Capacitor — Page 10 | NEET Physics

Q91

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

Q92

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}

Q93

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: $1.8 \times 10^3 \, \text{J/m}^3$

Q94

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}}

Q95

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

Q96

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

Q97

A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be

A

1/2

B

1

C

2

D

1/4

Correct Answer

Option A

Solution

Required ratio

= {{Energy\,\,\,stored\,\,in\,\,capacitor} \over {Workdone\,\,by\,\,the\,\,battery}} = {{{1 \over 2}C{V^2}} \over {C{e^2}}}

where

C=

Capacitance of capacitor

V=

Potential difference,

e=

emf

of battery

= {{{1 \over 2}C{e^2}} \over {C{e^2}}} = {1 \over 2}

\left( \, \right.

as

V=e

\left. \, \right)

Q98

If qf is the free charge on the capacitor plates and qb is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge qb an be expressed as :

A

{q_b} = {q_f}\left( {1 - {1 \over {\sqrt k }}} \right)

B

{q_b} = {q_f}\left( {1 - {1 \over k}} \right)

C

{q_b} = {q_f}\left( {1 + {1 \over {\sqrt k }}} \right)

D

{q_b} = {q_f}\left( {1 + {1 \over k}} \right)

Correct Answer

Option B

Solution

When a dielectric is inserted in a capacitor Due to free charge

\overrightarrow E = {\overrightarrow E _0}

only After dielectric

E' = {{{E_0}} \over k}

{q_B} = {q_f}\left( {1 - {1 \over k}} \right)