What is the electro-static energy of a sphere with uniform charge density?

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The discussion revolves around calculating the electro-static energy of a uniformly charged sphere, initially stated as (3/5)kQ/R. The user seeks clarification on integrating the potential over the volume, questioning whether to integrate from r=0 to r=infinity, given that charge density is zero outside the sphere. It is confirmed that integration should only occur from r=0 to r=R, as the potential outside the sphere contributes nothing. The user also realizes a mistake in the energy formula, correcting it to (3/5)kQ^2/R to ensure proper unit consistency. The conversation concludes with acknowledgment of the importance of careful calculations in physics.
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Homework Statement


hello
I have a question,it's not homework related..I know the electro-static energy of a sphere is:(3/5)kQ/R

I tried to calculate it today using the expression:
U = 1/2 * integral (phi * rho dV)
where: phi is the potential, rho is charge density(uniform), dV is the volume element.

it's not hard to calculate the potential everywhere in space,but my problem is when I integrate over volume elements (4*pi*r^2 dr) should I integrate over:
r=0 to r=infinity? it doesn't seem reasonable because rho=0 outside the sphere.

anyway I tried to do that,i.e integrating over r=0 to r=R using the potential inside the sphere,but I didnt get the correct answer.

I would love to get some help with this.

thanks.
 
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You're right: the integral should go over all of space (r = 0 to r = infinity), but since rho is 0 outside of the sphere, all of the integration that goes on outside the sphere amounts to 0. That's why it's OK to only integrate inside the sphere (r = 0 to r = R). So you should be set. If it's not coming out, check your math and your equations again. Is the sphere uniformly charged? Is it a hollow shell with the charge just on the outside? Make sure you know which situation you're looking at.

P.S. I'm guessing that you're equation for the energy in the sphere ( (3/5)kQ/R ) is off. There should be a Q^2 term in there to make the units work...right now you have the units of an electric potential...which is potential energy per unit charge...they're closely related, but there's a distinct difference.
 
Last edited:
of course I meant (3/5)kQ^2/R :)

and you were right,I had a stupid math mistake(I really hate those)

thanks alot.
 
Parallel said:
and you were right,I had a stupid math mistake(I really hate those)

Those will get you every time! Gotta love it :-p Glad it worked out though!
 
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