A spherical volume of radius R is filled with charge of unif

A spherical volume of radius R is filled with charge of uniform density rho. We want to know the potential energy U of this sphere of charge, that is the work done in assembling it. In the example in Section 1.15, we calculated U by integrating the energy density of the electric field; the result was U = (3/5)Q^2/4 pi epsilon_0 R. Derive U here by building up the sphere layer by layer, making use of the fact that the field outside a spherical distribution of charge is the same as if all the charge were at the center.

Solution

Consider the sphere is being made layer by layer,
at some instant, the radius of sphere is r and a layer of thickness dr is deposited on the sphere

charge of dr = J * A * dr (J = volume charge density, A = surface area of the layer)
charge of dr, dq = J * (pi * r^2) * dr
charge of the sphere, q = J * volume = j * (4/3) * pi * r^3

work done in bringing dq charge from infinity to the sphere\'s surface:

dw = k*q*dq/r = k*J *(4/3)*pi*r^3 *J * (pi * r^2 ) * dr /r = (4/3)k * j^2 * pi^2 * r^4 dr
integrating from 0 to R

U = (4/3)k*j^2*pi^2*R^5)/5
put j = Q/(4/3 pi R^3)
U = (3/5) Q^2 / (4 pi epslon r)