A parallel plate capacitor of capacitance C is charged to a poten… - Vidyadip

Question

A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.

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Answer

Energy stored in a capacitor $ = \frac{1}{2}\text{QV} = \frac{1}{2}\text{CV}^{2} = \frac{1}{2}\frac{\text{Q}^{2}}{\text{C}}$

Capacitance of the (parallel) combination = C+C=2C

Here, total charge, Q, remains the same

$\therefore\text{ initial energy } = \frac{1}{2}\frac{\text{Q}^{2}}{\text{C}}$

And final energy = $\frac{1}{2}\frac{\text{Q}^{2}}{2\text{C}}$

$\therefore\frac{\text{ final energy }}{\text{ initial energy }} = \frac{1}{2}$

Parallel or (ii) Series combination to remain constant (=V) and obtain the answers as (i) 2:1 or (ii) 1:2.

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