Capacitor — Page 7 | NEET Physics

Q61

A capacitance of 2

\mu

F is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1

\mu

F capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is:

A 2

B 16

C 32

D 24

Correct Answer

Option C

Solution

To get a capacitance of 2 μF arrangement of capacitors of capacitance 1μF as shown in figure 8 capacitors of 1μF in parallel with four such branches in series i.e., 32 such capacitors are required.

{1 \over {{C_{eq}}}} = {1 \over 8} + {1 \over 8} + {1 \over 8} + {1 \over 8}

$\Rightarrow$

{1 \over {{C_{eq}}}} = {1 \over 2}

$\Rightarrow$

{{C_{eq}} = 2}

Q62

A capacitor of capacitance

\mathrm{C}

is charged to a potential V. The flux of the electric field through a closed surface enclosing the positive plate of the capacitor is :

A Zero

B

\dfrac{C V}{\varepsilon_{0}}

C

\dfrac{C V}{2 \varepsilon_{0}}

D

\dfrac{2 C V}{\varepsilon_{0}}

Correct Answer

Option B

Solution

The electric field inside a parallel plate capacitor is uniform and given by $\mathbf{E}=\dfrac{V}{d} \hat{\mathbf{j}}$ where $d$ is the separation between the plates.

The electric flux through a closed surface enclosing only the positive plate of the capacitor is given by $\Phi_E = \oint_S \mathbf{E}\cdot d\mathbf{A}$ .

Since the electric field is perpendicular to the surface of the plate, the flux through the surface will be constant and given by $\Phi_E = E A$ , where $A$ is the area of the plate.

The area of the positive plate is $A = \dfrac{Q}{\varepsilon_0 V}$ , where $Q = C V$ is the charge on the positive plate.

Therefore, the electric flux through the surface is given by:

\Phi_E = \frac{Q}{\varepsilon_0 V} \frac{V}{d} = \frac{Q}{\varepsilon_0 d} = \frac{C V}{\varepsilon_0 d}

Thus, the answer is $\dfrac{C V}{\varepsilon_0}$ .

Q63

A capacitor with capacitance 5μF is charged to 5μC. If the plates are pulled apart to reduce the capacitance to 2μF, how much work is done ?

A 2.16 × 10–6 J

B 2.55 × 10–6 J

C 3.75 × 10–6 J

D 6.25 × 10–6 J

Correct Answer

Option C

Solution

Work done =

\Delta

U = Uf – Ui

= {{{q^2}} \over {2{C_r}}} - {{{q^2}} \over {2{C_i}}}

= {{{{\left( {5 \times {{10}^{ - 6}}} \right)}^2}} \over 2}.\left( {{1 \over {2 \times {{10}^{ - 6}}}} - {1 \over {5 \times {{10}^{ - 6}}}}} \right)

= {{15} \over 4} \times {10^{ - 6}} = 3.75{\rm{ }} \times {\rm{ }}{10^{-6}}{\rm{ }}J

Q64

If there are

n

capacitors in parallel connected to

V

volt source, then the energy stored is equal to

A

CV

B

{1 \over 2}nC{V^2}

C

C{V^2}

D

{1 \over {2n}}C{V^2}

Correct Answer

Option B

Solution

The equivalent capacitance of

n

identical capacitors of capacitance

C

is equal to

nC.

Energy stored in this capacitor

E = {1 \over 2}\left( {nC} \right){V^2} = {1 \over 2}nC{V^2}

Q65

Two metallic plates form a parallel plate capacitor. The distance between the plates is 'd'. A metal sheet of thickness

{d \over 2}

and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?

A 2 : 1

B 1 : 2

C 1 : 4

D 4 : 1

Correct Answer

Option A

Solution

{C_{eq}} = {{{\varepsilon _0}A} \over {d - {d \over 2} + {d \over {2k}}}} = {{{\varepsilon _0}A} \over {{d \over 2}}} = {{2{\varepsilon _0}A} \over d}

If

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

\Rightarrow {C_{eq}} = 2C

or

{{{C_{new}}} \over {{C_{old}}}} = {2 \over 1}

Q66

Using a battery, a 100 pF capacitor is charged to 60 V and then the battery is removed. After that, a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is 20 V , its capacitance is: (in pF )

A 600

B 100

C 400

D 200

Correct Answer

Option D

Solution

At first, a capacitor with capacitance $C_0 = 100$ pF is charged to a voltage $V_0 = 60$ V.

This means it stores a charge $Q = C_0 V_0$ .

Then, the battery is removed, so the total charge stays the same.

Now, a second uncharged capacitor, $C$ , is connected in parallel.

The two capacitors will now share the charge.

The final voltage across both capacitors is given in the question as 20 V.

The charge is now shared between both capacitors: Total charge = $(C_0 + C) \times 20$ But total charge stays the same as before, so: $(C_0 + C) \times 20 = C_0 \times 60$ Divide both sides by 20: $C_0 + C = 3 C_0$ So, $C = 2 C_0$ Since $C_0 = 100$ pF, the second capacitor $C$ is $2 \times 100 = 200$ pF.

Q67

An electron with kinetic energy K1 enters between parallel plates of a capacitor at an angle '

\alpha

' with the plates. It leaves the plates at angle '

\beta

' with kinetic energy K2. Then the ratio of kinetic energies K1 : K2 will be :

A

{{{{\cos }^2}\beta } \over {{{\cos }^2}\alpha }}

B

{{\cos \beta } \over {\cos \alpha }}

C

{{{{\sin }^2}\beta } \over {{{\cos }^2}\alpha }}

D

{{\cos \beta } \over {\sin \alpha }}

Correct Answer

Option A

Solution

$\because$

{v_1}\cos \alpha = {v_2}\cos \beta

{{{v_1}} \over {{v_2}}} = {{\cos \beta } \over {\cos \alpha }}

Then the ratio of kinetic energies

{{{k_1}} \over {{k_2}}} = {{{1 \over 2}m{v_1}^2} \over {{1 \over 2}m{v_2}^2}} = {\left( {{{{v_1}} \over {{v_2}}}} \right)^2} = {\left( {{{\cos \beta } \over {\cos \alpha }}} \right)^2}

$\Rightarrow$

{{{k_1}} \over {{k_2}}} = {{{{\cos }^2}\beta } \over {{{\cos }^2}\alpha }}

Q68

A parallel plate capacitor has 1μF capacitance. One of its two plates is given +2μC charge and the other plate, +4μC charge. The potential difference developed across the capacitor is:-

A 1V

B 5V

C 2V

D 3V

Correct Answer

Option A

Solution

Charges at inner plates are 1 $\mu$ C and –1 $\mu$ C. $\therefore$ Potential difference across capacitor =

{q \over c} = {{1\mu C} \over {1\mu F}} = {{1 \times {{10}^{ - 6}}C} \over {1 \times {{10}^{ - 6}}Farad}} = 1V

Q69

Let

C

be the capacitance of a capacitor discharging through a resistor

R.

Suppose

{t_1}

is the time taken for the energy stored in the capacitor to reduce to half its initial value and

{t_2}

is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio

{t_1}/{t_2}

will be

A

1

B

{1 \over 2}

C

{1 \over 4}

D

2

Correct Answer

Option C

Solution

Initial energy of capacitor,

{E_1} = {{q_1^2} \over {2C}}

Final energy of capacitor,

{E_2} = {1 \over 2}{E_1} = {{q_1^2} \over {4C}} = {\left( {{{{{{q_1}} \over {\sqrt 2 }}} \over {2C}}} \right)^2}

$\therefore$

{t_1}=

time for the charge to reduce to

{1 \over {\sqrt 2 }}

of its initial value and

{t_2} =

time for the charge to reduce to

{1 \over 4}

of its initial value We have,

{q_2} = {q_1}{e^{ - t/CR}}

\Rightarrow \ln \left( {{{{q_2}} \over {{q_1}}}} \right) = - {t \over {CR}}

$\therefore$

\ln \left( {{1 \over {\sqrt 2 }}} \right) = {{ - {t_1}} \over {CR}}...\left( 1 \right)

and

\ln \left( {{1 \over 4}} \right) = {{ - {t_2}} \over {CR}}\,\,...\left( 2 \right)

By

(1)

and

(2),

{{{t_1}} \over {{t_2}}} = {{\ln \left( {{1 \over {\sqrt 2 }}} \right)} \over {\ln \left( {{1 \over 4}} \right)}}

= {1 \over 2}{{\ln \left( {{1 \over 2}} \right)} \over {2\ln \left( {{1 \over 2}} \right)}} = {1 \over 4}

Q70

A parallel plate capacitor was made with two rectangular plates, each with a length of

l=3 \mathrm{~cm}

and breath of

\mathrm{b}=1 \mathrm{~cm}

. The distance between the plates is

3 \mu \mathrm{~m}

. Out of the following, which are the ways to increase the capacitance by a factor of 10 ? A.

l=30 \mathrm{~cm}, \mathrm{~b}=1 \mathrm{~cm}, \mathrm{~d}=1 \mu \mathrm{~m}

B.

l=3 \mathrm{~cm}, \mathrm{~b}=1 \mathrm{~cm}, \mathrm{~d}=30 \mu \mathrm{~m}

C.

l=6 \mathrm{~cm}, \mathrm{~b}=5 \mathrm{~cm}, \mathrm{~d}=3 \mu \mathrm{~m}

D.

l=1 \mathrm{~cm}, \mathrm{~b}=1 \mathrm{~cm}, \mathrm{~d}=10 \mu \mathrm{~m}

E.

l=5 \mathrm{~cm}, \mathrm{~b}=2 \mathrm{~cm}, \mathrm{~d}=1 \mu \mathrm{~m}

Choose the correct answer from the options given below:

A C only

B A only

C B and D only

D C and E only

Correct Answer

Option D

Solution

The capacitance of a parallel plate capacitor is given by

C = \frac{\epsilon_0 A}{d},

where:

\epsilon_0

is the permittivity of free space,

A

is the area of a plate, and

d

is the separation between the plates. A change in the capacitor’s dimensions will change its capacitance by a factor of

\frac{C_{\text{new}}}{C_{\text{initial}}} = \frac{A_{\text{new}}}{A_{\text{initial}}} \cdot \frac{d_{\text{initial}}}{d_{\text{new}}}.

The initial dimensions are: Length:

l = 3\,\text{cm} = 0.03\,\text{m},

Breadth:

b = 1\,\text{cm} = 0.01\,\text{m},

Hence, the original area is

A_{\text{initial}} = 0.03 \times 0.01 = 3 \times 10^{-4}\,\text{m}^2,

Plate separation:

d_{\text{initial}} = 3\,\mu\text{m} = 3 \times 10^{-6}\,\text{m}.

For each modification, we compare the new factor: Case C: New dimensions:

l = 6\,\text{cm} = 0.06\,\text{m}, \quad b = 5\,\text{cm} = 0.05\,\text{m}.

New area:

A_{\text{new}} = 0.06 \times 0.05 = 3 \times 10^{-3}\,\text{m}^2.

Ratio of areas:

\frac{A_{\text{new}}}{A_{\text{initial}}} = \frac{3 \times 10^{-3}}{3 \times 10^{-4}} = 10.

The plate separation is unchanged:

\frac{d_{\text{initial}}}{d_{\text{new}}} = \frac{3 \times 10^{-6}}{3 \times 10^{-6}} = 1.

Overall factor:

10 \times 1 = 10.

Case E: New dimensions:

l = 5\,\text{cm} = 0.05\,\text{m}, \quad b = 2\,\text{cm} = 0.02\,\text{m}.

New area:

A_{\text{new}} = 0.05 \times 0.02 = 1 \times 10^{-3}\,\text{m}^2.

Ratio of areas:

\frac{A_{\text{new}}}{A_{\text{initial}}} = \frac{1 \times 10^{-3}}{3 \times 10^{-4}} \approx 3.33.

New plate separation:

d_{\text{new}} = 1\,\mu\text{m} = 1 \times 10^{-6}\,\text{m},

so the ratio of separations is

\frac{d_{\text{initial}}}{d_{\text{new}}} = \frac{3 \times 10^{-6}}{1 \times 10^{-6}} = 3.

Overall factor:

3.33 \times 3 \approx 10.

Both Case C and Case E yield a capacitance that is 10 times the original value.

Thus, the correct modifications are those in Case C and Case E.