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

Answer:

$u = 0.057 J/m^3$

Explanation:

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

$u = \frac{1}{2}\epsilon_0 E^2$

also for sphere surface we know that

$E = \frac{V}{R}$

R = radius of sphere

V = potential of the surface

now we have

$u = \frac{1}{2}\epsilon_0 (\frac{V^2}{R^2})$

now from the above formula we have

$u = \frac{1}{2}(8.85 \times 10^{-12})(\frac{8500^2}{0.075^2})$

$u = 0.057 J/m^3$