A parallel plate capacitor has plate of length l , width w and...

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

${C}_{0} = \dfrac{A{ε}_{0}}{d} = \dfrac{wl{ε}_{0}}{d} . . . ( 1 )$

Energy stored in the capacitor before the insertion of dielectrics is given by,

${U}_{0} = \dfrac{1}{2}{C}_{0}{V}^{2} = \dfrac{1}{2}\left(\dfrac{wl{ε}_{0}}{d}\right){V}^{2} . . . ( 2 )$

Suppose that, the slab is inside the plate at $y$ distance, therefore, it becomes the combination of two capacitors one with air with length $( l - y )$ and another with dielectric with length $y$ as shown in diagram.

Now the capacitance of capacitor in parallel combination is given by,

$C = {C}_{1} + {C}_{2} . . . ( 3 )$

Where,

${C}_{1} = \dfrac{w ( l - y ) {ε}_{0}}{d} . . . ( 4 )$

${C}_{1} = \dfrac{wyK{ε}_{0}}{d} . . . ( 5 )$

Now energy stored in this combination of capacitor is given by

$U = \dfrac{1}{2} ( {C}_{1} + {C}_{2} ) {V}^{2} . . . ( 6 )$

Now according to the given condition,

$2{U}_{0} = U$

$⇒ 2\left(\dfrac{1}{2}{C}_{0}{V}^{2}\right) = \dfrac{1}{2}\left({C}_{1} + {C}_{2}\right){V}^{2}$

$⇒ 2{C}_{0} = {C}_{1} + {C}_{2}$

$⇒ 2\left(\dfrac{{ϵ}_{0}wl}{d}\right) = \dfrac{{ϵ}_{0}Kwy}{d} + \dfrac{{ϵ}_{0}w ( l - y )}{d}$

$⇒ 2l = Ky + l - y$

$⇒ l = 4y - y = 3y$

$⇒ y = \dfrac{l}{3}$