Dented Capacitor

Energy of a capacitor is $\frac{\text{Q}^2}{2\text{C}}$ . Thus change in capacitance can be calculated by calculating change in energy.

The energy in the capacitor increases because the charges have been pushed back in direction opposite to field. Further, an electrostatic field E has an energy $\frac{\epsilon_{o} \text{E}^2}{2}$ per unit volume, and an alternative view is that, when the capacitor is dented, the electric field exists in a volume where it was not previously present.

If the surface is dented only a little, the electric field near the surface can be taken same a the original one. Thus change in energy is solely dependent on change in volume.

Consider a sphere is transformed such that there is a $3 \%$ decrease in volume uniformly, the change in radius will be $1 \%$ . Since $C=4 \pi \epsilon_{o} R$ , change in capacitance is also $1 \%$ .

Since in a dented sphere, change in energy is same as the case considered above. Change in capcitance must be the same.

Final answer : $1 \%$