The plates of a parallel plate capacitor are separated by a distance 'd' with air as the medium between them. A dielectric slab of dielectric constant 3 is introduced between the plates so as to increase the capacity by 50 % . The thickness of the dielectric slab is

Step-by-step solution

Let C₀ = ₀ A d denote the initial capacitance of the parallel plate capacitor with plate area A and separation d . A dielectric slab of dielectric constant k = 3 and thickness t increases the capacitance, which becomes C = ₀ A d - t + t k . The problem states that C = 3 2 C₀ , so ₀ A d - t + t k = 3 2 ₀ A d . Common factors cancel, yielding 1 d - t + t 3 = 3 2d . Substituting k = 3 and simplifying the denominator gives d - t + t 3 = d - 2t 3 , so the equation becomes 1 d - 2t 3 = 3 2d . Cross-multiplying: 2d = 3 (d