Capacitor — Page 5 | JEE Physics

Q41

Given below are two statements: One is labeled as Assertion A and the other is labeled as Reason R. Assertion A : Two metallic spheres are charged to the same potential. One of them is hollow and another is solid, and both have the same radii. Solid sphere will have lower charge than the hollow one. Reason R : Capacitance of metallic spheres depend on the radii of spheres In light of the above statements, choose the correct answer from the options given below.

A Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is not the correct explanation of

\mathbf{A}

B Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

C

\mathbf{A}

is false but

\mathbf{R}

is true

D

\mathbf{A}

is true but

\mathbf{R}

is false

Correct Answer

Option C

Solution

The amount of charge on each sphere will be the same if they are charged to the same potential.

Whether the sphere is solid or hollow doesn't affect the amount of charge stored on the sphere as long as they have the same radii and are charged to the same potential.

Therefore, assertion A is false As we know, capacitance of spherical conductor

\mathrm{C}=4 \pi \varepsilon_0 \mathrm{R}

So, capacitance does not depend on its charge, it depends only on the radius of the conductor (R).

Therefore, Reason $\mathrm{R}$ is true.

Q42

Two identical capacitors have same capacitance

C

. One of them is charged to the potential

V

and other to the potential

2 \mathrm{~V}

. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is :

A

\dfrac{1}{4} \mathrm{CV}^2

B

\dfrac{3}{4} \mathrm{CV}^2

C

\dfrac{1}{2} \mathrm{CV}^2

D

2 \mathrm{CV}^2

Correct Answer

Option A

Solution

To solve the problem, let's start by considering the energy stored in each capacitor before they are connected together.

The energy stored in a capacitor is given by the formula:

E = \frac{1}{2} C V^2

Where $E$ is the energy, $C$ is the capacitance, and $V$ is the potential.

For the first capacitor charged to the potential $V$ , the energy stored is:

E_1 = \frac{1}{2} C V^2

For the second capacitor charged to the potential $2V$ , the energy stored is:

E_2 = \frac{1}{2} C (2V)^2 = \frac{1}{2} C \cdot 4V^2 = 2 CV^2

The total initial energy stored in the system is the sum of $E_1$ and $E_2$ :

E_{\text{initial}} = E_1 + E_2 = \frac{1}{2} CV^2 + 2 CV^2 = \frac{5}{2} CV^2

When the two capacitors are connected together, their potentials will become equal because they are identical capacitors.

Let's denote this final potential as $V_f$ .

The total charge before and after the connection remains constant because charge is conserved.

Therefore, we can write:

C \cdot V + C \cdot 2V = 2C \cdot V_f

Simplifying this equation gives us the final potential:

V + 2V = 2 V_f

3V = 2 V_f

V_f = \frac{3}{2} V

The final energy stored in the system when the capacitors are connected is now the total energy stored across both capacitors at the final potential $V_f$ :

E_{\text{final}} = 2 \cdot \frac{1}{2} C V_f^2 = 2 \cdot \frac{1}{2} C \left(\frac{3}{2} V\right)^2 = C \cdot \frac{9}{4} V^2

The decrease in energy $\Delta E$ of the combined system is the initial energy minus the final energy:

\Delta E = E_{\text{initial}} - E_{\text{final}}

\Delta E = \frac{5}{2} CV^2 - \frac{9}{4} CV^2

\Delta E = \frac{10}{4} CV^2 - \frac{9}{4} CV^2

\Delta E = \frac{1}{4} CV^2

The correct answer is: Option A: $\dfrac{1}{4} CV^2$

Q43

A parallel-plate capacitor of capacitance

40 \mu \mathrm{~F}

is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant

\mathrm{K}=2

. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are

A 8 mC and 2.0 J

B 4 mC and 0.2 J

C 2 mC and 0.2 J

D 2 mC and 0.4 J

Correct Answer

Option B

Solution

Given,

K = 2,\,C = 40\mu F,\,V = 100\,V

\Delta q = (KC)V - CV

(As

q = CV

)

= (K - 1)CV

= (2 - 1) \times 40 \times {10^{ - 6}} \times 100 = 4 \times {10^{ - 3}}C

\Delta q = 4\,mC

\Delta u = {1 \over 2}C'{V^2} - {1 \over 2}C{V^2}

= {1 \over 2}{V^2}(C' - C) = {1 \over 2}{V^2}(KC - C)

= {1 \over 2}{V^2}C(K - 1) = {1 \over 2} \times {10^4} \times 40 \times {10^{ - 6}}(2 - 1)

= \Delta u = 2 \times {10^{ - 1}} = 0.2\,J

\Rightarrow \Delta u = 0.2\,J

Q44

A fully charged capacitor has a capacitance

'C'

. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity

's'

and mass

'm'.

If the temperature of the block is raised by

'\Delta T',

the potential difference

'v'

across the capacitance is

A

{{mCAT} \over s}

B

\sqrt {{{2mCAT} \over s}}

C

\sqrt {{{2msAT} \over C}}

D

{{ms\Delta T} \over C}

Correct Answer

Option C

Solution

Applying conservation of energy,

{1 \over 2}C{V^2} = m.s\Delta T;\,\,\,

V = \sqrt {{{2m.s.\Delta T} \over C}}

Q45

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: $\dfrac{\mathrm{C}_2}{\mathrm{C}_1} = 3:1$

Q46

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

Q47

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)}}

Q48

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

Q49

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)}}

Q50

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