Question:
A parallel-plate capacitor having plate area $400 \mathrm{~cm}$-square and separation between the plates $\mathrm{I} .0 \mathrm{~mm}$ is connected to a power supply of $100 \mathrm{~V}$. A dielectric slab of thickness $0.5 \mathrm{~mm}$ and dielectric constant $5.0$ is inserted into the gap. (a) Find the increase in electrostatic energy, (b) If the power supply is now disconnected and the dielectric slab is taken out, find the further increase in energy, (c) Why does the energy increase in inserting the slab as well as in taking it out?
Solution:
(c) during insertion capacitance increases, more charge flows from battery and hence battery supplies energy During removal, energy increases, work has to be external agent to remove it.