A parallel plate capacitor has plate of length l , width w and separation of plates is d . It is connected to a battery of emf V . A dielectric slab of the same thickness d and of dielectric constant K = 4 is being inserted between the plates of the capacitor. At what length of the slab inside plates, will the energy stored in the capacitor be two times the initial energy stored?

Step-by-step solution

Given, Length of plate of parallel plate capacitor is l Width of plate is w Separation of plates is d . Emf of battery is V Thickness of dielectric slab is d Dielectric constant of slab is K   =   4 Capacitance of capacitor before the insertion of dielectric slab is C 0 = A ε 0 d = w l ε 0 d     . . . ( 1 ) Energy stored in the capacitor before the insertion of dielectrics is given by, U 0 = 1 2 C 0 V 2 = 1 2 w l ε 0 d V 2     . . . ( 2 ) Suppose that, the slab is inside