CBSE Class 12 · 2023 · CapacitanceShort Answer

A parallel plate capacitor (A) of capacitance

C

is charged by a battery to voltage

V

. The battery is disconnected and an uncharged capacitor (B) of capacitance

2 C

is connected across A. Find the ratio of (i) final charges on A and B, and (ii) total electrostatic energy stored in A and B finally and that stored in A initially.

Answer: (i) $\frac{Q _{A}^{'}}{Q _{B}^{'}} = \frac{1}{2}$ ;
(ii) $\frac{U _{final}}{U _{initial}} = \frac{1}{3}$ .

Solution

Initial charge on A:

Q = C V

On connecting B (capacitance

2 C

) they share charge at a common potential

V^{'}

C V = C V^{'} + 2 C V^{'} \Rightarrow V^{'} = \frac{V}{3}

(i) Final charges:

Q_{A}^{'} = C V^{'} = \frac{C V}{3}, Q_{B}^{'} = 2 C V^{'} = \frac{2 C V}{3}

\frac{Q _{A}^{'}}{Q _{B}^{'}} = \frac{1}{2}

(ii) Initial energy:

U_{i} = \frac{1}{2} C V^{2}

Final energy:

U_{f} = \frac{1}{2} C V^{'2} + \frac{1}{2} (2 C) V^{'2} = \frac{1}{2} (3 C) (\frac{V}{3})^{2} = \frac{C V ^{2}}{6}

\frac{U _{f}}{U _{i}} = \frac{C V ^{2} /6}{C V ^{2} /2} = \frac{1}{3}

CBSE Class 12 Board · 2023 · Subjective

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CBSE Class 12 · 2023 · CapacitanceShort Answer

Solution

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