Two identical parallel-plate capacitors A and B (plate area a, initial separation d) are connected in series across a battery of EMF V. The separation between the plates of capacitor B then increases at a constant rate v. The rate at which work is done ON the battery when the separation of

Question

Two identical parallel-plate capacitors A and B (plate area a, initial separation d) are connected in series across a battery of EMF V. The separation between the plates of capacitor B then increases at a constant rate v. The rate at which work is done ON the battery when the separation of B is 2d is given as epsilon₀ * a * V² * v / (n * d²). Find n.

Accepted Answer

4. C_A = epsilon₀*a/d, C_B = epsilon₀*a/s. In series: C_total = epsilon₀*a/(d+s). Q = V*C_total = V*epsilon₀*a/(d+s). Rate of change: dQ/dt = -V*epsilon₀*a*v/(d+s)². At s=2d: dQ/dt = -V*epsilon₀*a*v/(9d²). Power on battery = V*|dQ/dt| = epsilon₀*a*V²*v/(9d²). So n = 9.

Solution

Related JEE Advanced Physics questions