Non-conductve sphere with cavity -- find Electric field

  • #1
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I have a non conducting sphere with a charge ρ=A/r per uni vollume A is constant. suppose there is a cavity in the centre and within a particle of charge q. i want to find the E inside the sphere in respect with r.

Homework Equations




The Attempt at a Solution


for radius equal of the cavity i get ##E=kq/r^2## for r bigger than the radius of the sphere ##E=k(Q+q)/r^2## now only one case left
if r is bigger than the radius of the cavity and smaller than the radius of the sphere.
##E4πr^2=\frac{q+ρV}{ε_0} ##
where ρ is the density per unit vollume and V is the vollume
is my enclosed Q right? and my V is 4π(r^3-a^3) where a^3 is the small radius.
 
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Answers and Replies

  • #2
gneill
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is my enclosed Q right?
Since the charge density varies with radius you'll need to take that into account. Hint: an integration is required.
 
  • #3
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Since the charge density varies with radius you'll need to take that into account. Hint: an integration is required.
???
 
  • #4
gneill
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Think of the sphere being comprised of many thin spherical shells, each having a slightly different charge density according to their radii. No two shells have the same charge density due to your given expression for the charge density, ρ=A/r. The enclosed charge will be the sum of the charges in those shells "below" the radius of interest.
 
  • #5
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Think of the sphere being comprised of many thin spherical shells, each having a slightly different charge density according to their radii. No two shells have the same charge density due to your given expression for the charge density, ρ=A/r. The enclosed charge will be the sum of the charges in those shells "below" the radius of interest.
##E4πr^2=\frac{q+ \int ρVdr}{ε_0} ## is the right one?
 
  • #6
gneill
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##E4πr^2=\frac{q+ \int ρVdr}{ε_0} ## is the right one?
That's the idea, yes.
 
  • #7
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That's the idea, yes.
so i Keep the vollume inside the integral!
 
  • #8
gneill
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You want to integrate over the volume of interest, summing up the charges contained in each differential volume element. So really that V should be a dV(r). I assumed (my bad) that that was what you were implying with that V.
 

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