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Question

Physics Question on electrostatic potential and capacitance

A parallel plate capacitor having a plate separation on 2mm2\, mm is charged by connecting it to a 300V300\, V supply. The energy density is:

A

0.01J/m30.01\, J/m^3

B

0.1J/m30.1\, J/m^3

C

1.0J/m31.0 \,J/m^3

D

10J/m310 \,J/m^3

Answer

0.1J/m30.1\, J/m^3

Explanation

Solution

Energy density in the energy per unit volume.
The energy per unit volume or the energy density is given, by,
U=12ε0E2(1)U=\frac{1}{2} \varepsilon_{0} E^{2} \ldots(1)
Where ε0\varepsilon_{0} is permittivity of free space and EE is electric field.
Also E=Vd(2)E=\frac{V}{d} \ldots(2)
= Potential difference  Distance between the plates =\frac{\text { Potential difference }}{\text { Distance between the plates }}
From Eqs. (1) and (2), we have
U=12ε0(Vd)2U=\frac{1}{2} \varepsilon_{0}\left(\frac{V}{d}\right)^{2}
Given, V=300V=300 Volt,
d=2mmd=2\, m m
=2×103,m=2 \times 10^{-3} ,m,
ε0=8.85×1012C2/Nm2\varepsilon_{0}=8.85 \times 10^{-12} C^{2} / N m^{2}
U=12×8.85×1012×(3002×103)2\therefore U=\frac{1}{2} \times 8.85 \times 10^{-12} \times\left(\frac{300}{2 \times 10^{-3}}\right)^{2}
U=0.1J/m3U=0.1 J / m^{3}
Note: We can also say that if electric field E\vec{E} exists in some space, then the space is a store of energy whose amount per unit volume is equal to energy density.