A parallel-plate capacitor with a capacitance of 5 F is connected to a battery having an emf of 6 V . The separation between the plates is 2 mm . A dielectric slab with a thickness of 1 mm and a dielectric constant of 5 is introduced into the gap, occupying the lower half of it. Calculate the amount of charge that flows through the battery after the slab is inserted.

Step-by-step solution

Initial capacitance of the parallel-plate capacitor is given by: C_i = ₀ A d = 5 F The initial charge on the capacitor is: Q_i = C_i V = 5 6 = 30 C When a dielectric slab of thickness t and dielectric constant K is introduced into the gap, the new capacitance is: C_f = ₀ A d - t + t K Given d = 2 mm , t = 1 mm , and K = 5 , we have: C_f = ₀ A 2 - 1 + 1 5 = ₀ A 1.2 = 2 1.2 ₀ A 2 = 2 1.2 5 = 25 3 F The final charge on the capacitor is: Q_f = C_f V = 25 3 6 = 50 C The amount of charge that flows through the battery is