MHT CET202521 Apr 2025Evening 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 the plates is ( ₀= permittivity of free space)
Options
- A₀ AV ^2 3 ~d
- B₀ AV ^2 4 ~d
- C2 ₀ A V^2 d
- D3 ₀ A V^2 2 d
Correct answer
D. 3 ₀ A V^2 2 d
Step-by-step solution
The initial capacitance is C₁ = ₀ A d , so with potential difference V , the stored charge is Q = C₁V = ₀ A V d and the initial energy U₁ = 1 2 C₁ V^2 = 1 2 ₀ A d V^2 . After disconnecting the battery, charge Q remains constant while plate separation increases to 4d , giving new capacitance C₂ = ₀ A 4d . The final energy is then U₂ = Q^2 2C₂ = ( ₀ A V/d)^2 2( ₀ A/4d) = 4 ₀ A V^2 2d = 2 ₀ A V^2 d . The work done equals the energy change: W = U₂ - U₁ = 2 ₀ A V^2 d - 1 2 ₀ A V^2 d = 3 2 ₀ A V^2 d . D