MHT CET20228 Aug 2022Morning ShiftPhysicsCapacitanceActual
A parallel plate capacitor having plate area A and separation d is charged to a potential difference V . The charging battery is disconnected and the plates are pulled apart to four times the initial separation. The work required to increase the distance between plates is:
Options
- A₀ A V^2 4 d
- B2 ₀ A V^2 4 d
- C₀ A V^2 3 d
- D3 ₀ A V^2 2 d
Correct answer
D. 3 ₀ A V^2 2 d
Step-by-step solution
Initial energy: q^2 2 C The new capacitance is: C_n= C 4 Final energy after the change in capacitance is: 4 q^2 2 C Therefore, magnitude of work done is equal to the change in potential energy, W= 2 q^2 C - q^2 2 C = 3 C V^2 2 = 3 ₀ A V^2 2 d