A parallel plate capacitor having capacity C ₀ is charged to V ₀ . With battery disconnected, if the separation between the plates is doubled then the energy stored in it is E₁ . Instead if the separation between the plates is doubled, with battery in connection, the energy stored in it is E ₂ . Then the value of E₂ E₁ is

Step-by-step solution

When battery disconnected, charge is constant aligned & Q = Q ₀= C ₀ ~V ₀ & ~d ^ =2 ~d ( Here C ₀= ₀ ~A ~d ) & C ^ = ₀ ~A ~d ^ = ₀ ~A 2 ~d = C ₀ 2 & E ₁= Q ^2 2 c ^ & E ₁= Q ₀^2 2 c ₀ 2 = C ₀ ~V ₀^2 aligned When battery in connection then potential will same. aligned & V = V ₀ & C = C ₀ 2 & E ₂= 1 2 C ~V ^2= 1 2 ( C ₀ 2 ) V ₀^2= 1 4 C ₀ ~V ₀^2 & E ₂ E ₁ = 1 4 C ₀ ~V ₀^2 C ₀ ~V ^2 =0.25 aligned