A parallel plate capacitor of capacitance 12.5 pF is charged by a battery connected between its plates to potential difference of 12.0 ~V . The battery is now disconnected and a dielectric slab ( _ r =6 ) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is _____ 10⁻¹² ~J .

Step-by-step solution

Before inserting dielectric capacitance is given C ₀=12.5 pF and charge on the capacitor Q = C ₀ ~V After inserting dielectric capacitance will become _ r C ₀ . Change in potential energy of the capacitor aligned & = E _ i - E _ f & = Q ^2 2 C _ i - Q ^2 2 C _ f = Q ^2 2 C ₀ [1- 1 _ r ] & = ( C ₀ ~V )^2 2 C ₀ [1- 1 _ r ]= 1 2 C ₀ ~V ^2 [1- 1 _ r ] aligned Using C ₀=12.5 pF , V =12 ~V , _ r =6 aligned & = 1 2 (12.5) 12^2 [1- 1 6 ]= 1 2 (12.5) 12^2 5 6 & =750 pJ =750 10⁻¹² ~J aligned