Deduce the expression for the electrostatic energy stored in a capacitor of capacitance C and having charge Q. How will the (i) energy stored and (ii) the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant K?

Energy stored in a charged capacitor. The energy of a charged capacitor is measured by the total work done in charging the capacitor to a given potential. We know that capacitance is C = Q/V; where Q is the charge on the plates and V is the potential difference. When an additional amount of charge dQ is transferred from the negative to the positive plate, the work done is dW = VdQ = (Q/C)dQ. The total work stored as electrostatic potential energy U in the capacitor is given by the integral of this expression from 0 to Q:

U = ∫₀𝑄 (Q/C)dQ = (1/C) ∫₀𝑄 QdQ = (1/C) [Q²/2]₀𝑄 = Q²/2C

Alternatively, since U = QV and Q = CV, we can write U = (1/2)CV²

When a dielectric of dielectric constant K is filled fully between the plates, the capacitance increases by a factor of K: C' = KC. The charge stored in the capacitor, assuming the battery remains connected, will also increase to Q' = KC V. The energy stored will now be:

U' = (Q')²/2C' = (KCV)² / 2(KC) = KCV²/2 = KU

Therefore, the energy stored increases by a factor of K.

If the battery is disconnected before inserting the dielectric, the charge Q remains constant. The capacitance changes to C' = KC, and the voltage across the capacitor changes to V' = Q/C' = Q/KC = V/K. The energy stored is now:

U' = Q²/2C' = Q²/2KC = U/K

Therefore, the energy stored decreases by a factor of K.

The electric field inside the capacitor is given by E = V/d, where d is the separation between the plates. When a dielectric is inserted, the voltage across the capacitor decreases (if the battery is disconnected, as discussed above) thus decreasing the electric field. Specifically, E' = V'/d = (V/K)/d = E/K. Therefore the electric field decreases by a factor of K when the dielectric material is present.