Question

A parallel capacitor has plate of length $l$, width $w$ and separation of plates is $d$. It is connected to a battery of emf $\text{V}$. 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 in the capacitor be two times the initial energy stored?

• A. $2l/3$
• B. $l/3$
• C. $l/4$
• D. $l/2$

Answer 1

The correct option is B $l/3$

Capacitance before inserting the slab,

$C_i=\frac{{\epsilon }_{0}A}{d}=\frac{{\epsilon }_{0}lw}{d}(\therefore A=lw)$
Capacitance after inserting the dielectric slab,

$C_f=\frac{K{\epsilon }_0{A}_1}{d}+\frac{{\epsilon }_0{A}_2}{d}$
$C_f=\frac{K{\epsilon }_{0}wx}{d}+\frac{{\epsilon }_{0}w(l-x)}{d}$

According to question,
$2\times$Initial energy = Final energy
$\Rightarrow \left(\frac{1}{2}{C}_i{V}^2\right)2=\frac{1}{2}{C}_f{V}^2$
$\Rightarrow 2C_i=C_f$
$\Rightarrow 2\left(\frac{{\epsilon }_{0}wl}{d}\right)=\frac{{\epsilon }_{0}Kwx}{d}+\frac{{\epsilon }_0(l-x)}{d}$
$\Rightarrow 2l=Kx+(l-x)\Rightarrow 2l=4x+(l-x)\Rightarrow x=\frac{l}{3}$