Question

The plates of a parallel plate capacitor have an area of $90c{m}^2$ each and are separated by $2mm$. The capacitor is charged by connecting it to a $400V$ supply. Then the energy density of the energy stored $\left(inJ{m}^{-3}\right)$ in the capacitor is (Take ${ϵ}_0=8.8\times {10}^{-12}F{m}^{-1}$)

• A. $0.113$
• B. $0.117$
• C. $0.152$
• D. $\text{None of these}$

Answer 1

The correct option is D $\text{None of these}$

Area of the plates = A = $90c{m}^2$
Separation between the plates = d = $2mm$
We know that electric field $E$ = $\frac{V}{d}$ and energy density stored in the capacitor is given by $U$= $\frac{1}{2}{ϵ}_0{E}^2$
$U$= $\frac{1}{2}{ϵ}_0\left(\frac{V^2}{d^2}\right)$
putting the values of ${ϵ}_0$,$V$ and $d$ we get,
$U$=$\frac{1}{2}\times 8.8\times {10}^{-12}\times \left(\frac{{400}^2}{{0.002}^2}\right)J{m}^{-3}$
$⟹$ $U=0.176J{m}^{-3}$