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 while the capacitor remains connected to a battery (constant voltage V ), the electric field decreases due to polarization ( E = E₀/K ), and the capacitance increases ( C' = K C ). This happens because the dielectric reduces the field,