Potential Energy of an Atom Due to Its Charge Distribution

[Hoofbeat]

Could someone help me with this:

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Q. The nucleus of an atom can be considered to be a charge of +Ze uniformly distributed throughout a sphere of radius a. Show that the potential energy of a nucleus due to its charge is (3Z^2.e^2)/(20.pi.epsilon-0.a). What would the potential energy be if the charge was spread uniformly over the surface of the nucleus.
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I used Gauss' law and spherical coordinates (we can ignore all components other than R due to symmetry) to give the electric field (radial component):

Er = +Ze/(4.pi.epsilon-0.a^2)

Energy Density = 1/2.Epsilon-0.E^2

But we want total energy, thus:

Energy = 1/2.Epsilon-0.E^2.Volume

Energy = (Z^2.e^2)/(24.pi.Epsilon-0.a)

However, this is clearly wrong. Could someone tell me what I'm doing wrong?! Thanks

 

[Timbuqtu]

The electric field Er you computed is the electric field at the surface of the nucleus. But you want to find the total potential energy of the nucleus, which is the integral:

\int_{whole space} 1/2 \epsilon_0 E^2 dV

So you need to know the E every in space (use Gauss' law).

 

[Hoofbeat]

Thanks, I've now managed to solve the problem :D