# Gaussian solid sphere Q

by lonewolf219
Tags: gaussian, solid, sphere
 P: 176 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!
Emeritus
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P: 7,819
 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: 176 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?
Emeritus