Capacitor — Page 6 | JEE Physics

Q51

A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will :

A increase by 50%

B decrease by 15%

C increase by 25%

D increase by 33%

Correct Answer

Option A

Solution

U = {1 \over 2}(k{C_0}){V^2}

\Rightarrow {{U'} \over U} = 1.5

$\Rightarrow$ Energy increases by 50%

Q52

A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is :

A 508 pJ

B 692 pJ

C 560 pJ

D 600 pJ

Correct Answer

Option A

Solution

Initial energy of capacitor Ui =

{1 \over 2}

{{{v^2}} \over c}

=

{1 \over 2}

$\times$

{{120 \times 120} \over {12}}

= 600 J Since battery is disconnected so charge remain same. Final energy of capacitor Uf =

{1 \over 2}{{{v^2}} \over c}

=

{1 \over 2} \times {{120 \times 120} \over {12 \times 6.5}}

= 92 W + Uf = Ui W = 508 J

Q53

A parallel plate condenser with a dielectric of dielectric constant

K

between the plates has a capacity

C

and is charged to a potential

V

volt. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is

A zero

B

{1 \over 2}\,\left( {K - 1} \right)\,C{V^2}

C

{{C{V^2}\left( {K - 1} \right)} \over K}

D

\left( {K - 1} \right)\,C{V^2}

Correct Answer

Option A

Solution

First, let's consider the capacitor with the dielectric between its plates.

The charge on the capacitor is Q, its capacitance is C (which includes the effect of the dielectric), and the potential difference (voltage) across its plates is V.

According to the formula for the energy stored in a capacitor :

U = \frac{1}{2} CV^2

we find that the energy is equal to half of the product of the capacitance and the square of the voltage.

Now, consider the process where the dielectric slab is removed from the capacitor.

When the dielectric is removed, the capacitance of the capacitor decreases.

However, because the capacitor is not connected to anything that can supply or absorb charge, the charge Q on the capacitor stays the same.

Since the charge stays the same but the capacitance decreases, the voltage V across the capacitor must increase to keep Q = CV true.

The energy of the capacitor without the dielectric is still given by :

U = \frac{1}{2} CV^2

but now C is smaller and V is larger.

However, because both C and V2 change in such a way that their product remains constant, the energy of the capacitor doesn't change when the dielectric is removed.

Therefore, the energy of the capacitor before the dielectric is removed is the same as the energy of the capacitor after the dielectric is removed.

When the dielectric is reinserted, the process is just reversed, so again no net work is done.

So the total work done by the system in the process of removing the dielectric and then reinserting it is zero.

This is because work is the transfer of energy, and in this case, the energy of the system (the charged capacitor) does not change.

Therefore, no energy is transferred, so no work is done.

Q54

A 10

\mu

F capacitor is fully charged to a potential difference of 50 V. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is :

A 20

\mu

F

B 15

\mu

F

C 10

\mu

F

D 30

\mu

F

Correct Answer

Option B

Solution

Initially, Charge on capacitor 10 μF Q = CV = (10 μF) (50V) Q = 500 μC Final Charge on 10 μF capacitor Q = CV = (10 μF) (20V) Q = 200 μC From charge conservation, Charge on unknown capacitor Q = 500 μC – 200 μC = 300 μC $\Rightarrow$ Capacitance (C) =

{Q \over V}

=

{{300} \over {20}}

= 15 $\mu$ F

Q55

A parallel plate capacitor with area 200 cm2 and separation between the plates 1.5 cm, is connected across a battery of emf V. If the force of attraction between the plates is

25 \times {10^{ - 6}}N,

the value of V is approximately :

\left( {{ \in _o} = 8.85 \times {{10}^{ - 12}}{{{C^2}} \over {N.{m^2}}}} \right)

A 250 V

B 100 V

C 300 V

D 150 V

Correct Answer

Option A

Solution

Given area of Parallel plate capacitor, A = 200 cm2 Separation between the plates, d = 1.5 cm Force of attraction between the plates, F = 25 × 10–6 N F = QE $\Rightarrow$ F =

{{{Q^2}} \over {2A{ \in _0}}}

[ As E due to parallel plate =

{\sigma \over {2{ \in _0}}} = {Q \over {2A{ \in _0}}}

] Also we know, Q = CV =

{{{ \in _0}AV} \over d}

$\therefore$ F =

{{{{\left( {{ \in _0}AV} \right)}^2}} \over {{d^2} \times 2A{ \in _0}}}

$\Rightarrow$ V = d

\sqrt {{{2F} \over {{ \in _0}A}}}

$\Rightarrow$ V =

1.5 \times {10^{ - 2}}\sqrt {{{2 \times 25 \times {{10}^{ - 6}}} \over {8.85 \times {{10}^{ - 12}} \times 2 \times {{10}^{ - 2}}}}}

=

1.5 \times {10^{ - 2}}\sqrt {{{25} \over {8.85}}}

= 250 V

Q56

For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is

{3 \over 4}

d, where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance (C0) is given by the following relation :

A

C' = {{3 + K} \over {4K}}{C_0}

B

C' = {{4 + K} \over {3}}{C_0}

C

C' = {{4K} \over {K + 3}}{C_0}

D

C' = {{4} \over {3 + K}}{C_0}

Correct Answer

Option C

Solution

{C_0} = {{{ \in _0}A} \over d}

$\therefore$

{1 \over {C'}} = {1 \over {{C_1}}} + {1 \over {{C_2}}}

{1 \over {C'}} = {{(3d/4)} \over {{ \in _0}KA}} + {{(d/4)} \over {{ \in _0}A}}

{1 \over {C'}} = {d \over {4{ \in _0}A}}\left( {{{3 + K} \over K}} \right)

$\therefore$

C' = {{4K} \over {(K + 3)}}{C_0}

Q57

A parallel plate capacitor of capacitance 1 µF is charged to a potential difference of 20 V. The distance between plates is 1 µm. The energy density between plates of capacitor is :

A

1.8 \times 10^3

J/m3

B

2 \times 10^2

J/m3

C

2 \times 10^{-4}

J/m3

D

1.8 \times 10^5

J/m3

Correct Answer

Option A

Solution

The energy density ( $u$ ) between the plates of a capacitor is given by the formula:

u = \frac{1}{2} \varepsilon_0 E^2

where $\varepsilon_0$ is the permittivity of free space ( $8.85 \times 10^{-12} \, \text{F/m}$ ) and $E$ is the electric field between the plates.

The electric field $E$ is related to the potential difference $V$ and the separation $d$ between the plates by:

E = \frac{V}{d}

Given: Capacitance ( $C$ ) = 1 µF = $1 \times 10^{-6} \, \text{F}$ Potential Difference ( $V$ ) = 20 V Distance ( $d$ ) = 1 µm = $1 \times 10^{-6} \, \text{m}$ First, calculate the electric field $E$ :

E = \frac{V}{d} = \frac{20 \, \text{V}}{1 \times 10^{-6} \, \text{m}} = 2 \times 10^7 \, \text{V/m}

Now plug this into the formula for energy density:

u = \frac{1}{2} \times 8.85 \times 10^{-12} \, \text{F/m} \times (2 \times 10^7 \, \text{V/m})^2

Calculate $u$ :

u = \frac{1}{2} \times 8.85 \times 10^{-12} \times 4 \times 10^{14}

u = 2 \times 8.85 \times 10^2

u = 1770 \, \text{J/m}^3

This value is approximately $1.8 \times 10^3 \, \text{J/m}^3$ .

Therefore, the correct option is: Option A: $1.8 \times 10^3 \, \text{J/m}^3$

Q58

Two identical thin metal plates has charge

q_{1}

and

q_{2}

respectively such that

q_{1}>q_{2}

. The plates were brought close to each other to form a parallel plate capacitor of capacitance C. The potential difference between them is :

A

\dfrac{\left(q_{1}+q_{2}\right)}{C}

B

\dfrac{\left(q_{1}-q_{2}\right)}{C}

C

\dfrac{\left(q_{1}-q_{2}\right)}{2 C}

D

\dfrac{2\left(q_{1}-q_{2}\right)}{C}

Correct Answer

Option C

Solution

Charge on the left surface of plate

\mathrm{A} = {{\mathrm{Total\,charge}} \over 2}

= {{{q_1} + {q_2}} \over 2}

Let right surface of plate A has charge

= x

And total charge on plate

A = {q_1}

$\therefore$

q_1

= Charge on left surface of plate A + Charge on right surface of plate A

= {{{q_1} + {q_2}} \over 2} + x

\Rightarrow x = {q_1} - {{{q_1} + {q_2}} \over 2}

= {{2{q_1} - {q_1} - {q_2}} \over 2}

= {{{q_1} - {q_2}} \over 2}

Let potential difference between two plates

= V

For capacitor we know,

q = CV

$\therefore$

{{{q_1} - {q_2}} \over 2} = CV

\Rightarrow V = {{{q_1} - {q_2}} \over {2C}}

Q59

The material filled between the plates of a parallel plate capacitor has resistivity 200

\Omega

m. The value of capacitance of the capacitor is 2 pF. If a potential difference of 40 V is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is : (given the value of relative permittivity of material is 50)

A 9.0

\mu

A

B 9.0 mA

C 0.9 mA

D 0.9

\mu

A

Correct Answer

Option C

Solution

$\rho$ = 200

\Omega

m C = 2 $\times$ 10 $-$ 12 F V = 40 V K = 56

i = {q \over {\rho k{\varepsilon _0}}} = {{{q_0}} \over {\rho k{\varepsilon _0}}}{e^{ - {t \over {\rho k{\varepsilon _0}}}}}

{i_{\max }} = {{2 \times {{10}^{ - 12}} \times 40} \over {200 \times 50 \times 8.85 \times {{10}^{ - 12}}}}

= {{80} \over {{{10}^4} \times 8.85}}

= 903 $\mu$ A = 0.9 mA

Q60

A parallel plate capacitor is made by stacking

n

equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is

'C'

then the resultant capacitance is

A

\left( {n + 1} \right)C

B

\left( {n - 1} \right)C

C

nC

D

C

Correct Answer

Option B

Solution

As

n

plates are joined, it means

(n-1)

capacitor joined in parallel. $\therefore$ resultant capacitance

=(n-1)C