Question: Q. 3. 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 of stored initially in the single capacitor.
A [O.D. I, II, III 2014]
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Solution:
Ans. Energy stored in a capacitor
$$ U=\frac{1}{2} C V^{2} \text { or } \frac{1}{2} \frac{q^{2}}{C} \quad 1 / 2 $$
Capacitance of the (parallel) combination
$$ =C+C=2 C $$
Here, the total charge, $Q$, remains the same :
$\therefore \quad$ Initial energy $=\frac{1}{2} \cdot \frac{q^{2}}{C}$ and $\quad$ Final energy $=\frac{1}{2} \frac{q^{2}}{2 C}$
$\therefore \quad$ Ratio of energies $=\frac{\text { final energy }}{\text { initial energy }}$
$$ \begin{aligned} & =\frac{\frac{1}{4} \frac{q^{2}}{C}}{\frac{1}{2} \frac{q^{2}}{C}} \ & =\frac{2}{4}=\frac{1}{2} \end{aligned} $$
[CBSE Marking Scheme 2014]
