QBManager

Capacitance

165700 Two capacitors, one $4 pF$ and the other $6 pF$ , connected in parallel, are charged by a $100 V$ battery. The energy stored in the capacitors is

1

12 \times 10^{- 8} J

2

2.4 \times 10^{- 8} J

3

5.0 \times 10^{- 8} J

4

1.2 \times 10^{- 6} J

Explanation:

: Given, $C_{1} = 4 pF$ and $C_{2} = 6 pF$
$Voltage = 100 V$
Both capacitors connected in parallel.
Hence, equivalent capacitance,
$C = C_{1} + C_{2} = 4 pF + 6 pF = 10 pF$
$C = 10 \times 10^{- 12} F$
Energy stored in the capacitor $(E) = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times 10 \times 10^{- 12} \times (100)^{2}$
$= \frac{1}{2} \times 10 \times 10^{- 12} \times 10^{4}$
$E = 5.0 \times 10^{- 8} J$

[Manipal UGET-2017]

Capacitance

165701 Two conductors of the same material have their diameters in the ratio $1 : 2$ and their lengths in the ratio $2 : 1$ . If the temperature difference between their ends is the same then the ratio of amounts of heat conducted per second through them will be

1

4 : 1

2

1 : 4

3

8 : 1

4

1 : 8

Explanation:

: We know,
$H = \frac{Q}{t} = \frac{K \times A \times Δ t}{l}$
$H \propto \frac{A}{l}$
$\frac{H_{1}}{H_{2}} = \frac{A_{1}}{A_{2}} \times \frac{l_{2}}{l_{1}}$
$\frac{H_{1}}{H_{2}} = \frac{d_{1}^{2}}{d_{2}^{2}} \times \frac{l_{2}}{l_{1}}$ $[A = π r^{2} = \frac{π d^{2}}{4}]$
$\frac{H_{1}}{H_{2}} = {(\frac{1}{2})}^{2} \times \frac{1}{2}$
$\frac{H_{1}}{H_{2}} = \frac{1}{8}$
$∴ H_{1} : H_{2} = 1 : 8$

[Manipal UGET-2012]

Capacitance

165702 As shown in the figure below, if a capacitor $C$ is charged by connecting it resistance $R$ , then energy given by the battery will be

1

\frac{1}{2} {CV}^{2}

2 more than

\frac{1}{2} {CV}^{2}

3 less than

\frac{1}{2} {CV}^{2}

4 zero

Explanation:

: We know,
Energy stored in a capacitor
$E = \frac{1}{2} {CV}_{C}^{2}$
Where, $V_{C} =$ Voltage drop across capacitor.
Apply KVL in Circuit
$V = V_{R} + V_{C}$
We know, in steady state current through resistance $= 0$
$V_{R} = IR = 0$
Hence, $V = V_{C}$
From equation (i)
$E = \frac{1}{2} {CV}^{2}$

[CG PET -2018]

Capacitance

165703 The energy per unit volume for a capacitor having area $A$ and separation $d$ kept at potential different $V$ is given by

1

\frac{1}{2} ε_{o} \frac{V^{2}}{d^{2}}

2

\frac{1}{2 ε_{o}} \frac{V^{2}}{d^{2}}

3

\frac{1}{2} {CV}^{2}

4

\frac{Q^{2}}{2 C}

Explanation:

: We know,
Total energy Stored in a capacitor,
$E = \frac{1}{2} {CV}^{2}$
$or E = \frac{1}{2} (\frac{ε_{0} A}{d}) V^{2} (∵ C = \frac{ε_{0} A}{d})$
Volume between the plate $= A \times d$
Hence, Energy per unit Volume,
$\frac{E}{A d} = \frac{\frac{1}{2} (\frac{ε_{0} A}{d}) V^{2}}{A d} = \frac{1}{2} \frac{ε_{0} V^{2}}{d^{2}}$

[CG PET- 2017

Capacitance

165704 A $600 pF$ capacitor is charged by a $200 V$ supply. It is then disconnected from the supply and is connected to another uncharged $600 pF$ capacitor. Electrostatic energy lost in the process is

1

6 \times 10^{- 6} J

2

3 \times 10^{- 6} J

3

6 \times 10^{- 9} J

4

3 \times 10^{- 9} J

Explanation:

: Given that,
Capacitance of Capacitor $(C) = 600 pF = 600 \times 10^{- 12} F$ and potential difference $(V) = 200 V$

Electrostatic Energy stored $(E) = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times (600 \times 10^{- 12}) \times (200)^{2}$
$= \frac{1}{2} \times 600 \times 10^{- 12} \times 4 \times 10^{4}$
$= 1.2 \times 10^{- 5} J$
After disconnecting, A $600 pF$ capacitor added.
Then equivalent capacitance $(C^{'})$
$\frac{1}{C^{'}} = \frac{1}{C} + \frac{1}{C}$
$= \frac{1}{600} + \frac{1}{600}$
$\frac{1}{C^{'}} = \frac{1}{300}$
$C^{'} = 300 pF$
Then, new electrostatic energy,
$E^{'} = \frac{1}{2} C^{'} \times V^{2}$
$= \frac{1}{2} \times 300 \times 10^{- 12} \times (200)^{2}$
$= 0.6 \times 10^{- 5} J$
Loss in electrostatic energy $(E - E^{'})$
$= 1.2 \times 10^{- 5} - 0.6 \times 10^{5}$
$= 0.6 \times 10^{- 5}$
$= 6 \times 10^{- 6} J$

[CG PET -2016]

Capacitance

165700 Two capacitors, one $4 pF$ and the other $6 pF$ , connected in parallel, are charged by a $100 V$ battery. The energy stored in the capacitors is

1

12 \times 10^{- 8} J

2

2.4 \times 10^{- 8} J

3

5.0 \times 10^{- 8} J

4

1.2 \times 10^{- 6} J

Explanation:

: Given, $C_{1} = 4 pF$ and $C_{2} = 6 pF$
$Voltage = 100 V$
Both capacitors connected in parallel.
Hence, equivalent capacitance,
$C = C_{1} + C_{2} = 4 pF + 6 pF = 10 pF$
$C = 10 \times 10^{- 12} F$
Energy stored in the capacitor $(E) = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times 10 \times 10^{- 12} \times (100)^{2}$
$= \frac{1}{2} \times 10 \times 10^{- 12} \times 10^{4}$
$E = 5.0 \times 10^{- 8} J$

[Manipal UGET-2017]

Capacitance

165701 Two conductors of the same material have their diameters in the ratio $1 : 2$ and their lengths in the ratio $2 : 1$ . If the temperature difference between their ends is the same then the ratio of amounts of heat conducted per second through them will be

1

4 : 1

2

1 : 4

3

8 : 1

4

1 : 8

Explanation:

: We know,
$H = \frac{Q}{t} = \frac{K \times A \times Δ t}{l}$
$H \propto \frac{A}{l}$
$\frac{H_{1}}{H_{2}} = \frac{A_{1}}{A_{2}} \times \frac{l_{2}}{l_{1}}$
$\frac{H_{1}}{H_{2}} = \frac{d_{1}^{2}}{d_{2}^{2}} \times \frac{l_{2}}{l_{1}}$ $[A = π r^{2} = \frac{π d^{2}}{4}]$
$\frac{H_{1}}{H_{2}} = {(\frac{1}{2})}^{2} \times \frac{1}{2}$
$\frac{H_{1}}{H_{2}} = \frac{1}{8}$
$∴ H_{1} : H_{2} = 1 : 8$

[Manipal UGET-2012]

Capacitance

165702 As shown in the figure below, if a capacitor $C$ is charged by connecting it resistance $R$ , then energy given by the battery will be

1

\frac{1}{2} {CV}^{2}

2 more than

\frac{1}{2} {CV}^{2}

3 less than

\frac{1}{2} {CV}^{2}

4 zero

Explanation:

: We know,
Energy stored in a capacitor
$E = \frac{1}{2} {CV}_{C}^{2}$
Where, $V_{C} =$ Voltage drop across capacitor.
Apply KVL in Circuit
$V = V_{R} + V_{C}$
We know, in steady state current through resistance $= 0$
$V_{R} = IR = 0$
Hence, $V = V_{C}$
From equation (i)
$E = \frac{1}{2} {CV}^{2}$

[CG PET -2018]

Capacitance

165703 The energy per unit volume for a capacitor having area $A$ and separation $d$ kept at potential different $V$ is given by

1

\frac{1}{2} ε_{o} \frac{V^{2}}{d^{2}}

2

\frac{1}{2 ε_{o}} \frac{V^{2}}{d^{2}}

3

\frac{1}{2} {CV}^{2}

4

\frac{Q^{2}}{2 C}

Explanation:

: We know,
Total energy Stored in a capacitor,
$E = \frac{1}{2} {CV}^{2}$
$or E = \frac{1}{2} (\frac{ε_{0} A}{d}) V^{2} (∵ C = \frac{ε_{0} A}{d})$
Volume between the plate $= A \times d$
Hence, Energy per unit Volume,
$\frac{E}{A d} = \frac{\frac{1}{2} (\frac{ε_{0} A}{d}) V^{2}}{A d} = \frac{1}{2} \frac{ε_{0} V^{2}}{d^{2}}$

[CG PET- 2017

Capacitance

165704 A $600 pF$ capacitor is charged by a $200 V$ supply. It is then disconnected from the supply and is connected to another uncharged $600 pF$ capacitor. Electrostatic energy lost in the process is

1

6 \times 10^{- 6} J

2

3 \times 10^{- 6} J

3

6 \times 10^{- 9} J

4

3 \times 10^{- 9} J

Explanation:

: Given that,
Capacitance of Capacitor $(C) = 600 pF = 600 \times 10^{- 12} F$ and potential difference $(V) = 200 V$

Electrostatic Energy stored $(E) = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times (600 \times 10^{- 12}) \times (200)^{2}$
$= \frac{1}{2} \times 600 \times 10^{- 12} \times 4 \times 10^{4}$
$= 1.2 \times 10^{- 5} J$
After disconnecting, A $600 pF$ capacitor added.
Then equivalent capacitance $(C^{'})$
$\frac{1}{C^{'}} = \frac{1}{C} + \frac{1}{C}$
$= \frac{1}{600} + \frac{1}{600}$
$\frac{1}{C^{'}} = \frac{1}{300}$
$C^{'} = 300 pF$
Then, new electrostatic energy,
$E^{'} = \frac{1}{2} C^{'} \times V^{2}$
$= \frac{1}{2} \times 300 \times 10^{- 12} \times (200)^{2}$
$= 0.6 \times 10^{- 5} J$
Loss in electrostatic energy $(E - E^{'})$
$= 1.2 \times 10^{- 5} - 0.6 \times 10^{5}$
$= 0.6 \times 10^{- 5}$
$= 6 \times 10^{- 6} J$

[CG PET -2016]

Capacitance

165700 Two capacitors, one $4 pF$ and the other $6 pF$ , connected in parallel, are charged by a $100 V$ battery. The energy stored in the capacitors is

1

12 \times 10^{- 8} J

2

2.4 \times 10^{- 8} J

3

5.0 \times 10^{- 8} J

4

1.2 \times 10^{- 6} J

Explanation:

: Given, $C_{1} = 4 pF$ and $C_{2} = 6 pF$
$Voltage = 100 V$
Both capacitors connected in parallel.
Hence, equivalent capacitance,
$C = C_{1} + C_{2} = 4 pF + 6 pF = 10 pF$
$C = 10 \times 10^{- 12} F$
Energy stored in the capacitor $(E) = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times 10 \times 10^{- 12} \times (100)^{2}$
$= \frac{1}{2} \times 10 \times 10^{- 12} \times 10^{4}$
$E = 5.0 \times 10^{- 8} J$

[Manipal UGET-2017]

Capacitance

165701 Two conductors of the same material have their diameters in the ratio $1 : 2$ and their lengths in the ratio $2 : 1$ . If the temperature difference between their ends is the same then the ratio of amounts of heat conducted per second through them will be

1

4 : 1

2

1 : 4

3

8 : 1

4

1 : 8

Explanation:

: We know,
$H = \frac{Q}{t} = \frac{K \times A \times Δ t}{l}$
$H \propto \frac{A}{l}$
$\frac{H_{1}}{H_{2}} = \frac{A_{1}}{A_{2}} \times \frac{l_{2}}{l_{1}}$
$\frac{H_{1}}{H_{2}} = \frac{d_{1}^{2}}{d_{2}^{2}} \times \frac{l_{2}}{l_{1}}$ $[A = π r^{2} = \frac{π d^{2}}{4}]$
$\frac{H_{1}}{H_{2}} = {(\frac{1}{2})}^{2} \times \frac{1}{2}$
$\frac{H_{1}}{H_{2}} = \frac{1}{8}$
$∴ H_{1} : H_{2} = 1 : 8$

[Manipal UGET-2012]

Capacitance

165702 As shown in the figure below, if a capacitor $C$ is charged by connecting it resistance $R$ , then energy given by the battery will be

1

\frac{1}{2} {CV}^{2}

2 more than

\frac{1}{2} {CV}^{2}

3 less than

\frac{1}{2} {CV}^{2}

4 zero

Explanation:

: We know,
Energy stored in a capacitor
$E = \frac{1}{2} {CV}_{C}^{2}$
Where, $V_{C} =$ Voltage drop across capacitor.
Apply KVL in Circuit
$V = V_{R} + V_{C}$
We know, in steady state current through resistance $= 0$
$V_{R} = IR = 0$
Hence, $V = V_{C}$
From equation (i)
$E = \frac{1}{2} {CV}^{2}$

[CG PET -2018]

Capacitance

165703 The energy per unit volume for a capacitor having area $A$ and separation $d$ kept at potential different $V$ is given by

1

\frac{1}{2} ε_{o} \frac{V^{2}}{d^{2}}

2

\frac{1}{2 ε_{o}} \frac{V^{2}}{d^{2}}

3

\frac{1}{2} {CV}^{2}

4

\frac{Q^{2}}{2 C}

Explanation:

: We know,
Total energy Stored in a capacitor,
$E = \frac{1}{2} {CV}^{2}$
$or E = \frac{1}{2} (\frac{ε_{0} A}{d}) V^{2} (∵ C = \frac{ε_{0} A}{d})$
Volume between the plate $= A \times d$
Hence, Energy per unit Volume,
$\frac{E}{A d} = \frac{\frac{1}{2} (\frac{ε_{0} A}{d}) V^{2}}{A d} = \frac{1}{2} \frac{ε_{0} V^{2}}{d^{2}}$

[CG PET- 2017

Capacitance

165704 A $600 pF$ capacitor is charged by a $200 V$ supply. It is then disconnected from the supply and is connected to another uncharged $600 pF$ capacitor. Electrostatic energy lost in the process is

1

6 \times 10^{- 6} J

2

3 \times 10^{- 6} J

3

6 \times 10^{- 9} J

4

3 \times 10^{- 9} J

Explanation:

: Given that,
Capacitance of Capacitor $(C) = 600 pF = 600 \times 10^{- 12} F$ and potential difference $(V) = 200 V$

Electrostatic Energy stored $(E) = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times (600 \times 10^{- 12}) \times (200)^{2}$
$= \frac{1}{2} \times 600 \times 10^{- 12} \times 4 \times 10^{4}$
$= 1.2 \times 10^{- 5} J$
After disconnecting, A $600 pF$ capacitor added.
Then equivalent capacitance $(C^{'})$
$\frac{1}{C^{'}} = \frac{1}{C} + \frac{1}{C}$
$= \frac{1}{600} + \frac{1}{600}$
$\frac{1}{C^{'}} = \frac{1}{300}$
$C^{'} = 300 pF$
Then, new electrostatic energy,
$E^{'} = \frac{1}{2} C^{'} \times V^{2}$
$= \frac{1}{2} \times 300 \times 10^{- 12} \times (200)^{2}$
$= 0.6 \times 10^{- 5} J$
Loss in electrostatic energy $(E - E^{'})$
$= 1.2 \times 10^{- 5} - 0.6 \times 10^{5}$
$= 0.6 \times 10^{- 5}$
$= 6 \times 10^{- 6} J$

[CG PET -2016]

Capacitance

165700 Two capacitors, one $4 pF$ and the other $6 pF$ , connected in parallel, are charged by a $100 V$ battery. The energy stored in the capacitors is

1

12 \times 10^{- 8} J

2

2.4 \times 10^{- 8} J

3

5.0 \times 10^{- 8} J

4

1.2 \times 10^{- 6} J

Explanation:

: Given, $C_{1} = 4 pF$ and $C_{2} = 6 pF$
$Voltage = 100 V$
Both capacitors connected in parallel.
Hence, equivalent capacitance,
$C = C_{1} + C_{2} = 4 pF + 6 pF = 10 pF$
$C = 10 \times 10^{- 12} F$
Energy stored in the capacitor $(E) = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times 10 \times 10^{- 12} \times (100)^{2}$
$= \frac{1}{2} \times 10 \times 10^{- 12} \times 10^{4}$
$E = 5.0 \times 10^{- 8} J$

[Manipal UGET-2017]

Capacitance

165701 Two conductors of the same material have their diameters in the ratio $1 : 2$ and their lengths in the ratio $2 : 1$ . If the temperature difference between their ends is the same then the ratio of amounts of heat conducted per second through them will be

1

4 : 1

2

1 : 4

3

8 : 1

4

1 : 8

Explanation:

: We know,
$H = \frac{Q}{t} = \frac{K \times A \times Δ t}{l}$
$H \propto \frac{A}{l}$
$\frac{H_{1}}{H_{2}} = \frac{A_{1}}{A_{2}} \times \frac{l_{2}}{l_{1}}$
$\frac{H_{1}}{H_{2}} = \frac{d_{1}^{2}}{d_{2}^{2}} \times \frac{l_{2}}{l_{1}}$ $[A = π r^{2} = \frac{π d^{2}}{4}]$
$\frac{H_{1}}{H_{2}} = {(\frac{1}{2})}^{2} \times \frac{1}{2}$
$\frac{H_{1}}{H_{2}} = \frac{1}{8}$
$∴ H_{1} : H_{2} = 1 : 8$

[Manipal UGET-2012]

Capacitance

165702 As shown in the figure below, if a capacitor $C$ is charged by connecting it resistance $R$ , then energy given by the battery will be

1

\frac{1}{2} {CV}^{2}

2 more than

\frac{1}{2} {CV}^{2}

3 less than

\frac{1}{2} {CV}^{2}

4 zero

Explanation:

: We know,
Energy stored in a capacitor
$E = \frac{1}{2} {CV}_{C}^{2}$
Where, $V_{C} =$ Voltage drop across capacitor.
Apply KVL in Circuit
$V = V_{R} + V_{C}$
We know, in steady state current through resistance $= 0$
$V_{R} = IR = 0$
Hence, $V = V_{C}$
From equation (i)
$E = \frac{1}{2} {CV}^{2}$

[CG PET -2018]

Capacitance

165703 The energy per unit volume for a capacitor having area $A$ and separation $d$ kept at potential different $V$ is given by

1

\frac{1}{2} ε_{o} \frac{V^{2}}{d^{2}}

2

\frac{1}{2 ε_{o}} \frac{V^{2}}{d^{2}}

3

\frac{1}{2} {CV}^{2}

4

\frac{Q^{2}}{2 C}

Explanation:

: We know,
Total energy Stored in a capacitor,
$E = \frac{1}{2} {CV}^{2}$
$or E = \frac{1}{2} (\frac{ε_{0} A}{d}) V^{2} (∵ C = \frac{ε_{0} A}{d})$
Volume between the plate $= A \times d$
Hence, Energy per unit Volume,
$\frac{E}{A d} = \frac{\frac{1}{2} (\frac{ε_{0} A}{d}) V^{2}}{A d} = \frac{1}{2} \frac{ε_{0} V^{2}}{d^{2}}$

[CG PET- 2017

Capacitance

165704 A $600 pF$ capacitor is charged by a $200 V$ supply. It is then disconnected from the supply and is connected to another uncharged $600 pF$ capacitor. Electrostatic energy lost in the process is

1

6 \times 10^{- 6} J

2

3 \times 10^{- 6} J

3

6 \times 10^{- 9} J

4

3 \times 10^{- 9} J

Explanation:

: Given that,
Capacitance of Capacitor $(C) = 600 pF = 600 \times 10^{- 12} F$ and potential difference $(V) = 200 V$

Electrostatic Energy stored $(E) = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times (600 \times 10^{- 12}) \times (200)^{2}$
$= \frac{1}{2} \times 600 \times 10^{- 12} \times 4 \times 10^{4}$
$= 1.2 \times 10^{- 5} J$
After disconnecting, A $600 pF$ capacitor added.
Then equivalent capacitance $(C^{'})$
$\frac{1}{C^{'}} = \frac{1}{C} + \frac{1}{C}$
$= \frac{1}{600} + \frac{1}{600}$
$\frac{1}{C^{'}} = \frac{1}{300}$
$C^{'} = 300 pF$
Then, new electrostatic energy,
$E^{'} = \frac{1}{2} C^{'} \times V^{2}$
$= \frac{1}{2} \times 300 \times 10^{- 12} \times (200)^{2}$
$= 0.6 \times 10^{- 5} J$
Loss in electrostatic energy $(E - E^{'})$
$= 1.2 \times 10^{- 5} - 0.6 \times 10^{5}$
$= 0.6 \times 10^{- 5}$
$= 6 \times 10^{- 6} J$

[CG PET -2016]

Capacitance

165700 Two capacitors, one $4 pF$ and the other $6 pF$ , connected in parallel, are charged by a $100 V$ battery. The energy stored in the capacitors is

1

12 \times 10^{- 8} J

2

2.4 \times 10^{- 8} J

3

5.0 \times 10^{- 8} J

4

1.2 \times 10^{- 6} J

Explanation:

: Given, $C_{1} = 4 pF$ and $C_{2} = 6 pF$
$Voltage = 100 V$
Both capacitors connected in parallel.
Hence, equivalent capacitance,
$C = C_{1} + C_{2} = 4 pF + 6 pF = 10 pF$
$C = 10 \times 10^{- 12} F$
Energy stored in the capacitor $(E) = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times 10 \times 10^{- 12} \times (100)^{2}$
$= \frac{1}{2} \times 10 \times 10^{- 12} \times 10^{4}$
$E = 5.0 \times 10^{- 8} J$

[Manipal UGET-2017]

Capacitance

165701 Two conductors of the same material have their diameters in the ratio $1 : 2$ and their lengths in the ratio $2 : 1$ . If the temperature difference between their ends is the same then the ratio of amounts of heat conducted per second through them will be

1

4 : 1

2

1 : 4

3

8 : 1

4

1 : 8

Explanation:

: We know,
$H = \frac{Q}{t} = \frac{K \times A \times Δ t}{l}$
$H \propto \frac{A}{l}$
$\frac{H_{1}}{H_{2}} = \frac{A_{1}}{A_{2}} \times \frac{l_{2}}{l_{1}}$
$\frac{H_{1}}{H_{2}} = \frac{d_{1}^{2}}{d_{2}^{2}} \times \frac{l_{2}}{l_{1}}$ $[A = π r^{2} = \frac{π d^{2}}{4}]$
$\frac{H_{1}}{H_{2}} = {(\frac{1}{2})}^{2} \times \frac{1}{2}$
$\frac{H_{1}}{H_{2}} = \frac{1}{8}$
$∴ H_{1} : H_{2} = 1 : 8$

[Manipal UGET-2012]

Capacitance

165702 As shown in the figure below, if a capacitor $C$ is charged by connecting it resistance $R$ , then energy given by the battery will be

1

\frac{1}{2} {CV}^{2}

2 more than

\frac{1}{2} {CV}^{2}

3 less than

\frac{1}{2} {CV}^{2}

4 zero

Explanation:

: We know,
Energy stored in a capacitor
$E = \frac{1}{2} {CV}_{C}^{2}$
Where, $V_{C} =$ Voltage drop across capacitor.
Apply KVL in Circuit
$V = V_{R} + V_{C}$
We know, in steady state current through resistance $= 0$
$V_{R} = IR = 0$
Hence, $V = V_{C}$
From equation (i)
$E = \frac{1}{2} {CV}^{2}$

[CG PET -2018]

Capacitance

165703 The energy per unit volume for a capacitor having area $A$ and separation $d$ kept at potential different $V$ is given by

1

\frac{1}{2} ε_{o} \frac{V^{2}}{d^{2}}

2

\frac{1}{2 ε_{o}} \frac{V^{2}}{d^{2}}

3

\frac{1}{2} {CV}^{2}

4

\frac{Q^{2}}{2 C}

Explanation:

: We know,
Total energy Stored in a capacitor,
$E = \frac{1}{2} {CV}^{2}$
$or E = \frac{1}{2} (\frac{ε_{0} A}{d}) V^{2} (∵ C = \frac{ε_{0} A}{d})$
Volume between the plate $= A \times d$
Hence, Energy per unit Volume,
$\frac{E}{A d} = \frac{\frac{1}{2} (\frac{ε_{0} A}{d}) V^{2}}{A d} = \frac{1}{2} \frac{ε_{0} V^{2}}{d^{2}}$

[CG PET- 2017

Capacitance

165704 A $600 pF$ capacitor is charged by a $200 V$ supply. It is then disconnected from the supply and is connected to another uncharged $600 pF$ capacitor. Electrostatic energy lost in the process is

1

6 \times 10^{- 6} J

2

3 \times 10^{- 6} J

3

6 \times 10^{- 9} J

4

3 \times 10^{- 9} J

Explanation:

: Given that,
Capacitance of Capacitor $(C) = 600 pF = 600 \times 10^{- 12} F$ and potential difference $(V) = 200 V$

Electrostatic Energy stored $(E) = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times (600 \times 10^{- 12}) \times (200)^{2}$
$= \frac{1}{2} \times 600 \times 10^{- 12} \times 4 \times 10^{4}$
$= 1.2 \times 10^{- 5} J$
After disconnecting, A $600 pF$ capacitor added.
Then equivalent capacitance $(C^{'})$
$\frac{1}{C^{'}} = \frac{1}{C} + \frac{1}{C}$
$= \frac{1}{600} + \frac{1}{600}$
$\frac{1}{C^{'}} = \frac{1}{300}$
$C^{'} = 300 pF$
Then, new electrostatic energy,
$E^{'} = \frac{1}{2} C^{'} \times V^{2}$
$= \frac{1}{2} \times 300 \times 10^{- 12} \times (200)^{2}$
$= 0.6 \times 10^{- 5} J$
Loss in electrostatic energy $(E - E^{'})$
$= 1.2 \times 10^{- 5} - 0.6 \times 10^{5}$
$= 0.6 \times 10^{- 5}$
$= 6 \times 10^{- 6} J$

[CG PET -2016]

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