Question

In the figure shown, a parallel plate capacitor has a dielectric of width $d/2$ and dielectric constant $K=2.$ The other dimensions of the dielectric are same as that of the plates. The plates $P_1$ and $P_2$ of the capacitor have area 'A' each. The energy of the capacitor is :

• A. $\frac{{ϵ}_{0}A{V}^2}{3d}$
• B. $\frac{2{ϵ}_{0}A{V}^2}{d}$
• C. $\frac{3}{2}\frac{{ϵ}_{0}A{V}^2}{d}$
• D. $\frac{2{ϵ}_{0}A{V}^2}{3d}$

Answer 1

The correct option is D $\frac{2{ϵ}_{0}A{V}^2}{3d}$

Using

$Energy=\frac{1}{2}\times C_{equivalent}\times V^2$

We need to find the equivalent capacitance of the capacitor.

Using law for series combination of capacitors,

$\frac{1}{C_{net}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

where, $C_1=\frac{A{ϵ}_0}{x}$ ( x is the width between capacitor walls and dielectric on one side )

$C_2=\frac{A{ϵ}_0\kappa }{\left(\frac{d}{2}\right)}=\frac{4A{ϵ}_0}{d}$

$C_3=\frac{A{ϵ}_0}{(\frac{d}{2}-x)}$

therefore,

$\frac{1}{C_{net}}=\frac{x}{\left(A{ϵ}_0\right)}+\frac{(\frac{d}{2}-x)}{\left(A{ϵ}_0\right)}+\frac{d}{\left(4A{ϵ}_0\right)}=\frac{3d}{\left(4A{ϵ}_0\right)}\Rightarrow C_{net}=\frac{4A{ϵ}_0}{3d}$

Therefore energy $=\frac{2A{ϵ}_0{V}^2}{3d}$