Question

A parallel plate capacitor has a 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. $2l/3$
• B. $l/2$
• C. $l/4$
• D. $l/3$

Answer 1

The correct option is D

$l/3$

The explanation for the correct option:

Step 1: Given data:

where length $‘l’$, width $‘w’$ and separation of plates is $‘d’$

thickness of dielectric slab is also $‘d’$ and its dielectric constant is $k=4$.

$\frac{U_{initial}}{U_{final}}=2$

where, $U$ is the energy stored in the capacitor

Step 2: Finding the length at which initial energy will be twice of the stored energy:

Using the equation for energy stored in a capacitor, $U=\frac{1}{2}C{V}^2$…………$\left(1\right)$

Also using capacitance in parallel plate capacitor is given by $C=\epsilon (lw/d)$

where, $C$ is the capacitance

$\epsilon$ is the permittivity of free space

$A$is area of the plate

Comparing above equations and our given condition,

$\begin{array}{rcl}\frac{U_{initial}}{U_{final}}& =& 2\\ & \Rightarrow & \frac{C_1+C_2}{C_{stored}}=2\\ & \Rightarrow & \frac{kx+\left(l-x\right)}{l}=2\\ & \Rightarrow & 4x+l-x=2l\\ & \Rightarrow & x=\frac{l}{3}\end{array}$

Hence, option D is correct.