Capacitor — Page 3 | JEE Physics

Q21

A capacitor is connected to a 20 V battery through a resistance of 10

\Omega

. It is found that the potential difference across the capacitor rises to 2 V in 1

\mu

s. The capacitance of the capacitor is __________

\mu

F. Given :

\ln \left( {{{10} \over 9}} \right) = 0.105

A 9.52

B 0.95

C 0.105

D 1.85

Correct Answer

Option B

Solution

Given, the peak voltage of the battery, V0 = 20V The voltage of the battery, V = 2V Time, t = 1 $\mu$ s = 1 $\times$ 10 $-$ 6s Resistance of the capacitor, R = 10

\Omega

As we know that, V = V0(1 $-$ e $-$ t/RC) Substituting the values in the above equation, we get 2 = 20(1 $-$ e $-$ t/RC)

\Rightarrow {t \over {RC}} = \ln \left( {{{10} \over 9}} \right)

\Rightarrow C = {t \over {R\ln (10/9)}} = {{{{10}^{ - 6}}} \over {10 \times \ln (10/9)}} = 0.95

$\mu$ F $\therefore$ The capacitance of the capacitor is 0.95 $\mu$ F.

Q22

Consider the combination of 2 capacitors C1 and C2 with C2 > C1, when connected in parallel, the equivalent capacitance is

{{15} \over 4}

times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors,

{{{C_2}} \over {{C_1}}}

.

A

{{15} \over {11}}

B No Solutions

C

{{29} \over {15}}

D

{{15} \over {4}}

Correct Answer

Option B

Solution

When connected in parallel Ceq = C1 + C2 When in series

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

{C_1} + {C_2} = {{15} \over 4}\left( {{{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)

4{({C_1} + {C_2})^2} = 15{C_1}{C_2}

4{C_1}^2 + 4{C_2}^2 - 7{C_1}{C_2} = 0

dividing by

{C_1}^2

4{\left( {{{{C_2}} \over {{C_1}}}} \right)^2} - {{7{C_2}} \over {{C_1}}} + 4 = 0

Let

{{{C_2}} \over {{C_1}}} = x

4{x^2} - 7x + 4 = 0

{b^2} - 4ac = 49 - 64 < 0

$\therefore$ No solution exixts.

Q23

A parallel plate capacitor has plate area 40 cm

^2

and plates separation 2 mm. The space between the plates is filled with a dielectric medium of a thickness 1 mm and dielectric constant 5. The capacitance of the system is :

A

\mathrm{10\varepsilon_0~F}

B

\mathrm{24\varepsilon_0~F}

C

\mathrm{\dfrac{3}{10}\varepsilon_0~F}

D

\mathrm{\dfrac{10}{3}\varepsilon_0~F}

Correct Answer

Option D

Solution

This can be seen as two capacitors in series combination so

\begin{aligned} & \frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2} \\\\ & =\frac{1}{\frac{\mathrm{K} \in_0 \mathrm{~A}}{\mathrm{t}}}+\frac{1}{\frac{\in_0 \mathrm{~A}}{\mathrm{~d}-\mathrm{t}}} \end{aligned}

\begin{aligned} & =\frac{\mathrm{t}}{\mathrm{K} \in_0 \mathrm{~A}}+\frac{\mathrm{d}-\mathrm{t}}{\epsilon_0 \mathrm{~A}} \\\\ & =\frac{1 \times 10^{-3}}{5 \in_0 \times 40 \times 10^{-4}}+\frac{1 \times 10^{-3}}{\in_0 40 \times 10^{-4}} \\\\ & \frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{20 \in_0}+\frac{1}{4 \in_0} \\\\ & \mathrm{C}_{\mathrm{eq}}=\frac{20 \times 4 \in_0}{24}=\frac{10 \in_0}{3} \mathrm{~F} \end{aligned}

Q24

Two capacitors

{C_1}

and

{C_2}

are charged to

120

V

and

200

V

respectively. It is found that connecting them together the potential on each one can be made zero. Then

A

5{C_1} = 3{C_2}

B

3{C_1} = 5{C_2}

C

3{C_1} + 5{C_2} = 0

D

9{C_1} = 4{C_2}

Correct Answer

Option B

Solution

For potential to be made zero, after connection

120{C_1} = 200{C_2}

\left[ {\,\,} \right.

as

\left. {C = {q \over v}\,\,} \right]

\Rightarrow 3{C_1} = 5{C_2}

Q25

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}

Q26

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 $\dfrac{C V}{\varepsilon_0}$ .

Q27

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

Q28

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}

Q29

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}

Q30

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.