Capacitor — Page 8 | NEET Physics

Q71

A parallel plate capacitor with plates of area 1 m2 each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate is : (Take

\varepsilon

0 = 8.85

\times

10

-

12

{{{C^2}} \over {N - {m^2}}}

)

A 9.85

\times

10–10 C

B 8.85

\times

10–10 C

C 6.85

\times

10–10 C

D 7.85 × 10–10 C

Correct Answer

Option B

Solution

E = {\sigma \over {{ \in _0}}} = {Q \over {A\,{ \in _0}}}

Q = AE

\in

0 Q = (1) (100) (8.85 $\times$ 10 $-$ 12) Q = 8.85 $\times$ 10 $-$ 10C

Q72

A parallel plate capacitor is made of two circular plates separated by a distance

5

mm

and with a dielectric of dielectric constant

2.2

between them. When the electric field in the dielectric is

3 \times {10^4}\,V/m

the charge density of the positive plate will be close to:

A

6 \times {10^{ - 7}}\,\,C/{m^2}

B

3 \times {10^{ - 7}}\,\,C/{m^2}

C

3 \times {10^4}\,\,C/{m^2}

D

6 \times {10^4}\,\,C/{m^2}

Correct Answer

Option A

Solution

Electric field in presence of dielectric between the two plates of a parallel plate capacitor is given by,

E = {\sigma \over {K{\varepsilon _0}}}

Then, charge density

\sigma = K{\varepsilon _0}E

= 2.2 \times 8.85 \times {10^{ - 12}} \times 3 \times {10^4} \approx 6 \times {10^{ - 7}}\,\,C/{m^2}

Q73

A capacitor is discharging through a resistor R. Consider in time t1, the energy stored in the capacitor reduces to half of its initial value and in time t2, the charge stored reduces to one eighth of its initial value. The ratio t1/t2 will be

A 1/2

B 1/3

C 1/4

D 1/6

Correct Answer

Option D

Solution

For a discharging capacitor when energy reduces to half the charge would become

{1 \over {\sqrt 2 }}

times the initial value.

\Rightarrow {\left( {{1 \over 2}} \right)^{1/2}} = {e^{ - {t_1}/\tau }}

Similarly,

{\left( {{1 \over 2}} \right)^3} = {e^{ - {t_2}/\tau }}

\Rightarrow {{{t_1}} \over {{t_2}}} = {1 \over 6}

Q74

Capacitance of an isolated conducting sphere of radius R1 becomes n times when it is enclosed by a concentric conducting sphere of radius R2 connected to earth. The ratio of their radii

\left( {{{{R_2}} \over {{R_1}}}} \right)

is :

A

{n \over {n - 1}}

B

{{2n} \over {2n + 1}}

C

{{n + 1} \over n}

D

{{2n + 1} \over n}

Correct Answer

Option A

Solution

Initially

= {C_0} = 4\pi {\varepsilon _0}{R_1}

Finally

{{4\pi {\varepsilon _0}{R_1}{R_2}} \over {{R_2} - {R_1}}} = n{C_0} = 4\pi {\varepsilon _0}n{R_1}

$\Rightarrow$

{{{R_2}} \over {{R_2} - {R_1}}} = n

$\Rightarrow$

1 - {{{R_1}} \over {{R_2}}} = {1 \over n}

$\Rightarrow$

{{{R_1}} \over {{R_2}}} = {{n - 1} \over n}

$\Rightarrow$

{{{R_2}} \over {{R_1}}} = {n \over {n - 1}}

Q75

A slab of dielectric constant

\mathrm{K}

has the same cross-sectional area as the plates of a parallel plate capacitor and thickness

\dfrac{3}{4} \mathrm{~d}

, where

\mathrm{d}

is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be : (Given

\mathrm{C}_{0}

= capacitance of capacitor with air as medium between plates.)

A

\dfrac{4 K C_{0}}{3+K}

B

\dfrac{3 K C_{0}}{3+K}

C

\dfrac{3+K}{4 K C_{0}}

D

\dfrac{K}{4+K}

Correct Answer

Option A

Solution

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

C = {{{\varepsilon _0}A} \over {d - {{3d} \over 4} + {{3d} \over {4K}}}} = {{4{\varepsilon _0}AK} \over {3d + Kd}}

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

Q76

A capacitor C is fully charged with voltage V0. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance

{C \over 2}

. The energy loss in the process after the charge is distributed between the two capacitors is :

A

{1 \over 2}CV_0^2

B

{1 \over 4}CV_0^2

C

{1 \over 3}CV_0^2

D

{1 \over 6}CV_0^2

Correct Answer

Option D

Solution

Heat loss =

{1 \over 2}\left( {{{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)V_0^2

=

{1 \over 2}\left( {{{C \times {C \over 2}} \over {C + {C \over 2}}}} \right)V_0^2

=

{1 \over 6}CV_0^2

Q77

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.

Q78

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: $\dfrac{1}{4} CV^2$

Q79

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

Q80

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}}