

Ans : (C) SOLUTION : Since the electric field between the parallel charge plates is uniform and independent of the distance, neglecting the fringe effect, the effective area of the plate of area A 2 is A1. Thus the

A capacitance between the plates is C 0 1 d

(C)

A1 0

d

E

A2

EXAMPLE : 9 The charge flowing across the circuit on closing the key K is equal to

C V 2

(A) CV

(B)

(C)2CV

(D) Zero

SOLUTION : When the key K is kept open the charge drawn from the source is Q C 'V Where C' is the equivalent capacitance given by C'

C 2

C V 2

Therefore Q =

Whey the key K is closed, the capacitor 2 gets short circuited and the charge in the circuit

Q1 CV

Charge flowing is Q Q1 Q

(B)

C V 2

EXAMPLE : 10 The figure shows a spherical capacitor with inner sphere earthed. The capacitance of the system is (A)

4 0 ab ba

(C) 4 0 b a

[Type text]

(B)

4 0 b 2 ba

a

b

(D) None of these

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SOLUTION : Let V be potential of the outer sphere. Thus we can consider two capacitors, one between the outer sphere and inner sphere and the other between outer sphere and infinity. Thus,

C1 4 0

ab ba

C 2 4 0 b C 4 0

C

(B)

a

b

ab 4 0 b ba

C1

C2

4 0 b 2 ba

EXAMPLE : 11 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 (B) V

(A) V

Q C

(C) V

Q 2C

(D) V

Q , if V < C

CV SOLUTION : In the figure given below, left X and Y be positive and negative plates. After charging from the cell, the inner faces of X and Y have charges CV , as shown in (A). The outer surfaces have no charge. When charge Q is given to X, let the inner faces of X and Y have charges q Then, by the principle of charge conservation, the outer

faces have charges Q CV q for X and q CV for Y, as shown in (B). Now, the outer faces must have equal charges. Q CV q q CV or 2q 2CV Q or q CV

Q 2

Potential difference

q Q V C 2C

(C) EXAMPLE : 12 In an isolated parallel - plate capacitor of capacitance

[Type text]

Q1

Q3

Q2

Q4

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C, the four surfaces have charges Q1 , Q2 , Q3 and Q4 as shown. The potential difference between the plates is (A)

Q Q3 Q1 Q 2 Q 3 Q 4 (B) 2 2C 2C

(C)

Q 2 Q3 2C

(D)

Q1 Q 4 2C

SOLUTION : Plane conducting surfaces facing each other must have equal and opposite charge densities. Here, as the plate areas are equal, Q2 = - Q3 The charge on a capacitor means the charge on the inner surface of the positive plate-(in this case Q2) Potential difference between the plates = charge on the capacitor capacitance. Potential difference =

Q 2 2Q 2 Q 2 Q 2 Q 2 Q 3 C 2C 2C 2C

(C) ***

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SINGLE ANSWER OBJECTIVE TYPE QUESTIONS : LEVEL - 1 : 1.

2.

3.

A capacitor of capacitance C is charged to potential difference V 0 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) V0 B) V0 + (Q/C) C) V0 + (Q/2C) D) V0 - (Q/C) A dielectric of dielectric constant 3 fills up three fourths of the space between the plates of a parallel plate capacitor. The percentage of energy stored in the dielectric is A) 25% B) 50% C) 75% D) 100% A parallel plate capacitor having capacitance C 0 is connected to a battery of emf E. It is then disconnected from the battery and a dielectric slab of dielectric constant k completely filling the air gap of the capacitor is inserted in it. If U indicates the change in energy, then A) U = 0

B) U =

1 0E2 (k - 1) 2

C) U =

1 1 0E2 1 k 2

D)U

1 1 1 0E2 k 2 4.

5.

The work done in increasing the voltage across the plates of the capacitor from 5V to 10V is W. The work done in increasing the voltage from 10V to 15V will be A) W B) 4/3 W C) 5/3 W D) 2W A parallel plate capacitor of plate area A and plate separation 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 work done on the system, in questions, in the process of inserting the slab, then A) Q W

6.

7.

8.

0 AV d

0 AV 2 2d

B) Q

0 KAV d

C) E =

V Kd

D)

1 1 K

When the capacitance in an oscillator circuit of frequency f is increased nine times, the frequency of the oscillator is reduced to: A) f/9 B) f/6 C) f/4 D) f/3 64 small drops of water having the same charge & same radius are combined to form one big drop. The ratio of capacitance of big drop to small drop is: A) 4 : 1 B) 1 : 4 C) 2 : 1 D) 1 : 2 A parallel plate condenser is connected to a battery. The plates are pulled apart with a uniform speed. If x is the separation between the plates, then the time rate of charge of the electrostatic energy of the condenser is proportional to

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CAPACITOR

9.

A) x2 B) x C) 1/x D) 1/x2 A spherical condenser has inner and outer spheres of radii a and b respectively. The space between the two is filled with air. The difference between the capacities of two condensers formed when outer sphere is earthed and when inner sphere is earthed will be B) 4 0 a

A) zero a)]

10.

ab ba

D) 40

12.

13.

14.

15.

D) 4 0 a [b/(b -

The sphere shown in the figure are connected by a conductor. The capacitance of the system is: A) 4 0

11.

C) 4 0 b

B) 4 0 a

C) 4 0 b

a2 ba

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 the plates. The capacitor is given some charge. As X goes from 0 to 3d. A) the electric potential increases at first, then decreases and again increases B) the electric potential increases continuously C) the direction of the electric field remains the same D) the magnitude of the electric field remains the same Two similar conducting balls are placed near each other in air. The radius of each ball is r and the separation between the centers is d (d >> r). The capacitance of two balls system when they are connected by a wire is: A) 80r B) 40r C) 40r loge (r/d) D) 4 loge 0 (r/d) A hollow sphere of radius 2R is charged to V volt and another smaller sphere of radius R is charged to V/2 volt. Then the smaller sphere is placed inside the bigger sphere without changing the net charge on each sphere. The potential difference between the two spheres would be A) 3V/2 B) V/4 C) V/2 D) V A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates which results in: A) reduction of charges on the plates and increase of potential difference across the plates B) increase in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates C) decrease in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates D) none of the above Force acting on a charged particle kept between the plates of a charged condenser is F. If one of the plates of the condenser is removed, force acting on the same particle will become:

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CAPACITOR

16.

17.

A) zero B) F/2 C) F D) 2F A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with this capacitor are given by Q 0, V0, E0 and U0 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 in the previous ones as: A) Q > Q0 B) V > V0 C) E > E0 D) U > U0 Figure shows two capacitors connected in series and7joined to a battery. The graph shows the variation in potential as one moves from left to right on the branch containing the capacitors: A) C1 > C2 B) C1 = C2 C) C1 < C2 D) the information is not sufficient to decide the relation between C 1 and C2

LEVEL - II : 1.

2.

3.

4.

Identical charged 103 oil drops each of radius 0.9nm are combined to form a single drop. The electrical capacity of the drop is 1) 0.1 PF 2) 1 PF 3) 10 PF 4) 100 PF A parallel plate condenser is charged end isolated. The distance between the plate is increased by 2mm and a dielectric slab of thickness 3 mm is introduced between the plates. If the potential difference between the plates remains same, the dielectric constant of the dielectric slab is 1) 2 2) 3 3) 4 4) 5 Two identical parallel plate condensers are connected in series. A cell of e.m.f of 20V is connected between their ends. A dielectric slab of constant 4 is placed between the plates of one of the condensers. The potential difference across condenser with dielectric slab is 1) 4V 2) 10V 3) 16V 4) 18V Four identical parallel metal plates each of area A are placed with separation ‘d’ between adjacent plates as shown. The capacitance between the plates P and Q is

3 0 A d 2 0 A 3) d 1)

5.

6.

7.

2 0 A 3d 0 A 4) 2d 2)

Two spherical conductors of radii 3cm and 6cm are in contact. A charge 10 -9 is given to them. The potential of the smaller sphere is 1) 67V 2) 33V 3) 50V 4) 100V A parallel plate condenser is charged and isolated. The energy stored by the condenser is E. The separation of the plate is doubled and then the space is completely filled with a dielectric of constant 5. The energy stored by the condenser now is 1) E 2) 0.8E 3) 0.4E 4) 2.5E Two condensers charged to potentials 50V and 80V are connected in parallel, the common potential is 60V. The capacities of the condensers are in the ratio of 1) 2 : 1 2) 1 : 2 3) 3 : 4 4) 4 : 3

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CAPACITOR

8.

The capacitor of 4F charged to 50V is connected to another capacitor of 2F charged to 100V. The total energy of the combination is 1)

9.

4 x102 J 3

2)

3 x102 J 2

3) 3 x102 J

4)

8 x102 J 3

The radii of two charged metal spheres are 5cm and 10cm both having same charge 75c. If they are connected by a wire, the quantity of charge transferred through the wire is 1) 75c 2) 50c 3) 25c 4) 15c

10.

11.

Two identical capacitors have equivalent capacity of 2F when they are connected in series. If they are connected in parallel and charged to a potential of 200V, the energy stored in the system is 1) 18 x 10-4 J 2) 18 x 10-4 J 3) 0.16 J 4) 0.36 The capacity of a parallel plate condenser is C. When half the space between the plates is filled with a slab of dielectric constant K as shown in the figure. If the slab is removed from the condenser, then the capacity of the condenser becomes 1) 3)

12.

13.

14.

15.

16.

2KC K 1

K 1 C 2K

2) 4)

K 1 C 2K KC 2K 1

Two parallel plate capacitors C and 2C are connected in parallel and charged to P.D ‘V’. The battery is then disconnected and the region between the plates of the capacitor ‘C’ is completely filled with a material of dielectric constant ‘3’. The P.D. across the capacitors now becomes 1) V 2) 3V 3) 3V/5 4) 4V/5 A condenser of capacity 10F is connected between the terminals of a battery of potential difference 100V and is charged. The amount of work done by the battery to charge the condenser is 1) 0.05 J 2) 0.025 J 3) 0.1 J 4) 0.0125 J Two condensers of capacities C1 = 5 F – 100V and C2 = 10F – 100V are connected in series. The maximum potential difference that may be applied between them without damaging the condensers is 1) 200 V 2) 175 V 3) 150 V 4) 300 V ‘n’ identical charged spheres combine together to form a large sphere. The ratio of the potential of small sphere to potential of large sphere is 1 : 9. The ratio of the energy stored in small sphere to the energy stored in the large sphere is 1) 1 : 3 2) 1 : 27 3) 1 : 81 4) 1 : 243 The capacity of parallel plate condenser with air between the plates is 10F. If the space between the plates is completely filled with two dielectric slabs each of thickness is equal to half of the separation between the plates, whose dielectric constants are 3 and 2. What will be the percentage change in capacity of the condenser? 1) 133.3 % 2) 140 % 3) 166.6 % 4) 280 %

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17.

18.

19.

20.

21.

The work done in charging a capacitor from 5V to 10V is W. The work done in charging the capacitor from 10V to 15V is 1) 4 W/3 2) 5 W/3 3) 2 W/5 4) 5 W/2 The voltages across C1 and C2 are in the ratio 2 : 3. When C2 is completely filled with paraffin, the voltage ratio became 3 : 2. The dielectric constant of paraffin is 1) 2.25 2) 13/6 3) 27/8 4) 6 A parallel plate capacitor of capacity 5F and plate separation 6cm is connected to a 1V battery and is charged. A dielectric of dielectric constant 4 and thickness 4cm is introduced into the capacitor. The additional charge that flows into the capacitor from the battery is 1) 2C 2) 3C 3) 5C 4) 10C A metal sphere ‘A’ of radius a is charged to a potential ‘V’. What will be its potential if it is enclosed by a spherical conducting shell ‘B’ of radius b and the two are connected by a conducting wire? 1) V 2) (a/b) V 3) (b/a) V 4) zero The capacity of a parallel plate capacitor is 5F. When a glass plate of same area as the plates but thickness half of the distance between the plates is placed between the plates of the capacitor, its potential difference reduces to 2/5 of the original value. The dielectric constant of the glass is 1) 1.5 2) 2.5 3) 5 4) 2s

V0 E 0 d Hint:

V 22.

Qd A 0

Qd A 0

t 1 1 d 1 k

2 V t 1 3 1 1 1 1 ; 1 5 V0 d k 5 2 k

Two spheres A and B of radii 4cm and 6cm are given charges of 80C and 40C. If they are connected by a wire, the amount of charge flowing from one to other is 1) 20C from A to B 2) 16C from A to B 3) 24C from B to A 4) 32C from A to B Hint: Q = (C1 – C2)V = i.e. Q =

23.

C2 C1 C1V1 C2 V2 C2 C1 Q1 Q 2

r2 r1 Q1 Q2 r1 r2

C1 C2

C1 C2

2 x 120 24C from B to A 10

Three capacitor of capacitance 10F, 15F, 20F are in series with a cell.

The

charge drawn from the cell is 60C. If they are connected in parallel with the same cell, then the charge drawn from the cell is 1) 385 C 2) 485 C 3) 585 C 4) 685 C Hint: 24.

qp qs

Cp E Cs E

Cp Cs

qp 6 x

45 x13 585C 60

The plates of a parallel plate capacitor are horizontal and parallel. A thin conducting sheet P is initially placed parallel to both the plates and nearer to the lower plate. From t = 0 onwards, the sheet P is moved at constant speed vertically

[Type text]

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CAPACITOR

25.

upwards so that it is always parallel to the capacitor plates. At t = 20 milli seconds, it is nearer to the upper plates. Then during the time interval from t = 0 to t = 20 milli seconds, the capacity of the capacitor will 1) increase gradually 2) decrease gradually 3) remains constant 4) first increases and then decreases Hint: Since the thickness of conducting plate is constant through out, capacity remains constant A capacitor is charged to 200V. A dielectric slab of thickness 4mm is inserted. The distance between the plates is increased by 3.2mm to maintain the same potential difference. Find the dielectric constant of the slab 1) 3 2) 4 3) 5 4) 6 Hint: V1 = V2;

26.

27.

C0

C 1

t 1 1 d k

30.

1

t 1 d

C 1 6 0 k C 7

Between the plates of a parallel plate capacitor of capacity C, two parallel plates of the same material and area same as the plate of the original capacitor are placed. If the thickness of these plates is equal to I/5 th of the distance between the plates of the original capacitor, then the capacity of the new capacitor is 1) 5/3 C 2) 3/5 C 3) 3C/10 4) 10C/3 Hint: d1 = d, d2 =

29.

d 2 d1 1 1 d 2 t 1 k k 1 t

A capacitor is charged with a dielectric to V volts. If the dielectric of constant K is removed then ___ is true a) capacity decreases by k times b) electric field intensity decreased by k times c) potential increases by k times d) charge increases by k times 1) a, b, c 2) a, b, c, d 3) a, c 4) b, d The capacitance of a capacitor becomes 7/6 times its original value if a dielectric slab of thickness t = 2/3d is introduced in between the plates ‘d’ is the distance between the plates. The dielectric constant of dielectric field is 1) 14/11 2) 11/14 3) 7/11 4) 11/7 Hint:

28.

Qd1 Q A 0 A 0

C2 d1 5 5 3 2 C2 C d, t = d & C1 d 2 3 3 5 5

A parallel plate capacitor of capacity Co is charge to a potential V0. A) The energy stored in the capacitor when the battery is disconnected and the plate separation is double is E1 B) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is E2, the E1/E2 value is 1) 4 2) 3/2 3) 2 4) ½ A capacitor of capacity of 10F is charged to 40V and a second capacitor of capacity 15F is charged to 30V. If they are connected in a parallel the amount of charge that flows from the smaller capacitor to higher capacitor in C is 1) 30 2) 60 3) 200 4) 250

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CAPACITOR

C1V1 C2 V2 & Q = (V1 – V)C1 = 60C C1 C2

Hint: V = 31.

A parallel plate capacitor of capacity 100F is charged by a battery of 50 volts. The battery remains connected and if the plates of the capacitor are separated so that the distance between them becomes half the original distance, the additional energy given by the battery to the capacitor in joules is 1) 125 x 10-3 2) 12.5 x 10-3 3) 1.25 x 10-3 4) 0.125 x 10-3 2 Hint: V = ½(C2 – C1)V & C2 = 2C

32.

A parallel plate capacitor of capacity 5F and plate separation 6cm is connected to a I volt battery and is charged. A dielectric of dielectric constant 4 and thickness 4cm introduced into the capacitor. The additional charge that flows into the capacitor from the battery is 1) 2C 2) 3C 3) 5C 4) 10C

C0 V t 1 = 10C; Q = Q2 – Q1 Hint: Q1 C0 V = 5C; Q2 = C2V = 1 1 d k 33.

A 20F capacitor is charged to 5V end isolated. It is then connected in parallel with an uncharged 30F capacitor. The decrease in the energy of the system will be 1) 25 J 2) 100 J 3) 125 J 4) 150 J Hint: V

34.

C1V1 C2 1 & U1 C1V12 U U1 C1 C2 2 C1 C2

A dielectric of thickness 5cm and dielectric constant 10 is introduced in between the plates of a parallel plate capacitor having plate area 500 sq cm and separation between plates 10cm. The capacitance of the capacitor is (0=8.8 x 10-12 SI units) 1) 8 PF 2) 6PF 3) 4PF 4) 20PF

A 0 1 d t 1 k

Hint: C = 35.

A 4F capacitor is charged by 200V battery. It is then disconnected from the supply and is connected to another uncharged capacitor of 2F capacity. The loss of energy during this process is _____ 1) 0 2) 5.33 x 10-2 3) 4 x 10-2 4) 2.67 x 10-2 Hint: U =

36.

CV 1 C1C2 V1 V2 2 ; V2 1 1 2 C1 C2 C1 C2

Energy E is stored in a parallel plate capacitor C 1. An identical uncharged capacitor C2 is connected to it kept in contact with it fro a while and then disconnected. The energy stored in C2 is ___ 1) E/2 2) E/3 3) E/4 4) 0 Hint: U1 =

1 C1V 2 , Potential on second capacitor in contact with the first one V 2 = 2 2

V/2; U2 =

[Type text]

1 V E C 2 2 4

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CAPACITOR

37. 38. is

If the capacity of a spherical conductor is 1PF, then its diameter is 1) 9 x 10-15m 2) 9 x 10-3m 3) 9 x 10-5m 4) 18 x 10-7m A 700PF capacitor is charged by a 50V battery. The electrostatic energy stored by it 1) 17 x 10-5 J

2) 13.6 x 10-9 J

3) 9.5 x 10-9 J

4) 8.75 x 10-

7

J 39.

40.

Two equal capacitors are first connected in parallel and then in series. The ratio of the total capacities in the two cases will be 1) 2 : 1 2) 1 : 2 3) 4 : 1 4) 1 : 4 Two condensers of capacity 2C and C are joined in parallel and charged up to potential V. The battery is removed and the condensor of capacity ‘C’ is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be 1)

41.

3V K

C1 V1 C2 V2 C1 C2

V1 V2 2

3)

V K2

4)

V K

2)

C1V2 C 2 V1 C1 C2

4)

C1V2 C 2 V1 C1 C2

3)

A number of capacitors are connected as shown in figure. The equivalent capacity is given by 1) nC 2) n(n + 1)C 3) 2n(n + 1)C

43.

3V K2

Two condensers C1 and C2 in a circuit are joined as shown in figure. The potential at A is V 1 and that of B is V2. The potential of point D will be 1)

42.

2)

4)

n n 1 C 2

Three capacitors of capacitances 3F, 10F and 15F are connected in series to a voltage source of 100V. The charge on 15F is

44.

45.

46.

47.

1) 50C 2) 100C 3) 200C 4) 280C In a parallel plate capacitor of capacitance ‘C’, a metal sheet is inserted between the plates parallel to them. If the thickness of the sheet is half of the separation between the plates, the capacitance will be 1) C/2 2) 3C/4 3) 4C 4) 2C A 10 micro farad capacitor is charged to 500 V and then its plates are joined together through a resistance of 10 ohm. The heat produced in the resistance is 1) 500 J 2) 250 J 3) 125 J 4) 1.25 J A capacitor is charged to 200 volt. It has a charge of 0.1C. When it is discharged, energy liberated will be 1) 1 J 2) 10 J 3) 14 J 4) 20 J Half of the separation between two parallel plates of a capacitor is filled with a dielectric medium. The capacitance of the capacitor becomes 5/3 times its original value with full space dielectric. The dielectric constant of the medium K is

[Type text]

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CAPACITOR

48.

49.

1) K = 2 2) K = 3 3) K = 4 4) K = 5 A 10F capacitor is charged to a potential difference of 50 V and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes 20 V. The capacitance of second capacitor is 1) 10 F 2) 20 F 3) 30 F 4) 15 F Capacity of a spherical capacitor having two spheres of radii a and b (a > b) separated by a medium of dielectric constant K is given by

Kab in SI system ab 4 0 Kab 3) in SI system a b 1)

2) 4)

4 0 Kab in CGS system ab K a b ab

in CGS system

50.

A capacitor of capacity ‘C’ has charge Q. The stored energy is W. If the charge is increased to 2Q, the stored energy will be 1) 2W 2) W/2 3) 4W 4) W/4

51.

A D.C. potential of 100 volt is connected to the combination as shown in figure. The equivalent capacity between A and B will be equal to 1) 40 F 2) 20 F

52.

3) 30 F 4) 10 F The capacitance of four plates, each of area A arranged as shown in figure as

2 0 A d 4 0 A 3) d 1)

3 0 A d 5 0 A 4) d 2)

53.

Identify the wrong statement of the following a) the resultant capacity ‘C’ is less than the capacitance of smallest capacitor in series combination b) the resultant capacity ‘C’ is greater than greatest capacitance in parallel combination c) In series combination, charge on capacitor plates is inversely proportional to capacitance of capacitor 1) a is wrong 2) c is wrong 3) b is wrong 4) all are wrong 54. A parallel plate capacitor of area A, plate separation d with the electric capacity C0 is filled with three different dielectric materials with constants K1, K2 and K3 as in the figure. If these three are replaced by a single dielectric, its dielectric constant K is ___ 1)

1 1 1 1 K K1 K 2 2K 3

2)

3)

1 K1 K 2 2K 3 K K1 K 2

4) K

[Type text]

1 1 1 K K1 K 2 2K 3 K1K 3 K 2 K3 K1 K 3 K 2 K 3

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CAPACITOR

55.

56.

57.

58.

59.

60.

61.

62.

63.

Four metallic plates, each with a surface area of one side A, are placed at a distance d apart. The outer plates are connected to terminal X and the inner plates to terminal Y. The capacitance of system between X and Y is 1) 0A/d 2) 20A/d 3) 30A/d 4) 40A/d Four metallic plates, each with a surface area of one side A are placed at a distance d apart, these plats are connected as shown in figure. The capacitance of the system between X and Y is 1) 0A/d 2) 20A/d 3) 30A/d 4) 40A/d An infinite ladder is made as shown in figure using capacitors C1 = 1 F and C2 = 2F. The equivalent capacitance of the ladder, in F is 1) 1 2) 2 3) 0.75 4) 0.5 In the circuit segment shown VA - VB = 19V. The p.d. across 3 F capacitor is 1) 7V 2) 8V B A 3) 23V 4) 4V A parallel plate capacitor with a dielectric constant K = 3 filling the space between the plates is charged to a potential difference V. The battery is then disconnected and the dielectric slab is withdrawn and replaced by another dielectric slab having K = 2. The ratio of energy stored in the capacitor before and after replacing the dielectric slab by new one is 1) 3 : 2 2) 9 : 4 3) 4 : 9 4) 2 : 3 Two parallel capacitors of capacitance C and 2C are connected in parallel and charged to a potential difference V. The battery is then disconnected and the capacitor C is completely filled with a material of dielectric constant k. The potential difference across the capacitors is now 1) 2V/k 2) 3V/k 3) 3V/(k+2) 4) 2V/(k+3) Initially the capacitors C1 and C2 shown in figure have equal capacitances. If a dielectric plate (k = 2) is introduced in capacitor C2, then potential difference across its plates and charge 1) both will decrease 2) both will increase 3) p.d. will increase but charge will decrease 4) p.d. will decrease but charge will increase Five identical capacitor plates, each of area A, are arranged such that adjacent plates are at a distance d apart. The plates are connected to a source of emf V as shown in figure. The charge on plate 1 is q and that on 4 is q', where 1) q' = q 2) q' = 2q 3) q' = -2q 4) q' = 3q For the circuit shown in figure which of the following statements is true 1) With S1 closed, V1 = 15V, V2 = 20V 2) With S3 closed, V1 = V2 = 25V 3) With S1 and S2 closed, V1 = V2 = 0 4) With S1 and S3 closed, V1 = 30V, V2 = 20V

[Type text]

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CAPACITOR

64.

Two identical capacitors, having the same capacitance C. One of them is charged to a potential V1 and the other to V2. The negative ends of the capacitor are connected together. When the positive ends are also connected, the decrease in energy of the combined system is 1)

65.

66.

67.

1 C ( V12 V2 2 ) 4

69.

70.

72.

1 C ( V1 V2 ) 2 4

4)

1 C ( V1 V2 ) 2 4

2) (3/2)

CV 2

3) (25/6) CV 2

4) (9/2) CV 2

a composite capacitor having capacitance 8 F? 1) 4 Condensers in series 8 such groups in parallel 2) 2 Condensers in series and 16 such groups in parallel 3) 8 Condensers in series and 4 such groups in parallel 4) All of them in series. Find out the effective capacitance between points P and Q. It will be equal to 1) 9 F 2) 4.5 F 3) 1 F 4) 6 F An uncharged parallel plate capacitor having a dielectric of constant K is connected to a similar air filled capacitor charged to a potential V. The two share the charge and the common potential is V'. The dielectric constant K is

V V V V

2)

V V V

3)

V V V

4)

V V V

Two condensers each having capacitance C and breakdown voltage V are joined in series. The capacitance and the breakdown voltage of the combination will be 1) 2C and 2V 2) C/2 and V/2 3) 2C and V/2 4) C/2 and 2V Two identical metal plates are given positive charges Q 1 and Q2 (< Q1) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C, the potential difference between them is 1)

73.

3)

You are given thirty two capacitors each having capacity 4 F. How do you connect all of them to prepare

1) 71.

1 C (V13 V2 2 ) 4

Each edge of a cube (figure) made of wire contains a capacitor of capacitance C. Find the effective capacitance of this bank of capacitors between opposite corners A and G. 1) 5C/6 2) 4C/3 3) 3C/4 4) 6C/5 Consider the situation shown in figure. The capacitor A has charge q on it whereas B is uncharged. The charge appearing on the capacitor B a long time after switch is closed is 1) zero 2) q/2 3) q 4) 2q A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is similarly charged to potential differences 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the negative terminal of one is connected to the negative terminal of the other. The final energy of the configuration is 1) zero

68.

2)

(Q 1 Q 2 ) 2C

2)

(Q 1 Q 2 ) C

3)

(Q1 Q 2 ) C

4)

(Q1 Q 2 ) 2C

Three very large plates are given charges as shown in the figure. If the cross sectional area of each plate is the same, the final charge distribution on plate C

[Type text]

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CAPACITOR

74.

is: 1) +5 Q on the inner surface, +5 Q on the outer surface 2) +6 Q on the inner surface, +4 Q on the outer surface 3) +7 Q on the inner surface, +3 Q on the outer surface 4) +8 Q on the inner surface, +2 Q on the outer surface Two identical sheets of metallic foil are separated by d and capacitance of the system is C and is charged to a potential difference E. Keeping the charge constant, the separation is increased by l. Then the new capacitance and potential difference will be: 1)

75.

0 A ,E d

2)

0 A ,E (d l )

3)

0 A A l l , 1 E 4) 0 , 1 E (d l ) d d d

n conducting plates are placed face to face as shown in figure. Distance between any two plates is d. Area of the plates is A,

A A A A , , ...... (n 1) . The equivalent 2 4 8 2

d

capacitance of the system is 1)

0A

2)

2n d

0A

3)

n

(2 1)d

0A

4)

n

( 2 2) d

0A n

(2 1)d 76.

77.

The adjoining figure shows two identical parallel plate capacitors connected to a battery with switch S close(4). The switch is now opened and the plates are filled with a dielectric of dielectric constant 3. The ratio of the total electrostatic energy stored in both the capacitors before and after the introduction of the dielectric is 1) 2 : 5 2) 3 : 5 3) 5 : 2 4) 5 : 3 A parallel plate capacitor has two layers of dielectric as shown in figure. This capacitor is connected across a battery, then 2 1 the ratio of potential difference across the dielectric layers 1) 4/3 2) 1/2 3) 1/3 4) 3/2 Five conducting plates are placed parallel to each other. Separation between them is d and area of each plate is A. Plate number 1, 2 and 3 are connected with each other and at the same time through a cell of emf E. The charge on plate number 1 is 1) E0A/d 2) E0A/2d 3) 2 E0A/d 4) zero A capacitor of capacity C1, is charged by connecting it across a battery of emf V 0. The battery is then removed and the capacitor is connected in parallel with an uncharged capacitor of capacity C 2. The potential difference across this combination is:

k 2k 6

78.

79.

1)

C2 V0 C1 C 2

[Type text]

2)

C1 V0 C1 C 2

3)

C1 C 2 V0 C2

4)

C1 C 2 V0 C1

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CAPACITOR

80.

81. 82.

The capacitance of a parallel plate capacitor is 16F. When a glass slab is placed between the plates, the potential difference reduces to 1/8 th of the original value. What is dielectric constant of glass ? 1) 4 2) 8 3) 16 4) 32 The equivalent capacity across M and N in the given figure is: 1) 5C/3 2) 2/3C 3) C 4) 3/2C Three plates of common surface area A are connected as shown. The effective capacitance between points P and Q will be:

0A d 3 0A 3) 2 d 1)

2) 4)

3 0 A d

2 0 A d

83.

A dielectric slab of dielectric constant K = 5 is covered from all sides with a metallic foil. This system is introduced into the space of a parallel plate capacitor of capacitance 10 F. The slab fills almost the entire space between the plates, but does not touch the plates. The capacitance will become nearly: 1) 2) zero 3) 2 pF 4) 50 Pf

84.

A capacitor of capacitance 1 F can withstand the maximum voltage 6 kV while a capacitor of

85.

capacitance 2.0 F can withstand the maximum voltage 4 kV. If the two capacitors are connected in series, then the two capacitors combined can take up a maximum voltage of: 1) 2.4 kV 2) 5 kV 3) 9 kV 4) 10 kV A spherical conductor of radius 2m is charged to a potential of 120 V. It is now placed inside another hollow spherical conductor of radius 6 m. Calculate the potential of bigger sphere if the smaller sphere is made to touch the bigger sphere: 1) 20 V 2) 60 V 3) 80 V 4) 40 V

86.

A parallel plate capacitor has two layers of dielectrics as shown in figure. Then the ratio of potential difference across the dielectric layers when connected to the battery is: 1)

K1 K2

2)

K1a K 2b

3)

K 2b K1a

4)

K 2a K 1b

LEVEL - III : 1.

A parallel plate capacitor C is connected to a battery and it is charged to a potential difference of V. Another capacitor of capacitance 2C is similarly charged to a potential difference of 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to positive of the other. The final energy of the configuration is A) zero

2.

B) 3/2 CV2

C) 25/6 CV2

D) 9/2 CV2

The amount of heat liberated when a capacitor of C farads charged to a potential difference of V volts is discharged through a resistor of R ohms is H joules. The same capacitor is now charged to a potential difference of 2V and discharged through a

[Type text]

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CAPACITOR

resistor of 2 R ohms, then heat liberated is A) 4H 3.

B) 2H

C) H

D) H/2

A capacitor of capacity C1 is charged to a potential V0. The electrostatic energy stored in it is U0. It is connected to another uncharged capacitor of capacitance C2 in parallel. The energy dissipated in the process is A)

C2 U0 C1 C 2

B)

C1 U0 C1 C 2

C C

2 C) 1 C C 2 1

2

U0

D)

C1C 2 U0 2(C1 C 2 ) 4.

5.

In the circuit shown in figure, if the key is turned so that instead of 1, 2 terminals 1, 3 are connected, then the heat liberated in resistor R is A)

1 C(1 ~ 2 ) 2 2

B)

1 C(1 2 ) 2 2

C)

1 1 C12 C 2 2 2 2

D)

1 1 C12 C 2 2 2 2

Three capacitors C1, C2, C3 are connected as shown in figure to one another and to points P1, P2, P3 at potentials V1, V2 and V3. If the total charge on the capacitors is zero, the potential of the point O is A)

C1V1 C 2 V2 C 3 V3 C1 C 2 C 3

B) V1 + V2 + V3

D) V1

C) 0 6.

A parallel plate capacitor having capacitance C has two plates of same area A and thickness t. Figure shows the charges available on the four surfaces of the plates. The potential difference V between the two plates is given by A)

7.

1 q 2 q3 2 C

q2 q3 C

B)

C)

1 q2 q4 2 C

q2 q4 C

D)

The plates of a parallel plate capacitor are separated by d cm. A plate of thickness t cm with dielectric constant k 1 is inserted and the remaining space is filled with a plate of dielectric constant k2. If Q is the charge on the capacitor and area of plates is A cm2 each, then potential difference between the plates is A)

8.

C C2 C V2 3 V3 1 C3 C1 C2

Q t dt 0 A k1 k 2

B)

4Q t dt A k1 k 2

C)

k 4Q k 1 2 A t dt

D)

Q k1 d t 0 A t k 2

Find the capacitance of a system of three parallel plates each of area A separated

[Type text]

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CAPACITOR

by distances d1 and d2. The space between them is filled with dielectrics of relative dielectric constants 1 and 2. The dielectric constant of free space is 0

9.

A)

1 2 0 A 1 d 2 2 d1

C)

1 2 A A D) 1 2 0 ( 1 d 1 2 d 2 ) 0 (1 2 ) d1 d 2

1 C V12 V22 4

B)

1 C V12 V22 4

C)

1 C V1 V2 2 4

D)

1 C V1 V2 2 4

One plate of a capacitor is connected to a spring as shown in the 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)

11.

1 2 0 A 1 d1 2 d 2

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 A)

10.

B)

6 0 E 2 Ad

3

B)

4 0 AE 2 d

3

C)

4 0 AE 3 2d

3

D)

2 0 AE d2

A slab of copper of thickness b is inserted in between the plates of parallel plate capacitor as shown in figure. The separation of the plates is d. If b = d/2, then the ratio of capacities of the capacitor after and before inserting the slab will be: A)

2:1

B) 2 : 1

C) 1 : 1

D) 1 :

2 12.

13.

Consider a parallel plate capacitor of capacity 10 F with air filled in the gap between the plates. Now one-half of the space between the plates is filled with a dielectric of dielectric constant 4, as shown in the figure. The capacity of the capacitor changes to: A) 25 F

B) 20 F

C) 40 F

D) 5 F

k=4

Consider the arrangement of three plates X, Y and Z each of area A and separation d. The energy stored when the plates are fully charged is: A)

0 AV 2 2d

[Type text]

B)

0 AV 2 d

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CAPACITOR

C)

2 0 AV 2 d

D)

3 0 AV 2 2d

MULTIPLE ANSWER OBJECTIVE TYPE QUESTIONS : 1.

2.

In the arrangement shown in Fig, the charge on capacitor C 2 is 1 C and on capacitorC3 is 2 C. If the capacitance of capacitor C2 is 1 F, then A) p.d across C1 is 2 V

B) Charge on C1 is 3 C

C) Energy stored in system is 4.5 J battery is 4.5 J

D) Energy supplied by

In the arrangement shown in Fig, all capacitors have equal capacities equal to 1 F. If the emf of the battery is 2 V A) The charge on capacitor C1 is 1 C B) The potential difference across C2 is 1 V C) The energy stored in C3 is 2 J

3.

In a parallel plate capacitor, the area of each plate is A and the plate separation is d. The capacitor carries a charge q and the force of attraction between the two plates is F. Then A) F q2

4.

D) Energy supplied by battery is 4 J

B) F d

C) F 1/A

D) F 1/d

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 the plates. The capacitor if given some charge. As x goes from o to 3d A) the magnitude of electric field remains the same B) the direction of electric field remains the same C) the electric potential increases continuously D) the electric potential increases at first, then decreases and again increases

5.

In the circuit shown, some potential difference is applied between A and B. If C is joined to D, A) no charge will flow between C and D B) some charge will flow between C and D C) the equivalent capacitance between C and D will not change D) the equivalent capacitance between C and D will change

6.

The two plates X and Y of a parallel-plate capacitor of capacitance C are given a charge of amount Q each. X is now joined to the positive terminal and Y to the negative terminal of a cell of emf = Q/C A) Charge of amount Q will flow from the positive terminal to the negative terminal of the cell through the capacitor B) The total charge on the plate X will be 2Q C) The total charge on the plate Y will be zero D) The cell will supply C2 amount of energy

[Type text]

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CAPACITOR

7.

A parallel-plate capacitor is charged from a cell and then disconnected from the cell. The separation between the plates is now doubled A) The potential difference between the plates will become double B) The field between the plates will not change C) The energy of the capacitor doubles D) Some work will have to be done by an external agent on the plates

8.

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

9.

A parallel-plate air capacitor of capacitance C0 is connected to a cell of emf and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it A) The potential difference between the plates decreases K times B) The energy stored in the capacitor decreases K times C) The change in energy is

1 10.

1 C02 (K - 1) 2

D) The change in energy is

1 C 02 2

1 K

When a charged capacitor is connected with an uncharged capacitor, then flows A) the change always flows from the changed to the unchanged capacitor B) a steady state is obtained after which no further charge flow occurs C) the total potential energy stored in the capacitors remains conserved D) the charge conservation always holds true

MATCHING TYPE QUESTIONS : A parallel-plate air capacitor of capacitance C0 is connected to a cell of emf and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it A) The potential difference between the plates decreases K times B) The energy stored in the capacitor decreases K times C) The change in energy is

1 C02 (K - 1) 2

1. List – I

[Type text]

List – II

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CAPACITOR

a) energy stored in capacitor by external source b) field inside the dielectric when is placed in between plates of capacitor battery still connected c) charge on capacitor by placing dielectric and battery connection is removed

a) remain same

d) force between two parallel plates when separation is increased

d) decreases

b) half of energy supplied by battery

1Q 2C

c)

2. List – I a) Energy stored in capacitor

List – II e) 40R

b) Capacity of spherical capacitor c) Force between plates of capacitor d) Energy density of electric field

f)

q2 2 0 AK

g)

1 0E2 2

h) q2/2C

PREVIOUS YEAR IIT QUESTIONS : 1.

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 the plates. The capacitor is given some charge. As one goes from 0 to 3d [IIT-JEE 1998] (a) The magnitude of the electric field remains the same (b) The direction of the electric field remains the same (c) The electric potential increases continuously (d) The electric potential increases at first, then decreases and again increases

2.

A parallel plate capacitor is charged to a potential difference of 50 V. It is discharged through a resistance. After 1 second, the potential difference between plates becomes 40 V. Then [Roorkee 1999] (a) Fraction of stored energy after 1 second is 16/25 (b) Potential difference between the plates after 2 seconds will be 32 V (c) Potential difference between the plates after 2 seconds will be 20 V (d) Fraction of stored energy after 1 second is 4/5

3.

Five identical plates each of area A are joined as shown in the figure. The distance between the plates is d. The plates are connected to a potential difference of V volts. The charge on plates 1 – 1 2 3 4 5 and 4 will be [IIT 1984] V

AV 2 0 AV . (a) 0 d d (c)

0 AV 2 0 AV . d d

+

AV 2 0 AV . (b) 0 d d (d)

0 AV 2 0 AV . d d

3F

B

3F

1F 1F

[Type text]

20 A

1F

100V

10

Page C

CAPACITOR

4.

In the figure below, what is the potential difference between the point A and B and between B and C respectively in steady state [IIT 1979] (a) VAB VBC 100V

(b) VAB 75V, VBC 25 V

(c) VAB 25 V, VBC 75 V (d) VAB VBC 50 V 5.

6.

Figure given below shows two identical parallel plate capacitors connected to a battery with switch S closed. V A B The switch is now opened and the free space between the plate of capacitors is filled with a dielectric of dielectric constant 3. What will be the ratio of total electrostatic energy stored in both capacitors before and after the introduction of the dielectric [IIT 1983] (a) 3 : 1

(b) 5 : 1

(c) 3 : 5

(d) 5 : 3

A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is connected to another battery and is charged to potential difference 2V. The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is [IIT 1995] (a) Zero

7.

(b)

25CV 2 6

(c)

3CV 2 2

(d)

In an isolated parallel plate capacitor of capacitance C, the four surface have charges Q1 , Q2 , Q3 and Q4 as shown. The

Q3

Q1 Q2

potential difference between the plates is [IIT-JEE 1999]

8.

(a)

Q1 Q2 Q3 Q4 Q Q3 (b) 2 2C 2C

(c)

Q 2 Q3 2C

(d)

9CV 2 2

Q4

Q1 Q4 2C

For the circuit shown, which of the following statements is true [IIT-JEE 1999] (a) With S1 closed, V1 15V, V2 20 V

S1

(b) With S3 closed V1 V2 25V

V1=30V + –

V2=20V S3 + –

C1=2pF

C2=3pF

S2

(c) With S1 and S2 closed V1 V2 0 (d) With S1 and S3 closed, V1 30V, V2 20V 9.

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

q +

–

+

–

+

–

+

–

+

(a) Zero

[Type text]

(b) q / 2

–

A

[IIT-JEE (Screening) 2001] (c) q

s

B

(d)

Page

CAPACITOR

2q

10.

Point charge q moves from point P to point S along the path PQRS (figure shown) in a uniform electric field

E pointing co-parallel to the positive direction of the X axis. The coordinates of the points P , Q, R and S are (a, b, 0), (2a, 0, 0), (a, b, 0) and (0, 0, 0) respectively. The work done by the field in the above process is given by the expression [IIT 1989]

P S

Q

X

R

(b) qEa

(a) qEa (c)

E

(d) qE [(2a)2 b 2 ]

qEa 2

ASSERTION & REASON : Read the assertion and reason carefully to mark the correct option out of the options given below : (a) If both assertion and reason are true and the reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not the correct explanation of the assertion. (c) If assertion is true but reason is false. (d) If assertion is false but reason is true. 1.

Assertion : If three capacitors of capacitance C1 < C2 < C3 are connected in parallel then their equivalent capacitance Cp > Cs Reason

2.

:

1 1 1 1 C p C1 C2 C3

Assertion : If the distance between parallel plates of a capacitor is halved and dielectric constant is made three times, then the capacitor becomes 6 times. Reason material.

3. Assertion beam.

: Capacity of the capacitor does not depend upon the nature of the : Dielectric breakdown occurs under the influence of an intense light

Reason : Electromagnetic radiations exert pressure. 4. Assertion varied on it.

: The capacity of a given conductor remains same even if charge is

Reason : Capacitance depends upon nearly medium as well as size and shape of conductor. 5. Assertion : The whole charge of a conductor cannot be transferred to another isolated conductor.

[Type text]

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CAPACITOR

Reason 6.

: The total transfer of charge from one to another is not possible.

Assertion : Conductors having equal positive charge and volume, must also have same potential. Reason : Potential depends only on charge and volume of conductor.

7. Assertion : The lightening conductor at the top of high building has sharp pointed ends. Reason : The surface density of charge at sharp points is very high resulting in setting up of electric wind. 8.

Assertion : Circuit containing capacitors should be handled cautiously even when there is no current. Reason

9.

: The capacitors are very delicate and so quickly break down.

Assertion : The tyres of aircraft's are slightly conducting. Reason : If a conductor is connected to ground, the extra charge induced on conductor will flow to

ground. 10.

Assertion : A bird perches on a high power line and nothing happens to the bird. Reason

11.

: The level of bird is very high from the ground.

Assertion : Two capacitors are connected in series to a battery. If the dielectric medium is inserted between the plates of one of the capacitors, the energy stored in the system will increase

Reason : On inserting the dielectric medium, the capacity of the capacitor increases *****

K

E

Y

S

SINGLE ANSWER TYPE QUESTIONS LEVEL – I

LEVEL – II 1 2 2 1 3 4 1 3 6 1 4

2 2 2 2 3 4 2 4 6 2 3

3 1 2 3 3 4 3 3 6 3 4

[Type text]

4 2 2 4 3 4 4 4 6 4 3

5 4 2 5 3 4 5 4 6 5 4

6 3 2 6 3 4 6 2 6 6 1

7 1 2 7 1 4 7 3 6 7 2

8 1 2 8 1 4 8 3 6 8 1

9

1

1

1

1

1

1

1

1

1

1

2

3 2 9 3 4 9 3 6 9 1

0 3 3 0 2 5 0 3 7 0 4

1 2 3 1 1 5 1 4 7 1 4

2 3 3 2 3 5 2 2 7 2 4

3 3 3 3 3 5 3 2 7 3 3

4 3 3 4 1 5 4 4 7 4 3

5 4 3 5 4 5 5 2 7 5 3

6 2 3 6 3 5 6 3 7 6 2

7 2 3 7 2 5 7 2 7 7 4

8 1 3 8 4 5 8 2 7 8 4

9 3 3 9 3 5 9 4 7 9 2

0 2 4 0 4 6 0 3 8 0 2

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CAPACITOR

8 1 1

8 2 4

8 3 1

8 4 3

8 5 1

8 6 4

LEVEL – III 1

2

3

4

5

6

7

8

9

10

11

12

13

B

A

A

B

A

A

B

A

C

B

B

B

B

MULTIPLE ANSWER OBJECTIVE TYPE QUESTIONS 1

2

3

4

5

6

7

8

9

10

ABC

ABD

AC

BC

AC

ABCD

ABCD

AD

ABD

ABD

MATCH THE COLUMN : 01. a-bc, b-a, c-a, d-a

02. a – h, b –

e, c – f, d – g PRIVIOUS YEAR IIT QUESTIONS 1

2

3

4

5

6

7

8

9

10

BC

AB

C

C

C

C

C

D

A

B

ASSERTION & REASON 1

2

3

4

5

6

7

8

9

10

11

C

B

B

A

D

D

A

C

B

C

C

***

[Type text]

Page

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