A parallel plate capacitor of capacity C₀ is charged to a potential V₀ . (i) The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is E₁ . (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is E₂ . Then, E₁ / E₂ value is:

Step-by-step solution

Capacitance of parallel plate capacitor, C₀= ₀ A d where, A= area of the plates, d= separation between the plates Charge stored in the capacitor, Q=C₀ V₀ When battery is disconnected, then charge remains same. So, energy, E₁= 1 2 Q^2 C C= capacitance when plate separation is doubled So, C₁= C₀ 2 E₁= 1 2 Q^2 C₀ / 2 = Q^2 C₀ = C₀^2 V₀^2 C₀ =C₀ V₀^2 When battery is connected, then Energy, E₂= 1 2 C V₀^2 where, E₂= 1 2 C₀ 2 V₀^2= 1 4 (C₀ V₀^2 ) E₁ E₂ = C₀ V₀^2 1 4 C₀ V₀^2 = 1 4 E₁: E₂=4: 1