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Electrostatic Potentials and Capacitance

268154 A capacitoris connected with a battery and stores energy U. A fter removing the battery, it is connected with another similar capacitor in parallel. The new stored energy in each capacitor will be

1

\frac{U}{2}

2

U

3

\frac{U}{4}

4

\frac{3 U}{2}

Explanation:

$V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}; U = \frac{1}{2} (C_{1} + C_{2}) V^{2}$

Electrostatic Potentials and Capacitance

268155 A parallel plate condenser with a dielectric of dielectric constant $K$ between the plates has a capacity $C$ and is charged to a potential $V$ volts. The dielectric slab is slowly removed from between the plates and then reinserted.
The net work done by the system in this process is

1

\frac{1}{2} (K - 1) C V^{2}

2

C V^{2} (K - 1) / K

3

(K - 1) C V^{2}

4 zero

Explanation:

On introduction and removal and again on introduction, the capacity and potential remain sameSo, net work done by the system in this process.
$W = U_{f} - U_{i}; \frac{1}{2} C V^{2} - \frac{1}{2} C V^{2} = 0$

Electrostatic Potentials and Capacitance

268156 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 $Δ T$ , the potential difference $V$ across the capacitor is

1

\sqrt{\frac{2 m C Δ T}{s}}

2

\frac{m C Δ T}{s}

3

\frac{m Δ Δ T}{C}

4

\sqrt{\frac{2 m Δ Δ T}{C}}

Explanation:

$E = (\frac{1}{2}) C V^{2}$ The energy stored in capacitor is lost in form of heat energy. $H = m Δ Δ T$
$\begin{array}{r} ∴ m s Δ T = (\frac{1}{2}) C V^{2}; C = \frac{q_{1} + q_{2}}{V_{A} - V_{B}} = \frac{q_{1} + q_{2}}{\frac{q_{2}}{C_{2}} + \frac{q_{1}}{C_{1}}} \\ ∴ C = \frac{2 C_{1} C_{2} + C_{3} (C_{1} + C_{2})}{C_{1} + C_{2} + 2 C_{3}}; ∴ V = \sqrt{\frac{2 m Δ Δ T}{C}} \end{array}$

Electrostatic Potentials and Capacitance

268157 A parallel plate capacitor of capacity $100 μ F$ is charged by a battery at 50 volts. The battery remains connected and if the plates of the capacitor are separated so that the distance between them is halved the original distance, the additional energy gives by the battery to the capacitor in J oules is

1

125 \times 10^{- 3}

2

12.5 \times 10^{- 3}

3

1.25 \times 10^{- 3}

4

0.125 \times 10^{- 3}

Explanation:

$E_{1} = \frac{1}{2} C_{1} V^{2}; E_{2} = \frac{1}{2} C_{2} V^{2}; C_{2} = 2 C_{1}$

Electrostatic Potentials and Capacitance

268154 A capacitoris connected with a battery and stores energy U. A fter removing the battery, it is connected with another similar capacitor in parallel. The new stored energy in each capacitor will be

1

\frac{U}{2}

2

U

3

\frac{U}{4}

4

\frac{3 U}{2}

Explanation:

$V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}; U = \frac{1}{2} (C_{1} + C_{2}) V^{2}$

Electrostatic Potentials and Capacitance

268155 A parallel plate condenser with a dielectric of dielectric constant $K$ between the plates has a capacity $C$ and is charged to a potential $V$ volts. The dielectric slab is slowly removed from between the plates and then reinserted.
The net work done by the system in this process is

1

\frac{1}{2} (K - 1) C V^{2}

2

C V^{2} (K - 1) / K

3

(K - 1) C V^{2}

4 zero

Explanation:

On introduction and removal and again on introduction, the capacity and potential remain sameSo, net work done by the system in this process.
$W = U_{f} - U_{i}; \frac{1}{2} C V^{2} - \frac{1}{2} C V^{2} = 0$

Electrostatic Potentials and Capacitance

268156 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 $Δ T$ , the potential difference $V$ across the capacitor is

1

\sqrt{\frac{2 m C Δ T}{s}}

2

\frac{m C Δ T}{s}

3

\frac{m Δ Δ T}{C}

4

\sqrt{\frac{2 m Δ Δ T}{C}}

Explanation:

$E = (\frac{1}{2}) C V^{2}$ The energy stored in capacitor is lost in form of heat energy. $H = m Δ Δ T$
$\begin{array}{r} ∴ m s Δ T = (\frac{1}{2}) C V^{2}; C = \frac{q_{1} + q_{2}}{V_{A} - V_{B}} = \frac{q_{1} + q_{2}}{\frac{q_{2}}{C_{2}} + \frac{q_{1}}{C_{1}}} \\ ∴ C = \frac{2 C_{1} C_{2} + C_{3} (C_{1} + C_{2})}{C_{1} + C_{2} + 2 C_{3}}; ∴ V = \sqrt{\frac{2 m Δ Δ T}{C}} \end{array}$

Electrostatic Potentials and Capacitance

268157 A parallel plate capacitor of capacity $100 μ F$ is charged by a battery at 50 volts. The battery remains connected and if the plates of the capacitor are separated so that the distance between them is halved the original distance, the additional energy gives by the battery to the capacitor in J oules is

1

125 \times 10^{- 3}

2

12.5 \times 10^{- 3}

3

1.25 \times 10^{- 3}

4

0.125 \times 10^{- 3}

Explanation:

$E_{1} = \frac{1}{2} C_{1} V^{2}; E_{2} = \frac{1}{2} C_{2} V^{2}; C_{2} = 2 C_{1}$

Electrostatic Potentials and Capacitance

268154 A capacitoris connected with a battery and stores energy U. A fter removing the battery, it is connected with another similar capacitor in parallel. The new stored energy in each capacitor will be

1

\frac{U}{2}

2

U

3

\frac{U}{4}

4

\frac{3 U}{2}

Explanation:

$V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}; U = \frac{1}{2} (C_{1} + C_{2}) V^{2}$

Electrostatic Potentials and Capacitance

268155 A parallel plate condenser with a dielectric of dielectric constant $K$ between the plates has a capacity $C$ and is charged to a potential $V$ volts. The dielectric slab is slowly removed from between the plates and then reinserted.
The net work done by the system in this process is

1

\frac{1}{2} (K - 1) C V^{2}

2

C V^{2} (K - 1) / K

3

(K - 1) C V^{2}

4 zero

Explanation:

On introduction and removal and again on introduction, the capacity and potential remain sameSo, net work done by the system in this process.
$W = U_{f} - U_{i}; \frac{1}{2} C V^{2} - \frac{1}{2} C V^{2} = 0$

Electrostatic Potentials and Capacitance

268156 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 $Δ T$ , the potential difference $V$ across the capacitor is

1

\sqrt{\frac{2 m C Δ T}{s}}

2

\frac{m C Δ T}{s}

3

\frac{m Δ Δ T}{C}

4

\sqrt{\frac{2 m Δ Δ T}{C}}

Explanation:

$E = (\frac{1}{2}) C V^{2}$ The energy stored in capacitor is lost in form of heat energy. $H = m Δ Δ T$
$\begin{array}{r} ∴ m s Δ T = (\frac{1}{2}) C V^{2}; C = \frac{q_{1} + q_{2}}{V_{A} - V_{B}} = \frac{q_{1} + q_{2}}{\frac{q_{2}}{C_{2}} + \frac{q_{1}}{C_{1}}} \\ ∴ C = \frac{2 C_{1} C_{2} + C_{3} (C_{1} + C_{2})}{C_{1} + C_{2} + 2 C_{3}}; ∴ V = \sqrt{\frac{2 m Δ Δ T}{C}} \end{array}$

Electrostatic Potentials and Capacitance

268157 A parallel plate capacitor of capacity $100 μ F$ is charged by a battery at 50 volts. The battery remains connected and if the plates of the capacitor are separated so that the distance between them is halved the original distance, the additional energy gives by the battery to the capacitor in J oules is

1

125 \times 10^{- 3}

2

12.5 \times 10^{- 3}

3

1.25 \times 10^{- 3}

4

0.125 \times 10^{- 3}

Explanation:

$E_{1} = \frac{1}{2} C_{1} V^{2}; E_{2} = \frac{1}{2} C_{2} V^{2}; C_{2} = 2 C_{1}$

Electrostatic Potentials and Capacitance

268154 A capacitoris connected with a battery and stores energy U. A fter removing the battery, it is connected with another similar capacitor in parallel. The new stored energy in each capacitor will be

1

\frac{U}{2}

2

U

3

\frac{U}{4}

4

\frac{3 U}{2}

Explanation:

$V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}; U = \frac{1}{2} (C_{1} + C_{2}) V^{2}$

Electrostatic Potentials and Capacitance

268155 A parallel plate condenser with a dielectric of dielectric constant $K$ between the plates has a capacity $C$ and is charged to a potential $V$ volts. The dielectric slab is slowly removed from between the plates and then reinserted.
The net work done by the system in this process is

1

\frac{1}{2} (K - 1) C V^{2}

2

C V^{2} (K - 1) / K

3

(K - 1) C V^{2}

4 zero

Explanation:

On introduction and removal and again on introduction, the capacity and potential remain sameSo, net work done by the system in this process.
$W = U_{f} - U_{i}; \frac{1}{2} C V^{2} - \frac{1}{2} C V^{2} = 0$

Electrostatic Potentials and Capacitance

268156 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 $Δ T$ , the potential difference $V$ across the capacitor is

1

\sqrt{\frac{2 m C Δ T}{s}}

2

\frac{m C Δ T}{s}

3

\frac{m Δ Δ T}{C}

4

\sqrt{\frac{2 m Δ Δ T}{C}}

Explanation:

$E = (\frac{1}{2}) C V^{2}$ The energy stored in capacitor is lost in form of heat energy. $H = m Δ Δ T$
$\begin{array}{r} ∴ m s Δ T = (\frac{1}{2}) C V^{2}; C = \frac{q_{1} + q_{2}}{V_{A} - V_{B}} = \frac{q_{1} + q_{2}}{\frac{q_{2}}{C_{2}} + \frac{q_{1}}{C_{1}}} \\ ∴ C = \frac{2 C_{1} C_{2} + C_{3} (C_{1} + C_{2})}{C_{1} + C_{2} + 2 C_{3}}; ∴ V = \sqrt{\frac{2 m Δ Δ T}{C}} \end{array}$

Electrostatic Potentials and Capacitance

268157 A parallel plate capacitor of capacity $100 μ F$ is charged by a battery at 50 volts. The battery remains connected and if the plates of the capacitor are separated so that the distance between them is halved the original distance, the additional energy gives by the battery to the capacitor in J oules is

1

125 \times 10^{- 3}

2

12.5 \times 10^{- 3}

3

1.25 \times 10^{- 3}

4

0.125 \times 10^{- 3}

Explanation:

$E_{1} = \frac{1}{2} C_{1} V^{2}; E_{2} = \frac{1}{2} C_{2} V^{2}; C_{2} = 2 C_{1}$

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