Question

A sphere of radius $1cm$ has potential of $8000V$, then the energy density near its surface is

• A. $64\times {10}^{5}J/m^3$
• B. $2.83J/m^3$
• C. $8\times {10}^{3}J/m^3$
• D. $32J/m^3$

Answer 1

The correct option is B $2.83J/m^3$

Given radius of sphere, $r=1cm$

potential $=8000V$

We have to find the energy density near its surface.

So, here we use formula for energy density $=\frac{1}{2}{\epsilon }_0{E}^2$

${\epsilon }_0=$ permeability in air $=8.85\times {10}^{-12}$

$E=$ electric field $=\frac{V}{r}$

So, energy density $=\frac{1}{2}\times 8.85\times {10}^{-12}\times {\left(\frac{8000}{{10}^{-2}}\right)}^2$

$=283.5\times {10}^{-2}$

$=2.83J/m^3$