Question

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.

Short answer

The energy density in the electric field near the surface of sphere is $0.11\mathrm{J}/\mathrm{m}$.

Step-by-step solution

The given data

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 $V=0$.

Understanding the concept of the energy density

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)

Calculation of the energy density

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 $0.11\mathrm{J}/{\mathrm{m}}^3$.