A parallel-plate capacitor with plate separation d has an initial capacitance of 60 F . It is connected to a 10 V battery. While the battery remains connected, a dielectric slab of thickness d 2 and dielectric constant K = 2 is inserted between the plates, parallel to them. The extra charge drawn from the battery due to the insertion of the slab is

Step-by-step solution

The initial capacitance is C₀ = ₀ A d = 60 F . The initial charge on the capacitor is Q₀ = C₀ V = 60 F 10 V = 600 C . When a dielectric slab of thickness t = d 2 is inserted, the new equivalent capacitance is given by C_ eq = ₀ A d - t + t K . Substituting t = d 2 and K = 2 : C_ eq = ₀ A d - d 2 + d 4 = ₀ A 3d 4 = 4 3 ( ₀ A d ) = 4 3 C₀ . Calculating the value: C_ eq = 4 3 60 F = 80 F . Since the battery remains connected, the voltage V is constant at 10 V . The new charge on the capacitor is Q' = C_ eq V = 80 F 10