Capacitor — Page 9 | NEET Physics

Q81

The distance between two plates of a capacitor is

\mathrm{d}

and its capacitance is

\mathrm{C}_{1}

, when air is the medium between the plates. If a metal sheet of thickness

\dfrac{2 d}{3}

and of the same area as plate is introduced between the plates, the capacitance of the capacitor becomes

\mathrm{C}_{2}

. The ratio

\dfrac{\mathrm{C}_{2}}{\mathrm{C}_{1}}

is

A 1 : 1

B 3 : 1

C 2 : 1

D 4 : 1

Correct Answer

Option B

Solution

When a metal sheet of thickness ( $\dfrac{2d}{3}$ ) is introduced between the plates of a capacitor, it divides the capacitor into two separate capacitors.

The metal sheet acts as a new plate in each capacitor, and because the metal is a conductor, it is at the same potential as the plates on either side.

The capacitance of the original capacitor with air between the plates is given by: $\mathrm{C}_1 = \dfrac{\epsilon_0 \mathrm{A}}{\mathrm{d}}$ where ( $\epsilon_0$ ) is the permittivity of free space, ( $\mathrm{A}$ ) is the area of one of the plates, and ( $\mathrm{d}$ ) is the distance between the plates.

When the metal sheet of thickness ( $\dfrac{2d}{3}$ ) is introduced, the distance between the plates is reduced by ( $\dfrac{2d}{3}$ ), so the capacitance of each of the new capacitors is: $\mathrm{C}' = \dfrac{\epsilon_0 \mathrm{A}}{\mathrm{d}-\dfrac{2d}{3}}$ Simplifying, we get: $\mathrm{C}' = \dfrac{3\epsilon_0 \mathrm{A}}{\mathrm{d}}$ Since the two capacitors are in parallel with each other, the total capacitance is: $\mathrm{C}_2 = 2\mathrm{C}' = 2 \times \dfrac{3\epsilon_0 \mathrm{A}}{\mathrm{d}} = 3\mathrm{C}_1$ So the ratio of the capacitances is: $\dfrac{\mathrm{C}_2}{\mathrm{C}_1} = 3:1$

Q82

Voltage rating of a parallel plate capacitor is 500V. Its dielectric can withstand a maximum electric field of 106 V/m. The plate area is 10–4 m2. What is the dielectric constant is the capacitance is 15 pF? (given

\varepsilon

0 = 8.86 × 10–12 C2/Nm2)

A 8.5

B 4.5

C 3.8

D 6.2

Correct Answer

Option A

Solution

A = 10–4 m2 Emax = 106 V/m C = 15 $\mu$ F

C = {{k{\varepsilon _0}A} \over d};{{Cd} \over {{\varepsilon _0}A}} = k

k = {{15 \times {{10}^{ - 12}} \times 500 \times {{10}^{ - 6}}} \over {8.86 \times {{10}^{ - 12}} \times {{10}^4}}}

=

{{15 \times 5} \over {8.86}} = 8.465

k $\approx$ 8.5

Q83

The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery of voltage, V. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is :-

A V

B

{V \over {K + n}}

C

{{(n+1)V} \over {K + n}}

D

{{nV} \over {K + n}}

Correct Answer

Option C

Solution

Initially Q = CV(1 + n) $\therefore$ Ceq = (K + n)C $\therefore$

V = {{CV\left( {1 + n} \right)} \over {\left( {K + n} \right)C}} = {{V\left( {1 + n} \right)} \over {\left( {K + n} \right)}}

Q84

A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be.

A 5 N

B 10 N

C 20 N

D Zero

Correct Answer

Option A

Solution

E between two plates is

{\sigma \over {{\varepsilon _0}}}

and due to one plate is

{\sigma \over {2{\varepsilon _0}}}

so the force will be halved So new force F = 5 N

Q85

A parallel plate capacitor with plate area 'A' and distance of separation 'd' is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as :

\varepsilon (x) = {\varepsilon _0} + kx

, for

\left( {0 < x \le {d \over 2}} \right)

\varepsilon (x) = {\varepsilon _0} + k(d - x)

, for

\left( {{d \over 2} \le x \le d} \right)

A

{\left( {{\varepsilon _0} + {{kd} \over 2}} \right)^{2/kA}}

B

{{kA} \over {2\ln \left( {{{2{\varepsilon _0} + kd} \over {2{\varepsilon _0}}}} \right)}}

C 0

D

{{kA} \over 2}\ln \left( {{{2{\varepsilon _0}} \over {2{\varepsilon _0} - kd}}} \right)

Correct Answer

Option B

Solution

Taking an element of width dx at a distance x(x < d/2) from left plate

dc = {{({\varepsilon _0} + kx)A} \over {dx}}

Capacitance of half of the capacitor

{1 \over C} = \int\limits_0^{d/2} {{1 \over {dc}} = {1 \over A}\int\limits_0^{d/2} {{{dx} \over {{\varepsilon _0} + kx}}} }

{1 \over C} = {1 \over {kA}}\ln \left( {{{{\varepsilon _0} + kd/2} \over {{\varepsilon _0}}}} \right)

Capacitance of second half will be same

{C_{eq}} = {C \over 2} = {{kA} \over {2\ln \left( {{{2{\varepsilon _0} + kd} \over {2{\varepsilon _0}}}} \right)}}

Q86

A parallel plate capacitor with air between the plates has capacitance of

9

pF.

The separation between its plates is

'd'.

The space between the plates has dielectric constant

{k_1}

=3

and thickness

{d \over 3}

while the other one has dielectric constant

{k_2} = 6

and thickness

{{2d} \over 3}

. Capacitance of the capacitor is now

A

1.8

pF

B

45

pF

C

40.5

pF

D

20.25

pF

Correct Answer

Option C

Solution

The given capacitance is equal to two capacitances connected in series where

{C_1} = {{{k_1}{ \in _0}A} \over {d/3}} = {{3{k_1}{ \in _0}A} \over d}

= {{3 \times 3{ \in _0}A} \over d} = {{9{ \in _0}A} \over d}

and

{C_2} = {{{k_2}{ \in _0}A} \over {2d/3}} = {{3{k_2}{ \in _0}A} \over {2d}}

= {{3 \times 6{ \in _0}A} \over {2d}} = {{9{ \in _0}A} \over d}

The equivalent capacitance

{C_{eq}}

is

{1 \over {C{}_{eq}}} = {1 \over {{C_1}}} + {1 \over {{C_2}}}

= {d \over {9{ \in _0}A}} + {d \over {9{ \in _0}A}}

= {{2d} \over {9{ \in _0}A}}

$\therefore$

{C_{eq}} = {9 \over 2}{{{\varepsilon _0}A} \over d} = {9 \over 2} \times 9pF = 40.5pF

Q87

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%

Q88

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

Q89

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.

Q90

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