Question 103 of 180

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?

A$\dfrac{2l}{3}$B$\dfrac{l}{3}$C$\dfrac{l}{4}$D$\dfrac{l}{2}$

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Answer: (B) $\dfrac{l}{3}$

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
${C}_{0}=\dfrac{A{\epsilon }_{0}}{d}=\dfrac{wl{\epsilon }_{0}}{d}...(1)$
Energy stored in the capacitor before the insertion of dielectrics is given by,
${U}_{0}=\dfrac{1}{2}{C}_{0}{V}^{2}=\dfrac{1}{2}\left(\dfrac{wl{\epsilon }_{0}}{d}\right){V}^{2}...(2)$
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,
${C}_{1}=\dfrac{w(l-y){\epsilon }_{0}}{d}...(4)$
${C}_{1}=\dfrac{wyK{\epsilon }_{0}}{d}...(5)$
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,
$2{U}_{0}=U$
$\Rightarrow 2\left(\dfrac{1}{2}{C}_{0}{V}^{2}\right)=\dfrac{1}{2}\left({C}_{1}+{C}_{2}\right){V}^{2}$
$\Rightarrow 2{C}_{0}={C}_{1}+{C}_{2}$
$\Rightarrow 2\left(\dfrac{{ϵ}_{0}wl}{d}\right)=\dfrac{{ϵ}_{0}Kwy}{d}+\dfrac{{ϵ}_{0}w(l-y)}{d}$
$\Rightarrow 2l=Ky+l-y$
$\Rightarrow l=4y-y=3y$
$\Rightarrow y=\dfrac{l}{3}$