A parallel-plate capacitor having plate area A and plate separation d is charged to a potential difference V , after which the battery is disconnected. A dielectric slab with dielectric constant K is subsequently inserted between the plates of the capacitor to completely fill the space. Determine the work done on the system during the insertion of the slab.

Step-by-step solution

Initial capacitance C_i = ₀ A d Initial charge on the capacitor Q = C_i V Initial energy of the system U_i = 1 2 C_i V^2 = ₀ A V^2 2d Since the battery is disconnected, the charge Q remains constant. When the dielectric slab is inserted, the final capacitance becomes C_f = K C_i = K ₀ A d Final energy of the system U_f = Q^2 2 C_f = (C_i V)^2 2 K C_i = C_i V^2 2K = ₀ A V^2 2Kd The work done on the system by the external agent is equal to the change in the potential energy of the system. W = U = U_f - U_i W = ₀ A V^