Question

A parallel plate capacitor having a plate separation on $2\, mm$ is charged by connecting it to a $300\, V$ supply. The energy density is:

• A. $0.01\, J/m^3$
• B. $0.1\, J/m^3$
• C. $1.0 \,J/m^3$
• D. $10 \,J/m^3$

Answer 1

Answer: (B)

Energy density in the energy per unit volume.
The energy per unit volume or the energy density is given, by,
$U=\frac{1}{2} \varepsilon_{0} E^{2} \ldots(1)$
Where $\varepsilon_{0}$ is permittivity of free space and $E$ is electric field.
Also $E=\frac{V}{d} \ldots(2)$
$=\frac{\text { Potential difference }}{\text { Distance between the plates }}$
From Eqs. (1) and (2), we have
$U=\frac{1}{2} \varepsilon_{0}\left(\frac{V}{d}\right)^{2}$
Given, $V=300$ Volt,
$d=2\, m m $
$=2 \times 10^{-3}
,m$,
$\varepsilon_{0}=8.85 \times 10^{-12} C^{2} / N m^{2}$
$\therefore U=\frac{1}{2} \times 8.85 \times 10^{-12} \times\left(\frac{300}{2 \times 10^{-3}}\right)^{2}$
$U=0.1 J / m^{3}$
Note: We can also say that if electric field $\vec{E}$ exists in some space, then the space is a store of energy whose amount per unit volume is equal to energy density.