
#1
Feb1512, 11:52 AM

P: 165

Question:
A solid sphere of radius R has a nonuniform 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 nonuniform, 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! 



#2
Feb1512, 12:17 PM

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PF Gold
P: 7,396

As you state, the volume element (differential), dV, is given by dV = (4πr^{2})dr. The amount of charge, dQ, in the volume element, dV, is dQ = ρ(r)∙dV = ρ(r)∙4πr^{2}∙dr , where ρ(r) is the volume charge density as a function of r and is given by ρ(r) = A∙r^{2} for some constant A. Thus dQ is given by: dQ = 4∙A∙π∙r^{4}∙drIntegrate that over the sphere to find the total charge. 



#3
Feb1512, 02:28 PM

P: 165

Thanks Sammy. So Guauss' Law is not used to find a nonuniform charge? If the sphere was a uniform charge, could we rearrange the variables of Gauss' Law to solve for q enclosed?




#4
Feb1512, 05:50 PM

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P: 7,396

gaussian solid sphere QAll they're asking you to do is find the total charge of the sphere. 


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