Capacitor — Page 4 | NEET Physics

Q31

A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulting handle. As a result the potential difference between the plates

A increases

B decreases

C does not charge

D becomes zero

Correct Answer

Option A

Solution

If we increase the distance between the plates its capacity decreases resulting in higher potential as we know Q = CV.

Since Q is constant (battery has been disconnected), on decreasing C, V will increase.

Q32

A network of four capacitors of capacity equal to C 1 = C, C 2 = 2C, C 3 = 3C and C 4 = 4C are connected to a battery as shown in the figure. The ratio of the charges on C 2 and C 4 is

A 4/7

B 3/22

C 7/4

D 22/3

Correct Answer

Option B

Solution

Equivalent capacitance for three capacitors (C 1 , C 2 & C 3 ) in series is given by

{1 \over {{C_{eq.}}}} = {1 \over {{C_1}}} + {1 \over {{C_2}}} + {1 \over {{C_3}}}

= {{{C_2}{C_3} + {C_3}{C_1} + {C_1}{C_2}} \over {{C_1}{C_2}{C_3}}}

\Rightarrow {C_{eq.}} = {{{C_1}{C_2}{C_3}} \over {{C_2}{C_3} + {C_3}{C_1} + {C_1}{C_2}}}

\Rightarrow {C_{eq.}} = {{2\left( {2C} \right)\left( {3C} \right)} \over {C\left( {2C} \right) + \left( {2C} \right)\left( {3C} \right) + \left( {3C} \right)C}} = {6 \over {11}}C

$\Rightarrow$ Charge on capacitors (C 1 , C 2 & C 3 ) in series

= {C_{eq}}V = {{6C} \over {11}}V

Charge on capacitor

{C_4} = {C_4}V = 4CV

{{Charg e\,on\,{C_2}} \over {Charg e\,on\,{C_4}}} = {{{{6C} \over {11}}V} \over {4CV}} = {6 \over {11}} \times {1 \over 4} = {3 \over {22}}

Q33

Three capacitors each of capacity 4

\mu

F are to be connected in such a way that the effective capacitance is

6\mu F.

This can be done by

A connecting all of them in series

B connecting them in parallel

C connecting two in series and one in parallel

D connecting two in parallel and one in series

Correct Answer

Option C

Solution

For series,

C' = {{{C_1} \times {C_2}} \over {{C_1} + {C_2}}} = {{4 \times 4} \over {4 + 4}} = 2\mu F

For parallel

{C_{eq}} = C' + {C_3} = 2 + 4 = 6\mu F

Q34

A capacitor of capacity C 1 charged upto V volt and then connected to an uncharged capacitor of capacity C 2 . The final potential difference across each will be

A

{{{C_1}V} \over {{C_1} + {C_2}}}

B

{{{C_2}V} \over {{C_1} + {C_2}}}

C

\left( {1 - {{{C_2}} \over {{C_1}}}} \right)V

D

\left( {1 + {{{C_2}} \over {{C_1}}}} \right)

Correct Answer

Option A

Solution

Charge Q = C 1 V Total capacity of combination (parallel) C = C 1 + C 2

P.D. = {Q \over C} = {{{C_1}V} \over {{C_1} + {C_2}}}

Q35

Energy per unit volume for a capacitor having area A and separation d kept at potential difference V is given by

A

{1 \over 2}{\varepsilon _0}{{{V^2}} \over {{d^2}}}

B

{1 \over {2{\varepsilon _0}}}{{{V^2}} \over {{d^2}}}

C

{1 \over 2}C{V^2}

D

{{{Q^2}} \over {2C}}.

Correct Answer

Option A

Solution

Energy stored per unit volume

= {1 \over 2}{\varepsilon _0}{E^2} = {1 \over 2}{\varepsilon _0}{\left( {{V \over d}} \right)^2} = {1 \over 2}{\varepsilon _0}{{{V^2}} \over {{d^2}}}

\left(\because {E = {V \over d}} \right)

Q36

A capacitor is charged with a battery and energy stored is U. After disconnecting battery another capacitor of same capacity is connected in parallel to the first capacitor. Then energy stored in each capacitor is

A

U/2

B U/4

C 4U

D 2U

Correct Answer

Option B

Solution

Let q be the charge on each capacitor. $\therefore$ Energy stored,

U = {1 \over 2}C{V^2} = {1 \over 2}{{{q^2}} \over C}

Now, when battery is disconnected and another capacitor of same capacity is connected in parallel to the first capacitor, then voltage across each capacitor,

V = {q \over {2C}}

$\therefore$ Energy stored =

{1 \over 2}C{\left( {{q \over {2C}}} \right)^2} = {1 \over 4}.{1 \over 2}{{{q^2}} \over C} = {1 \over 4}U.

Q37

A capacitor has air as dielectric medium and two conducting plates of area

12 \mathrm{~cm}^2

and they are

0.6 \mathrm{~cm}

apart. When a slab of dielectric having area

12 \mathrm{~cm}^2

and

0.6 \mathrm{~cm}

thickness is inserted between the plates, one of the conducting plates has to be moved by

0.2 \mathrm{~cm}

to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given

\epsilon_0=8.834 \times 10^{-12} \mathrm{~F} / \mathrm{m}

)

A 1.50

B 0.66

C 1.33

D 1

Correct Answer

Option A

Solution

\begin{aligned} & \frac{\in_0 A}{d}=\frac{\epsilon_0 A}{\frac{d}{k}+\frac{0.2}{1}} \\ & \Rightarrow \quad 0.6-0.2=\frac{0.6}{k} \\ & \quad k=\frac{0.6}{0.4}=1.5 \end{aligned}

Q38

Capacitance (in

F

) of a spherical conductor with radius

1

m

is

A

1.1 \times {10^{ - 10}}

B

{10^{ - 6}}

C

9 \times {10^{ - 9}}

D

{10^{ - 3}}

Correct Answer

Option A

Solution

For an isolated sphere, the capacitance is given by

C = 4\pi \,{ \in _0}\,r

= {1 \over {9 \times {{10}^9}}} \times 1

= 1.1 \times {10^{ - 10}}F

Q39

Two capacitors of capacities 2C and C are joined in parallel and charged up to potential V. The battery is removed and the capacitor of capacity C is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be :

A

{V \over {K + 2}}

B

{V \over K}

C

{{3V} \over {K + 2}}

D

{{3V} \over K}

Correct Answer

Option C

Solution

{V_C} = {{2CV + CV} \over {KC + 2C}}

= {{3V} \over {K + 2}}

Q40

If the charge on a capacitor is increased by 2 C, the energy stored in it increases by 44%. The original charge on the capacitor is (in C)

A 10

B 20

C 30

D 40

Correct Answer

Option A

Solution

Let initially the charge is q so

{1 \over 2}{{{q^2}} \over C} = {U_i}

And

{1 \over 2}{{{{(q + 2)}^2}} \over C} = {U_f}

Given

{{{U_f} - {U_i}} \over {{U_i}}} \times 100 = 44

{{{{(q + 2)}^2} - {q^2}} \over q} = .44

\Rightarrow q = 10C