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. Capacitance C = (Q/V) . With constant charge Q , inserting a dielectric ( K > 1 ) increases C by reducing V ( C = (K ε₀ A/d) ). As the slab is withdrawn, the effective dielectric constant decreases (less area has K ), lowering C . Since Q is constant, V