A parallel plate capacitor has a plate separation of 5 mm . A dielectric slab of thickness 3 mm and dielectric constant 2 is inserted between the plates, leaving the remaining space filled with air. The dielectric strength of air is 3 10^6 V/m and that of the dielectric slab is 1.2 10^6 V/m . The maximum potential difference that can be applied across the capacitor without causing breakdown in either medium is :

Step-by-step solution

Let E_a be the electric field in the air and E_d be the electric field in the dielectric slab. Since the electric displacement must be continuous across the boundary, we have: ₀ E_a = k ₀ E_d E_a = 2E_d We must check which medium reaches its breakdown limit first. The maximum allowed fields are E_a 3 10^6 V/m and E_d 1.2 10^6 V/m . If the air reaches its limit ( E_a = 3 10^6 V/m ), the field in the slab would be E_d = 1.5 10^6 V/m , which exceeds its dielectric strength. Thus, the dielectric slab breaks down first.