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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 :
some $+ve$ work is done by external agent
energy of capacitor will decrease
energy of capacitor will increase
none of the above
Solution
For parallel plates capacitor, $C=\frac{A \epsilon_{0}}{d}, V=\frac{Q d}{A \epsilon_{0}}$
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.
