Practice question

**Effect of dielectric** is to reduce effective field due to polarization, bound surface charges opposite to free charges, net field E = E₀ - E_p = E₀/K. Capacitance C = Q/V = K ε₀ A/d for full filling, partial filling C = ε₀ A/(d - t + t/K) for slab thickness t. When a dielectric slab (with K > 1 ) is inserted into a disconnected capacitor, the charge Q remains constant. The capacitance increases ( C' = K C ), and the potential difference decreases ( V' = V/K ). The energy stored is given by U = (Q²/2C) , so with increased C