A parallel plate capacitor has plate area ' A ' and separation between plates is 'd'. It is charged to a potential difference of V₀ volt. The charging battery is then disconnected and plates are pulled apart three times the initial distance. The work done to increase the distance between the plates is ( ₀= . permittivity of free space )

Step-by-step solution

Let the initial capacitance be C ₀= ₀ ~A ~d Let the charge on the capacitor be Q _ initial = C ₀ ~V ₀ Plate separation is increased by 3 times i.e., d ^ =3 ~d C_ final = ₀ A 3 d = 1 3 ( ₀ A d )= C₀ 3 Let Q _ final be the final charge on the capacitor and V_ final be the final potential on the capacitor. Q _ final = C _ fiñal V _ final = 1 3 C ₀ ~V _ fial As the capacitor is isolated, aligned & Q _ final = Q _ initial, & aligned C ₀ ~V ₀ & = 1 3 C ₀ ~V _ fial V _ final =3 ~V ₀ & Work done & = Final P.E - Initial P.E