A parallel plate capacitor has an initial plate separation of 20 mm . Two different dielectric slabs are inserted simultaneously between the plates: one of thickness 4 mm and dielectric constant 2 , and another of thickness 12 mm and dielectric constant 3 . To restore the capacitance to its original value, one of the plates must be moved apart by a distance x . The value of x is :

Step-by-step solution

Let the initial capacitance be C₀ = ₀ A d . When multiple dielectric slabs of thicknesses t₁, t₂ and dielectric constants k₁, k₂ are inserted, and the plate separation is increased by x , the new capacitance is: C = ₀ A d + x - (t₁ + t₂) + ( t₁ k₁ + t₂ k₂ ) Since the capacitance is restored to its original value, C = C₀ : d = d + x - t₁ - t₂ + t₁ k₁ + t₂ k₂ Simplifying this, we get the required displacement x : x = t₁ (1 - 1 k₁ ) + t₂ (1 - 1 k₂ ) Substitute the given values t₁ = 4 mm , k₁ = 2, t₂ = 12 mm , k₂ = 3 :