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Capacitance

165643 In a parallel plate capacitor, if $10^{12}$ electrons pass from one plate to another, a potential difference of $10 V$ is developed across the plates. The capacitance of the capacitor is

1

0.16 \times 10^{- 8} F

2

1.6 \times 10^{- 8} F

3

16 \times 10^{- 8} F

4

0.8 \times 10^{- 8} F

Explanation:

: Given that,
Charge, $Q =$ ne $= 1 \times 10^{12} \times 1.6 \times 10^{- 19}$
$Q = 1.6 \times 10^{- 7} C$
Potential difference, $V = 10 V$
We know that,
$∴ C = \frac{Q}{V}$
$C = \frac{1.6 \times 10^{- 7}}{10} = 1.6 \times 10^{- 8} F$

[AP EAMCET-07.07.2022

Capacitance

165644 In the given figure, the capacitors $C_{1}, C_{3}, C_{4}, C_{5}$ have a capacitance $4 μ F$ each. If the capacitor $C_{2}$ has a capacitance $10 μ F$ , then effective capacitance between $A$ and $B$ will be

1

2 μ F

2

6 μ F

3

4 μ F

4

8 μ F

Explanation:

: The given network can be replaced,
$\frac{C_{4}}{C_{3}} = \frac{C_{1}}{C_{5}}$
No charge flow through $C_{2}$ . So, it is not considered -
Hence, capacitors $C_{1}$ and $C_{5}$ are in series.

$\frac{1}{C^{'}} = \frac{1}{C_{1}} + \frac{1}{C_{5}}$
$C^{'} = \frac{C_{1} C_{5}}{C_{1} + C_{5}}$
$C^{'} = \frac{4 \times 4}{4 + 4}$
$C^{'} = 2 μ F$
Similarly $C_{4}$ and $C_{3}$ are in series.
Therefore,
$C^{''} = \frac{C_{3} C_{4}}{C_{3} + C_{4}} = \frac{4 \times 4}{4 + 4} = 2 μ F$
Also, $C^{'}$ and $C^{''}$ are in parallel configuration.
Hence, effective capacitance between $A$ and $B$ is,
$C_{AB} = 2 + 2 = 4 μ F$

[Karnataka CET - 2004

Capacitance

165645 Capacity of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value when a dielectric slab
of thickness $t = \frac{d}{3}$ is inserted between the plates of the capacitor, of separation ' $d$ '. Dielectric constant of the slab is

1 8

2 4

3 6

4 2

Explanation:

: Given,
Thickness, $t = d / 3$
Capacity, $C^{'} = 4 / 3 C$
We know that,
$C_{air} = \frac{A ε_{0}}{d}$
$C^{'} = \frac{A ε_{o}}{(d - t) + \frac{t}{K}} = \frac{4}{3} C$
$\frac{A ε_{o}}{(d - t) + \frac{t}{K}} = \frac{4}{3} \frac{A ε_{o}}{d}$
$K = \frac{4 t}{4 t - d}$
$K = \frac{4 \times \frac{d}{3}}{\frac{4 \times d}{3} - d}$
$K = \frac{\frac{4 d}{3}}{\frac{d}{3}}$
$K = \frac{4 d}{d}$
$K = 4$

[AP EAMCET-07.09.2021

Capacitance

165646 A capacitor of capacitance $4 μ F$ is charged to a potential of $100 V$ . It is then disconnected from the battery and connected in parallel with another capacitor $C_{2}$ . If their common potential is 40 volts, then the value of $C_{2}$ is

1

2 μ F

2

3 μ F

3

5 μ F

4

6 μ F

Explanation:

: Given that, $C_{1} = 4 μ F, V = 100 V, V^{'} = 40 V$ Charge on first capacitor,
$Q = C_{1} V$
$Q = 4 \times 10^{- 6} \times 100$
$Q = 400 \times 10^{- 6}$
$C_{1} + C_{2} = \frac{Q}{V^{'}}$
$C_{2} = (10 \times 10^{- 6}) - (4 \times 10^{- 6}) [∵ μ = 10^{- 6}]$
$C_{2} = 6 \times 10^{- 6} F$
$C_{2} = 6 μ F$

[AP EAMCET-03.09.2021

Capacitance

165643 In a parallel plate capacitor, if $10^{12}$ electrons pass from one plate to another, a potential difference of $10 V$ is developed across the plates. The capacitance of the capacitor is

1

0.16 \times 10^{- 8} F

2

1.6 \times 10^{- 8} F

3

16 \times 10^{- 8} F

4

0.8 \times 10^{- 8} F

Explanation:

: Given that,
Charge, $Q =$ ne $= 1 \times 10^{12} \times 1.6 \times 10^{- 19}$
$Q = 1.6 \times 10^{- 7} C$
Potential difference, $V = 10 V$
We know that,
$∴ C = \frac{Q}{V}$
$C = \frac{1.6 \times 10^{- 7}}{10} = 1.6 \times 10^{- 8} F$

[AP EAMCET-07.07.2022

Capacitance

165644 In the given figure, the capacitors $C_{1}, C_{3}, C_{4}, C_{5}$ have a capacitance $4 μ F$ each. If the capacitor $C_{2}$ has a capacitance $10 μ F$ , then effective capacitance between $A$ and $B$ will be

1

2 μ F

2

6 μ F

3

4 μ F

4

8 μ F

Explanation:

: The given network can be replaced,
$\frac{C_{4}}{C_{3}} = \frac{C_{1}}{C_{5}}$
No charge flow through $C_{2}$ . So, it is not considered -
Hence, capacitors $C_{1}$ and $C_{5}$ are in series.

$\frac{1}{C^{'}} = \frac{1}{C_{1}} + \frac{1}{C_{5}}$
$C^{'} = \frac{C_{1} C_{5}}{C_{1} + C_{5}}$
$C^{'} = \frac{4 \times 4}{4 + 4}$
$C^{'} = 2 μ F$
Similarly $C_{4}$ and $C_{3}$ are in series.
Therefore,
$C^{''} = \frac{C_{3} C_{4}}{C_{3} + C_{4}} = \frac{4 \times 4}{4 + 4} = 2 μ F$
Also, $C^{'}$ and $C^{''}$ are in parallel configuration.
Hence, effective capacitance between $A$ and $B$ is,
$C_{AB} = 2 + 2 = 4 μ F$

[Karnataka CET - 2004

Capacitance

165645 Capacity of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value when a dielectric slab
of thickness $t = \frac{d}{3}$ is inserted between the plates of the capacitor, of separation ' $d$ '. Dielectric constant of the slab is

1 8

2 4

3 6

4 2

Explanation:

: Given,
Thickness, $t = d / 3$
Capacity, $C^{'} = 4 / 3 C$
We know that,
$C_{air} = \frac{A ε_{0}}{d}$
$C^{'} = \frac{A ε_{o}}{(d - t) + \frac{t}{K}} = \frac{4}{3} C$
$\frac{A ε_{o}}{(d - t) + \frac{t}{K}} = \frac{4}{3} \frac{A ε_{o}}{d}$
$K = \frac{4 t}{4 t - d}$
$K = \frac{4 \times \frac{d}{3}}{\frac{4 \times d}{3} - d}$
$K = \frac{\frac{4 d}{3}}{\frac{d}{3}}$
$K = \frac{4 d}{d}$
$K = 4$

[AP EAMCET-07.09.2021

Capacitance

165646 A capacitor of capacitance $4 μ F$ is charged to a potential of $100 V$ . It is then disconnected from the battery and connected in parallel with another capacitor $C_{2}$ . If their common potential is 40 volts, then the value of $C_{2}$ is

1

2 μ F

2

3 μ F

3

5 μ F

4

6 μ F

Explanation:

: Given that, $C_{1} = 4 μ F, V = 100 V, V^{'} = 40 V$ Charge on first capacitor,
$Q = C_{1} V$
$Q = 4 \times 10^{- 6} \times 100$
$Q = 400 \times 10^{- 6}$
$C_{1} + C_{2} = \frac{Q}{V^{'}}$
$C_{2} = (10 \times 10^{- 6}) - (4 \times 10^{- 6}) [∵ μ = 10^{- 6}]$
$C_{2} = 6 \times 10^{- 6} F$
$C_{2} = 6 μ F$

[AP EAMCET-03.09.2021

Capacitance

165643 In a parallel plate capacitor, if $10^{12}$ electrons pass from one plate to another, a potential difference of $10 V$ is developed across the plates. The capacitance of the capacitor is

1

0.16 \times 10^{- 8} F

2

1.6 \times 10^{- 8} F

3

16 \times 10^{- 8} F

4

0.8 \times 10^{- 8} F

Explanation:

: Given that,
Charge, $Q =$ ne $= 1 \times 10^{12} \times 1.6 \times 10^{- 19}$
$Q = 1.6 \times 10^{- 7} C$
Potential difference, $V = 10 V$
We know that,
$∴ C = \frac{Q}{V}$
$C = \frac{1.6 \times 10^{- 7}}{10} = 1.6 \times 10^{- 8} F$

[AP EAMCET-07.07.2022

Capacitance

165644 In the given figure, the capacitors $C_{1}, C_{3}, C_{4}, C_{5}$ have a capacitance $4 μ F$ each. If the capacitor $C_{2}$ has a capacitance $10 μ F$ , then effective capacitance between $A$ and $B$ will be

1

2 μ F

2

6 μ F

3

4 μ F

4

8 μ F

Explanation:

: The given network can be replaced,
$\frac{C_{4}}{C_{3}} = \frac{C_{1}}{C_{5}}$
No charge flow through $C_{2}$ . So, it is not considered -
Hence, capacitors $C_{1}$ and $C_{5}$ are in series.

$\frac{1}{C^{'}} = \frac{1}{C_{1}} + \frac{1}{C_{5}}$
$C^{'} = \frac{C_{1} C_{5}}{C_{1} + C_{5}}$
$C^{'} = \frac{4 \times 4}{4 + 4}$
$C^{'} = 2 μ F$
Similarly $C_{4}$ and $C_{3}$ are in series.
Therefore,
$C^{''} = \frac{C_{3} C_{4}}{C_{3} + C_{4}} = \frac{4 \times 4}{4 + 4} = 2 μ F$
Also, $C^{'}$ and $C^{''}$ are in parallel configuration.
Hence, effective capacitance between $A$ and $B$ is,
$C_{AB} = 2 + 2 = 4 μ F$

[Karnataka CET - 2004

Capacitance

165645 Capacity of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value when a dielectric slab
of thickness $t = \frac{d}{3}$ is inserted between the plates of the capacitor, of separation ' $d$ '. Dielectric constant of the slab is

1 8

2 4

3 6

4 2

Explanation:

: Given,
Thickness, $t = d / 3$
Capacity, $C^{'} = 4 / 3 C$
We know that,
$C_{air} = \frac{A ε_{0}}{d}$
$C^{'} = \frac{A ε_{o}}{(d - t) + \frac{t}{K}} = \frac{4}{3} C$
$\frac{A ε_{o}}{(d - t) + \frac{t}{K}} = \frac{4}{3} \frac{A ε_{o}}{d}$
$K = \frac{4 t}{4 t - d}$
$K = \frac{4 \times \frac{d}{3}}{\frac{4 \times d}{3} - d}$
$K = \frac{\frac{4 d}{3}}{\frac{d}{3}}$
$K = \frac{4 d}{d}$
$K = 4$

[AP EAMCET-07.09.2021

Capacitance

165646 A capacitor of capacitance $4 μ F$ is charged to a potential of $100 V$ . It is then disconnected from the battery and connected in parallel with another capacitor $C_{2}$ . If their common potential is 40 volts, then the value of $C_{2}$ is

1

2 μ F

2

3 μ F

3

5 μ F

4

6 μ F

Explanation:

: Given that, $C_{1} = 4 μ F, V = 100 V, V^{'} = 40 V$ Charge on first capacitor,
$Q = C_{1} V$
$Q = 4 \times 10^{- 6} \times 100$
$Q = 400 \times 10^{- 6}$
$C_{1} + C_{2} = \frac{Q}{V^{'}}$
$C_{2} = (10 \times 10^{- 6}) - (4 \times 10^{- 6}) [∵ μ = 10^{- 6}]$
$C_{2} = 6 \times 10^{- 6} F$
$C_{2} = 6 μ F$

[AP EAMCET-03.09.2021

Capacitance

165643 In a parallel plate capacitor, if $10^{12}$ electrons pass from one plate to another, a potential difference of $10 V$ is developed across the plates. The capacitance of the capacitor is

1

0.16 \times 10^{- 8} F

2

1.6 \times 10^{- 8} F

3

16 \times 10^{- 8} F

4

0.8 \times 10^{- 8} F

Explanation:

: Given that,
Charge, $Q =$ ne $= 1 \times 10^{12} \times 1.6 \times 10^{- 19}$
$Q = 1.6 \times 10^{- 7} C$
Potential difference, $V = 10 V$
We know that,
$∴ C = \frac{Q}{V}$
$C = \frac{1.6 \times 10^{- 7}}{10} = 1.6 \times 10^{- 8} F$

[AP EAMCET-07.07.2022

Capacitance

165644 In the given figure, the capacitors $C_{1}, C_{3}, C_{4}, C_{5}$ have a capacitance $4 μ F$ each. If the capacitor $C_{2}$ has a capacitance $10 μ F$ , then effective capacitance between $A$ and $B$ will be

1

2 μ F

2

6 μ F

3

4 μ F

4

8 μ F

Explanation:

: The given network can be replaced,
$\frac{C_{4}}{C_{3}} = \frac{C_{1}}{C_{5}}$
No charge flow through $C_{2}$ . So, it is not considered -
Hence, capacitors $C_{1}$ and $C_{5}$ are in series.

$\frac{1}{C^{'}} = \frac{1}{C_{1}} + \frac{1}{C_{5}}$
$C^{'} = \frac{C_{1} C_{5}}{C_{1} + C_{5}}$
$C^{'} = \frac{4 \times 4}{4 + 4}$
$C^{'} = 2 μ F$
Similarly $C_{4}$ and $C_{3}$ are in series.
Therefore,
$C^{''} = \frac{C_{3} C_{4}}{C_{3} + C_{4}} = \frac{4 \times 4}{4 + 4} = 2 μ F$
Also, $C^{'}$ and $C^{''}$ are in parallel configuration.
Hence, effective capacitance between $A$ and $B$ is,
$C_{AB} = 2 + 2 = 4 μ F$

[Karnataka CET - 2004

Capacitance

165645 Capacity of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value when a dielectric slab
of thickness $t = \frac{d}{3}$ is inserted between the plates of the capacitor, of separation ' $d$ '. Dielectric constant of the slab is

1 8

2 4

3 6

4 2

Explanation:

: Given,
Thickness, $t = d / 3$
Capacity, $C^{'} = 4 / 3 C$
We know that,
$C_{air} = \frac{A ε_{0}}{d}$
$C^{'} = \frac{A ε_{o}}{(d - t) + \frac{t}{K}} = \frac{4}{3} C$
$\frac{A ε_{o}}{(d - t) + \frac{t}{K}} = \frac{4}{3} \frac{A ε_{o}}{d}$
$K = \frac{4 t}{4 t - d}$
$K = \frac{4 \times \frac{d}{3}}{\frac{4 \times d}{3} - d}$
$K = \frac{\frac{4 d}{3}}{\frac{d}{3}}$
$K = \frac{4 d}{d}$
$K = 4$

[AP EAMCET-07.09.2021

Capacitance

165646 A capacitor of capacitance $4 μ F$ is charged to a potential of $100 V$ . It is then disconnected from the battery and connected in parallel with another capacitor $C_{2}$ . If their common potential is 40 volts, then the value of $C_{2}$ is

1

2 μ F

2

3 μ F

3

5 μ F

4

6 μ F

Explanation:

: Given that, $C_{1} = 4 μ F, V = 100 V, V^{'} = 40 V$ Charge on first capacitor,
$Q = C_{1} V$
$Q = 4 \times 10^{- 6} \times 100$
$Q = 400 \times 10^{- 6}$
$C_{1} + C_{2} = \frac{Q}{V^{'}}$
$C_{2} = (10 \times 10^{- 6}) - (4 \times 10^{- 6}) [∵ μ = 10^{- 6}]$
$C_{2} = 6 \times 10^{- 6} F$
$C_{2} = 6 μ F$

[AP EAMCET-03.09.2021

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