Capacitor Questions — JEE Physics

Q1

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}

Q2

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

Q3

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}}

Q4

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

Q5

The total charge on the system of capacitors

C_{1}=1 \mu \mathrm{F}, C_{2}=2 \mu \mathrm{F}, \mathrm{C}_{3}=4 \mu \mathrm{F}

and

\mathrm{C}_{4}=3 \mu \mathrm{F}

connected in parallel is : (Assume a battery of

20 \mathrm{~V}

is connected to the combination)

A

200 \,\mu \mathrm{C}

B 200 C

C

10 \,\mu \mathrm{C}

D 10 C

Correct Answer

Option A

Solution

Equivalent

C = \sum {{C_i}}

= 10\,\mu F

$\Rightarrow$ Charge

Q = CV

= 200\,\mu C

Q6

Two capacitors, each having capacitance

40 \,\mu \mathrm{F}

are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant

\mathrm{K}

such that the equivalence capacitance of the system became

24 \,\mu \mathrm{F}

. The value of

\mathrm{K}

will be :

A 1.5

B 2.5

C 1.2

D 3

Correct Answer

Option A

Solution

{{40K \times 40} \over {40K + 40}} = 24

40K = 24(K + 1)

40K = 24K + 24

16K = 24

K = {{24} \over {16}} = {3 \over 2} = 1.5

Q7

Which one of the following is the correct dimensional formula for the capacitance in F ?

\mathrm{M}, \mathrm{L}, \mathrm{T}

and

C

stand for unit of mass, length, time and charge,

A

[\mathrm{F}]=\left[\mathrm{CM}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2\right]

B

[\mathrm{F}]=\left[\mathrm{C}^2 \mathrm{M}^{-2} \mathrm{~L}^2 \mathrm{~T}^2\right]

C

[\mathrm{F}]=\left[\mathrm{C}^2 \mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2\right]

D

[\mathrm{F}]=\left[\mathrm{CM}^{-2} \mathrm{~L}^{-2} \mathrm{~T}^{-2}\right]

Correct Answer

Option C

Solution

The capacitance (in farads) is defined as the ratio of charge to potential difference.

Let's go through the steps: Capacitance is given by:

C = \frac{Q}{V}

where: $Q$ is the charge with dimensional symbol $C$ . $V$ is the potential difference.

Voltage (potential difference) is defined as energy per unit charge:

V = \frac{W}{Q}

where: $W$ is energy with dimensions:

[W] = [M L^2 T^{-2}]

Therefore, the dimensions of voltage are:

[V] = \frac{[ML^2T^{-2}]}{[C]}

Now, substitute this back into the expression for capacitance:

[C] = \frac{[Q]}{[ML^2T^{-2}]/[C]} = \frac{[C]^2}{ML^2T^{-2}}

Simplifying gives:

[C] = [C^2 M^{-1}L^{-2}T^2]

Comparing with the given options, we see that Option C is:

[F] = [C^2 M^{-1}L^{-2}T^2]

Thus, the correct dimensional formula for the capacitance in farads is given by Option C.

Q8

A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor

A decreases

B remains unchanged

C becomes infinite

D increases

Correct Answer

Option B

Solution

The capacitancce of parallel plate capacitor in which a metal plate of thickness

t

is inserted is given by

C = {{{\varepsilon _0}A} \over {d - t}}.\,\,\,\,\,

Here

t \to 0\,\,\,\,\,\,

$\therefore$

C = {{{\varepsilon _0}A} \over d}

Q9

The work done in placing a charge of

8 \times {10^{ - 18}}

coulomb on a condenser of capacity

100

micro-farad is

A

16 \times {10^{ - 32}}\,\,joule

B

3.1 \times {10^{ - 26}}\,\,joule

C

4 \times {10^{ - 10}}\,\,joule

D

32 \times {10^{ - 32}}\,\,joule

Correct Answer

Option D

Solution

The work done is stored as the potential energy. The potential energy stored in a capacitor is given by

U = {1 \over 2}{{{Q^2}} \over C}

= {1 \over 2} \times {{{{\left( {8 \times {{10}^{ - 18}}} \right)}^2}} \over {100 \times {{10}^{ - 6}}}}

= 32 \times {10^{ - 32}}J

Q10

Match with .

List - I		List - II
(a)	Capacitance, C	(i)	${M^1}{L^1}{T^{ - 3}}{A^{ - 1}}$
(b)	Permittivity of free space, ${\varepsilon _0}$	(ii)	${M^{ - 1}}{L^{ - 3}}{T^4}{A^2}$
(c)	Permeability of free space, ${\mu _0}$	(iii)	${M^{ - 1}}{L^{ - 2}}{T^4}{A^2}$
(d)	Electric field, E	(iv)	${M^1}{L^1}{T^{ - 2}}{A^{ - 2}}$

A (a)

\to

(iii), (b)

\to

(ii), (c)

\to

(iv), (d)

\to

(i)

B (a)

\to

(iii), (b)

\to

(iv), (c)

\to

(ii), (d)

\to

(i)

C (a)

\to

(iv), (b)

\to

(ii), (c)

\to

(iii), (d)

\to

(i)

D (a)

\to

(iv), (b)

\to

(iii), (c)

\to

(ii), (d)

\to

(i)

Correct Answer

Option A

Solution

q = CV

[C] = \left[ {{q \over V}} \right] = {{(A \times T)} \over {M{L^2}{T^{ - 2}}}}

= {M^{ - 1}}{L^{ - 2}}{T^4}{A^2}

[E] = \left[ {{F \over q}} \right] = {{ML{T^{ - 2}}} \over {AT}}

= ML{T^{ - 3}}{A^{ - 1}}

F = {{{q_1}{q_2}} \over {4\pi { \in _0}{r^2}}}

[{ \in _0}] = {M^{ - 1}}{L^{ - 3}}{T^4}{A^2}

Speed of light

c = {1 \over {\sqrt {{\mu _0}{ \in _0}} }}

{\mu _0} = {1 \over {{ \in _0}{c^2}}}

[{\mu _0}] = {1 \over {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}]{{[L{T^{ - 1}}]}^2}}}

= [{M^1}{L^1}{T^{ - 2}}{A^{ - 2}}]