A charged isolated metal sphere of diameter 10 cmhas a potential of 8000 Vrelative to V = 0at infinity. Calculate the energy density in the electric field near the surface of the sphere.

a) Diameter of the sphere,$\mathrm{d}=10\mathrm{cm}\mathrm{or}{10}^{-1}\mathrm{m}$

b) Potential of sphere,$V=8000V$

c) Potential at infinity is $V=0$.

The quantity of energy that is contained in a system or area of space per unit volume is known as the energy density. We use the electric field formula and substitute this value in the energy density formula; we can calculate the energy density of the sphere on the surface of the sphere.

Formulae:

The electric field of a capacitor,

$E=\frac{2v}{d}$ ...(i)

The energy density between the capacitors in the field,$u=\frac{1}{2}{\epsilon }_0{E}^2$ ...(ii)

Substituting the given values and equation (i) in equation (ii), we can get the energy density in the electric field near the surface of the sphere as follows:

(${\mathrm{\epsilon }}_0=8.85\times {10}^{-12}\frac{C^2}{N.m^2}$is permittivity of free space)

role="math" localid="1661340099858" $$\begin{gathered} u=\frac{1}{2}{\mathrm{\epsilon }}_0{\left(\frac{2\mathrm{v}}{\mathrm{d}}\right)}^2 \\ =\frac{1}{2}\left(8.85\times {10}^{-12}\frac{{\mathrm{C}}^2}{\mathrm{N}.{\mathrm{m}}^2}\right){\left(\frac{2\times 8000\mathrm{v}}{0.1\mathrm{m}}\right)}^2 \\ =0.11\mathrm{J}/{\mathrm{m}}^3 \end{gathered}$$

Hence, the value of the energy density is $0.11\mathrm{J}/{\mathrm{m}}^3$.