CBSE Class 12 · 2019 · CapacitanceLong Answer

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(a) Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.
(b) A parallel plate capacitor is charged by a battery to a potential difference

V

. It is disconnected from the battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor.

Answer: (a) $U = \frac{Q ^{2}}{2 C} = \frac{1}{2} C V^{2} = \frac{1}{2} Q V$ .
(b) The ratio is $1 : 2$ (final energy is half the initial).

Solution

(a) Connected to the battery, charge is driven onto the plates until the plate p.d. equals the emf.
The work to transfer

d q

at potential

\frac{q}{C}

\frac{q}{C} d q

; integrating from

0

Q

U = \frac{Q ^{2}}{2 C} = \frac{1}{2} C V^{2}

(b) Charge is conserved; the common potential is:

V^{'} = \frac{C V}{2 C} = \frac{V}{2}

U_{f} = \frac{1}{2} (2 C) (\frac{V}{2})^{2} = \frac{1}{4} C V^{2}, U_{i} = \frac{1}{2} C V^{2}

\frac{U _{f}}{U _{i}} = \frac{1}{2}

CBSE Class 12 Board · 2019 · Subjective

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CBSE Class 12 · 2019 · CapacitanceLong Answer

Solution

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