# gaussian solid sphere Q

by lonewolf219
Tags: gaussian, solid, sphere
 P: 163 Question: A solid sphere of radius R has a non-uniform charge distribution of p=Ar^2, where A is constant. Find total charge Q within the volume of the sphere. p=roe p=Q/dV EdA=qenclosed/Enaught Can you use Gauss' Law for this problem when sphere is solid? If so, how? Since p is non-uniform, we must integrate dq, correct? The answer to this question in the book is 4/5pieAr^5. But how to get the answer? I think dq=Ar^2dV. But to have dV=4pier^2 is incorrect since the charge is not on the surface of the sphere (it is not a conductor), am I wrong? Some help would be appreciated!!!! Thanks!
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 Quote by lonewolf219 Question: A solid sphere of radius R has a non-uniform charge distribution of p=Ar^2, where A is constant. Find total charge Q within the volume of the sphere. p=roe p=Q/dV EdA=qenclosed/Enaught Can you use Gauss' Law for this problem when sphere is solid? If so, how? Since p is non-uniform, we must integrate dq, correct? The answer to this question in the book is 4/5pieAr^5. But how to get the answer? I think dq=Ar^2dV. But to have dV=4pier^2 is incorrect since the charge is not on the surface of the sphere (it is not a conductor), am I wrong? Some help would be appreciated!!!! Thanks!
Gauss's Law gives the Electric field flux through a surface, which in very symmetric cases can give the Electric field itself. It won't help with this problem.

As you state, the volume element (differential), dV, is given by dV = (4πr2)dr.

The amount of charge, dQ, in the volume element, dV, is dQ = ρ(r)∙dV = ρ(r)∙4πr2∙dr , where ρ(r) is the volume charge density as a function of r and is given by ρ(r) = A∙r2 for some constant A. Thus dQ is given by:
dQ = 4∙A∙π∙r4∙dr
Integrate that over the sphere to find the total charge.
 P: 163 Thanks Sammy. So Guauss' Law is not used to find a non-uniform charge? If the sphere was a uniform charge, could we rearrange the variables of Gauss' Law to solve for q enclosed?
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## gaussian solid sphere Q

 Quote by lonewolf219 Thanks Sammy. So Guauss' Law is not used to find a non-uniform charge? If the sphere was a uniform charge, could we rearrange the variables of Gauss' Law to solve for q enclosed?
Indeed, for this case, Gauss;s Law Could be used to find the magnitude of the electric field if that's what was asked for.

All they're asking you to do is find the total charge of the sphere.

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