Capacitors Solved Sheet for IIT

Q.1

In four options options below below,, all all the the four circui circuits ts are arranged arranged in order of equi equival valent ent capaci capacitance tance between between points A & B. Select the correct corr ect order. Assume Assume all capacitors are of equal capacitance. capacitance.

(1)

Q.2 Q.2

[Sol: Q.3

[Sol:

(2)

(3)

(4)

(A) C1 > C2 > C3 > C4 (B*) C1 > C3 > C2 > C4 (C) C1< C2 < C3 < C4 (D) C1 < C3 < C2 < C4 Rows Rows of capac capaciitors contai containi ning ng 1, 2, 2, 4, 8, ..... ......... ....∞ capacitors each of capacitance 2µF are connected in parallel as shown in figure. The potential diff difference erence across AB is is 10 Volts. Volts. (A*) Total capacitance across AB is 4µF. (B) Charge on each capacitor will be same. (C*) Charge on the capacitor in the first row is more than that on any other capacitor. (D) Total energy stored in capacitors capacitors is 50 µJ.

 1 + 1 + 1 + 1 + ..................... C C C  = 4µF Ceq = C + + + + ................................. = C  2 4 8  2 4 8  Capaci Capacitan tance ce of 1st 1st row is max maxim imum um,, hence hence charg chargee stored wi will be als also o maxi maximum mum. (A, C) ] In the the circu circuit it,, both paral parallel plate plate capa capaci citors tors are are iden identi tical cal.. Column olumn I Column II I I indicates effect indicates indicates action done do ne on capacitor 1 and Column on capacitor 2. Select correct alternative. (A) (B) (C)

Column I Plates are moved further apart. Plate area increased Left plate is earthed

(D)

It's Plates are short circuited

Column I I (P) No effect (Q) Potential difference increases (R) Amount of charge on right plate decreases (S) Amount of charge on left plate increases [Ans. (A) R, (B) Q,S, (C) P, (D) Q, S ]

q q + =V C1 C2 V1 + V2 = V V2 = V – V1 when plates are moved apart ⇒ C1 ⇒ ↓ ⇒ V1 ↑ ⇒ V2 ↓ C1C2 Ceq = C1 + C2

↓

⇒ q = Ceq V ↓

right plate has –ve charge ⇒ when –q ¯ ⇒ q increases

]

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Q.4 Q.4

Three Three identi dentica call large arge con conduc ducti ting ng plate platess P1, P2 & P3 each having area of either face of plate eauql to A are placed parallel parallel to each other at very short distance d. A charge –2Q is given to plate P1, no charge is given to plate P2 & +4Q is given to plate P3. Now plate P2 & P3 are connected co nnected by thin conducting conducting wire. Then find the total amount of heat (in Joules) produced. (here Q = 2 × 10 –3 co ul.)

[Sol:

∈0 d

A

= 0.1µ coul/volt,

[Ans: 180 J]

1 (3Q)2 Vi = × 2 + Uoutside the capacitor 2 C when they are connected c onnected the t he charge distribution will will be as: 2

1 (3Q) Uf = 2 C Heat produced ∴

2

Q.5

Q.6

1 (3Q) = = 180 Joule. ] 2 C Two Two identical identical capacitors capacitors are connected connected in in series series as shown shown in in the figu figure. re. A diel dielectri ectricc slab slab (κ > 1) is placed between the plates of the capacitor B and the battery remains remains connected. Which Which of the following following statement(s) is/are correct following following the insertion of the dielectric?

(A*) The charge supplied by the battery increases. (B*) The capacitance capa citance of the system increases. (C) The electric field in the capacitor B increases. (D) The electrostatic potential po tential energy decreases. Two larg largee conducti conducting ng pla plates tes havi having ng surf surface ace charge charge den densi siti ties es +σ and – σ, respectively, are fixed ‘d’ distance apart. A small test charge q of mass m is attached to two identical springs as shown in the adjacent figure. The charge q is now released from rest with springs in natural length. Then q will [neglect gravity]





[Sol: [Sol:

(D) remai r emain n stationary A = deform deformati ation on in in equi equili libri brium um state state.. σ

2KA =

∈0

·q

∴

q A = K 2 ∈0 σ

(B)

Springs are connected in parallel K eq = 2K angular frequency frequenc y = Q.7

2 K m

(A)

The circui circuitt was in in the the shown state from from a long long time time.. Now the swi switch tch S is is closed. closed. Find Find the charge charge (in µC) that flows through thro ugh the switch S (from (fro m A to B) after closing it.

[Ans: 50 µC]

[Sol:

CAB =

2× 4 8 4 = = 2+ 4 6 3

4 200 × 50 = 3 3 When switch 'S' is closed, the circuit becomes as

q = CV =

q2 2

=

q1 4 q1 = 2q2

+

q1 = 50 4

∴ q2 2

2q1 200 = 50 q1 = = 100 ∴ 4 2 q2 = 50 charge flown = q1 – q2 = 100 – 50 = 50 µC ] F or Proble Problem m 8 to 10





[Sol [Sol::

char charge ge woul would d be sam same on on cap capac aciitors tors conne connect cted ed in seri series es,,

Q.9 Q.9

If the the aare reaa of the the pl plates ates of capa capaci citor torss are are A1, A2 & A3 respectively then (A) A1 < A2 > A3 (B*) A1 > A3 > A2 (C) A1 = A2 = A3 (D) A1 > A2 > A3 The The gra graph ph show showss that that E1 (electric field field in space between plates of capacitor 1) < E3< E 2 & hence, V1 < V3 < V2 as C1V1 = C2V2 = C3V3 (a1 = a2 = a3) giving us A1V1 = A2V2 = A3V3 we get, A2 < A3 < A1 ]

[Sol [Sol::

Q.10 [Sol [Sol::

Q.11 Q.11

]

If the magni magnitude tude of elec electric tric fiel field d between between thei theirr plates plates are E1, E 2 & E3 respectively then (A) E1 > E2 > E3 (B*) E1 < E 2 > E3 (C) E1 > E2 < E3 (D) E1 < E2 < E3 Elec Electr triic fi field eld α slope of V/x graph & hence, E1 < E3 < E2 or E 1 < E2 > E3] An isola isolated ted metal metalli licc object object is is charged charged in in vacuum vacuum to potential potential v0, its electrostatic electros tatic energy being being W0. It is then disconnected disconnecte d from the source sour ce of potential, po tential, its charge being left left unchanged and is im immersed mersed in a large volume of dielectric, with dielectric constant k. The electrostatic energy will be W0 W0 (C) (D) W0 k 2k Energy Energy of the metal metalli licc object equals equals work work done in chargi charging ng the object object against against electrostati electrostaticc forces. For the object completely surrounded in a medium of dielectric constant 'K', force or electric field

(A) kW0 [Sol:

q1 = q2 = q3

(B*)

W0 .] K Q.12 The plates plates of a parallel parallel plate plate capacitor capacitor are charged upto 100 volt. A 2 mm thick thick plate is is inserted inserted between between the plates, plates, then to main maintai tain n the same same potential potential diff difference, erence, the distanc distancee between between the capacitor capacitor plates is increased increased by 1.6 mm. mm. The dielectric dielectric constant of the plate is (A*) 5 (B) 1.25 (C) 4 (D) 2.5

everywhere everywhere gets get s reduced by the factor 'K' & hence, changed energy would be

[Sol:

C=

∈0

A

q = C × 100

d New capacitance

C1 =

∈0

=

∈0 d

A

× 100

A

 (d + 1.6 ) + t  − 1 1  k



where t = 2 mm q 100 = C = 1

∈0 A ×100  d  C1

Solving

K=5 ]





Q.13

The fig figure ure shows shows a battery battery with with emf emf 15 V in in a circui circuitt with with R 1 = 30 Ω, R 2 = 10 Ω , R 3 = 20 Ω and capacitance (C) = 10µF. The switch S is initially in the open position and is then closed at time t = 0. What will be the final steady state charge on capacitor (A*) 75µC (B) 50µC (C) 10µC

[Sol:

i=

(D) none of these

15 1 = amp. 60 4

1 × 30 = 7.5 volt 4 q = eV = 10 × 7.5 = 75 µe VAB =

Q.14

[Sol:

(A (A)

]

A point point charge charge is pla placed ced at x in in front front of an earthed earthed metal metal sheet y. y. P and and Q are two points between x and y as shown. If the electric field field strength at P and Q be E P and EQ. Which Which one of the following following statements is correct (A) EP = EQ (B) EP > EQ (C*) EP < EQ (D) EP = EQ = 0 EQ > E P (C) ]

Question No. 15 & 16 (2 que questions) stions) Two identical conducting very large plate P1 & P 2 having charges +4Q & +6Q are placed very closed to each other at separation 'd'. The plate area of either face of the t he plate is A. Then Q.15 Q.15 Potenti Potential al dif differen ference ce betwe between en plates plates P1 & P2 is

[Sol: [Sol:

Q.16

Qd

(A) VP1

− VP = A ∈ 0

(C) VP1

− VP =

(B*) VP1

2

2

5Qd A ∈0

(D) VP1

− VP = 2

− VP = 2

− Qd A ∈0

− 5Qd A ∈0

Potenti Potential al dif differen ference ce betwe between en the the plates plates

−Q  ∈0 A     d 

VP1

− VP =

VP1

− Qd − VP = ∈ A 0

2

2

]

If plate P1 & plate P2 are connected by thin wire then amount of heat produced prod uced is





Q.17 Q.17

In the diagra diagram m shown, shown, in steady steady state (A) Charge on 1 µf capacitor is 1 µC (B*) Charge on 2 µf capacitor is 2 µC (C*) Charge on 3 µf is zero (D) Potential Po tential diff difference erence across 2µf capacitor is 2 volt

Q.18

Which Which of the following following is suffi sufficient cient condition condition for findin finding g the electric electric flux flux surface?

Q.19

ΦE through a closed

(A) If the magnitude of E is known everywhere on the surface sur face (B*) If the total to tal charge inside inside the surface is specified specified (C) If the t he total charge outside o utside the surface is specified specified (D) Only if the location of each point charge inside the surface is specified The figu figure re shows a conductin conducting g sphere sphere 'A' 'A' of radius radius 'a' which which is is surrounded by a neutral conducting spherical shell B of radius 'b'(>a). Initially switches S1, S2 and S3 are open and sphere 'A' carries a charge Q. First the switch 'S1' is closed to connect the shell B with the ground and then opened. Now the switch 'S2' is is closed so that the t he sphere 'A' is is grounded and then t hen S2 is opened. Finally, the switch 'S 3' is closed closed to connect the spheres together. to gether. Find the heat (in Joule) which is is produced produ ced after closing the switch S3. [Consi [Co nsider der b = 4 cm, a = 2 cm and Q = 8 µC]

[Ans. 1.8] [Sol. When When outer surface surface is is grounded charg chargee '–Q' resides resides on the inne innerr surface surface of sphere sphere 'B' Now sphere A is connected to eart h potential on its its surface becomes zero. Let the charge on the t he surface A becomes becomes q kq kQ a – =0 q= Q ⇒ a b b Consider the figure. In this t his position position energy stored E1 =

 a Q 8πε 0a  b  1

2

1 a  Q2 + + 4πε b  b Q  (− Q) 8πε0 b  0 

when 'S3' is closed, total charge will appear on the outer surface of shell 'B'. In this position energy stored E2 =

2

 a − 1 Q 2   8πε0 b  b  1

Heat produced = E1 – E 2 = Q.20 Q.20

Q 2a ( b − a ) 8πε 0 b 3

= 1.8 ]

In the the fi figure gure show shown n, C1 = 11µF and C2 = 5 µF, then at steady st eady state: (A*) the potential difference difference across C1 is 5V







[Sol [Sol::

At stea steady dy stat statee I(3) + I(2) = 15 I=3 KVL C→ D → E → a → b → c V/C – I(3) +

q q –7+ = V/C 11 5

q q + = 7 + 3×3 11 5 q = 55 µC KVL = a → b

= 16

q q 55 55 = V b ⇒ Va – V b = 7 – = 7 – = = –4V 5 5 5 5 P.d. across C1 Va – 7 +

q 55 = = 5V 11 11 P.d. across C2 q = 11V 5 p.d. across acro ss terminal terminal = 15 – I(2) = 15 – 3 × 2 = 9V ] Q.21 Q.21

On a paral paralllel plate plate air air capaci capacitor of capaci capacitan tance ce C0 having plate separation d following steps are performed in the order as given in in column column I. (a) Capa Capaci citor tor is is char charge ged d by conne connecti cting ng it acro across ss a batte battery ry of EMF EMF V0. (b) Diel Dielectr ectric ic of diel dielectr ectric ic constan constantt k and and thickn thickness ess d is is ins inserte erted d (c) (c) Capa Capaci cito torr is is di discon sconn necte ected d fro from m bat batte tery ry (d) (d) Sepa Separa rati tion on betw betwee een n pl plates ates is is doub doublled





[Sol. (A)

(B)

Q C0V0

C C0

C0 V0

C0

2

2

C0 V0

C0

2

2

V0

C0 V0

KC 0

V0 (K + 1)

2

K + 1

2K

C0

0

2

0

C0

2

2

C0

C0

2

2

V0

C0 V0

KC 0

V0 (K + 1)

2

K + 1 KAε 0

2K

V0

0 KC0V0 KC0V0 KC0V0

(D)

V0

C0

V0

C)

V V0

d KC0 KC0 KC 0

C0V0 KC0V0

K + 1 C0 KC0

KC0 V0

KC 0

K + 1

K + 1

V0

0 V0 V0 (K+1)V0 V0 V0 V0

Capacitors Solved Sheet for IIT

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