A spherical volume of radius R is filled with charge of uniform density rho. We

Question

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.

Explanation / Answer

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)

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