A parallel plate capacitor has an initial plate separation of 15 mm . When a dielectric slab of dielectric constant 5 and unknown thickness t is inserted between the plates, one of the conducting plates has to be moved apart by 4 mm to keep the capacitance same as in the previous case. The thickness t of the dielectric slab is :

Step-by-step solution

Let the initial capacitance be C₀ = ₀ A d . When the dielectric slab of thickness t and dielectric constant k is inserted, and the plate separation is increased by x , the new capacitance is: C = ₀ A d + x - t + t k Since the capacitance remains unchanged, C = C₀ : d = d + x - t + t k Simplifying this, we get: x = t (1 - 1 k ) Substitute the given values x = 4 mm and k = 5 : 4 = t (1 - 1 5 ) 4 = t ( 4 5 ) t = 5 mm Answer: 5.0 mm