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…

Given,

Length of plate of parallel plate capacitor is $l$l

Width of plate is $w$w

Separation of plates is $d$d.

Emf of battery is $V$V

Thickness of dielectric slab is $d$d

Dielectric constant of slab is $K=4$K=4

Capacitance of capacitor before the insertion of dielectric slab is

$${C}_{0}=\frac{A{\epsilon }_{0}}{d}=\frac{wl{\epsilon }_{0}}{d}...(1)$$C0​=dAϵ0​​=dwlϵ0​​...(1)

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

$${U}_{0}=\frac{1}{2}{C}_{0}{V}^{2}=\frac{1}{2}(\frac{wl{\epsilon }_{0}}{d}){V}^{2}...(2)$$U0​=21​C0​V2=21​(dwlϵ0​​)V2...(2)

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

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

$$C={C}_{1}+{C}_{2}...(3)$$C=C1​+C2​...(3)

Where,

$${C}_{1}=\frac{w(l-y){\epsilon }_{0}}{d}...(4)$$C1​=dw(l−y)ϵ0​​...(4)

$${C}_{1}=\frac{wyK{\epsilon }_{0}}{d}...(5)$$C1​=dwyKϵ0​​...(5)

Now energy stored in this combination of capacitor is given by

$$U=\frac{1}{2}({C}_{1}+{C}_{2}){V}^{2}...(6)$$U=21​(C1​+C2​)V2...(6)

Now according to the given condition,

$$2{U}_{0}=U$$2U0​=U

$$\Rightarrow 2(\frac{1}{2}{C}_{0}{V}^{2})=\frac{1}{2}({C}_{1}+{C}_{2}){V}^{2}$$⇒2(21​C0​V2)=21​(C1​+C2​)V2

$$\Rightarrow 2{C}_{0}={C}_{1}+{C}_{2}$$⇒2C0​=C1​+C2​

$$\Rightarrow 2(\frac{{\epsilon }_{0}wl}{d})=\frac{{\epsilon }_{0}Kwy}{d}+\frac{{\epsilon }_{0}w(l-y)}{d}$$⇒2(dϵ0​wl​)=dϵ0​Kwy​+dϵ0​w(l−y)​

$$\Rightarrow 2l=Ky+l-y$$⇒2l=Ky+l−y

$$\Rightarrow l=4y-y=3y$$⇒l=4y−y=3y

$$\Rightarrow y=\frac{l}{3}$$⇒y=3l​