How can inserting a dielectric between capacitor plates increase the stored energy?
This question can be interpreted in 2 different ways according to which characteristics of the system are considered to be constant. Case 1 – The capacitor has fixed charge (Q) – i.e. it has been charged to voltage V, and then disconnected from the source of charging current. We need an expression for energy (E) in terms of relative permittivity (e). Starting with a well-known equation for stored energy – E = 0.5 C V^2, and using an equation for V, C and Q V = Q/C, we get E = 0.5 Q^2/C The capacitance C is proportional to relative permittivity of the dielectric, so we can write E = 0.5 Q^2/Ca*e where Ca is the capacitance with air as a dielectric. It is clear that inserting a dielectric of permittivity e will reduced the stored energy in inverse proportion to e. Where has the energy gone ? The answer to this is that as the dielectric material is inserted, the capacitor will exert an electrostatic force on it, and actually pull it in between the plates. So the capacitor works like a kin
