A parallel plate capacitor has an initial capacitance C when there is vacuum between its plates. A dielectric slab of thickness equal to half the plate separation and having dielectric constant K is then inserted between the plates. If the new capacitance of the system is 4C 3 , the value of K is :

Step-by-step solution

Let the plate area be A and the separation be d . The initial capacitance is C = ₀ A d . When a dielectric slab of thickness t = d 2 and dielectric constant K is inserted, the new capacitance is given by: C_ new = ₀ A d - t + t K Substituting t = d 2 : C_ new = ₀ A d - d 2 + d 2K = ₀ A d 2 + d 2K = 2 ₀ A d (1 + 1 K ) = 2K K+1 C Given that C_ new = 4C 3 , we have: 2K K+1 = 4 3 6K = 4K + 4 2K = 4 K = 2 Answer: 2