Question

A charged isolated metal sphere of diameter 15 cm has a potential of 8500 V relative to V = 0 at infinity. Calculate the energy density in the electric field near the surface of the sphere.

Answer 1

`u = 0.057 J/m^3`

Step-by-step explanation:

Energy density near the surface of the sphere is given by the formula

`u = (1)/(2)\epsilon_0 E^2`

also for sphere surface we know that

`E = (V)/(R)`

R = radius of sphere

V = potential of the surface

now we have

`u = (1)/(2)\epsilon_0 ((V^2)/(R^2))`

now from the above formula we have

`u = (1)/(2)(8.85 * 10^(-12))((8500^2)/(0.075^2))`

`u = 0.057 J/m^3`