Advanced Gauss's Law Question

  • #1
blackhawk97
2
0
A solid sphere of radius [tex]R[/tex] has a non-uniform volume charge density [tex]\rho(r)[/tex] and a constant surface charge density [tex]\sigma[/tex]. If the field inside the sphere is uniform and radially atuned, and the field a distance [tex]2R[/tex] away from the center is zero, find [tex]\rho[/tex] and [tex]\sigma[/tex] in terms of [tex]R[/tex], [tex]r[/tex] (distance from the center of the sphere), and [tex]Q_\text{volume}[/tex] (the charge associated with [tex]\rho[/tex], but not with [tex]\sigma[/tex]).

Homework Equations


Gauss's Law

The Attempt at a Solution


I'm not sure how to proceed, but I think the solution should begin by find the total charge on the sphere (ie., adding the integral of the charge calculable from the surface charge density with the integral of the charge calculable from the volume charge density). Am I on the right track?
 

Answers and Replies

  • #2
Matterwave
Science Advisor
Gold Member
3,967
327
So, inside the sphere, what does Gauss's law say? Outside the sphere? Try drawing Gaussian surfaces which are spheres inside and outside the sphere of charge.
 
  • #3
blackhawk97
2
0
I know how to do the first part, i think... (finding [tex]\sigma[/tex]). Basically you just use Gauss's Law

[tex]\oint{E \cdot dA}=\frac{Q_\text{enclosed}}{\epsilon_0}[/tex]

... except you set the [tex]r[/tex] in that equation equal to [tex]2R[/tex], so you can ultimately set the expression for the [tex]E[/tex]-field equal to zero and... yeah.

But the part about [tex]\rho[/tex] still has me stumped. Can anyone offer a bit more help?

Also, correction to the problem: the field inside the sphere is not radially atuned, it is directed radially outwards.
 
  • #4
betel
318
0
So can you express [tex]\sigma[/tex] in terms of [tex]Q_{total}[/tex]?

For the [tex]\rho[/tex] part you have to take Gauss surfaces inside the sphere as Matterwave said.
 

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