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. Find the capacitance of the new combination.

Step-by-step solution

The capacitance of the original parallel-plate capacitor is given by: C₀ = ₀ A d = 5 F When a dielectric slab of thickness t and dielectric constant K is introduced between the plates, the new capacitance is: C = ₀ A d - t + t K Given d = 2 mm , t = 1 mm , and K = 5 , we substitute these values into the formula: C = ₀ A 2 - 1 + 1 5 = ₀ A 1.2 From the initial condition, we have ₀ A = C₀ d = 5 2 = 10 F mm . Substituting this into the expression for the new capacitance: C = 10 1.2 = 100 12 = 25 3 8.3 F