Question

A capacitor $C$ is charged to a potential difference $V$ and battery is disconnected. Now if the capacitor plates are brought close slowly by some distance:

• A. some $+ve$ work is done by external agent
• B. energy of capacitor will decrease
• C. energy of capacitor will increase
• D. none of the above

Answer 1

The correct option is B energy of capacitor will decrease

For parallel plates capacitor, $C=\frac{A{ϵ}_0}{d},V=\frac{Qd}{A{ϵ}_0}$
Thus capacitance is inversely proportional to separation (d) between the plate and potential is proportional to the separation (d).
After disconnect the battery, the capacitor would maintain its charge indefinitely and if there some leakage current, capacitance will discharge very slowly.
As the energy $U=\frac{1}{2}C{V}^2$, so the energy is proportional to separation (d) between the plates. Thus when the plates are brought close , the energy of the capacitor will decrease.