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Charge enclosed in a sphere

  1. Oct 4, 2011 #1
    1. The problem statement, all variables and given/known data

    A solid insulating sphere of radius R is charged so that the charge density is proportional to the distance to the center, p(r)=kr, where k is a constant.

    Find Q(r), the charge enclosed by a concentric spherical surface of radius r<= R. What is the total charge of the sphere?

    2. Relevant equations



    3. The attempt at a solution

    I thought that Q=[Int from 0 to r] kr dV, and in spherical coordinates for radius, dV = r^2 dr.

    So I have Q=[int from 0 to r] kr^3 dr = k [int, 0, r] r^3 dr = kr^4 / 4
     
  2. jcsd
  3. Oct 4, 2011 #2

    ehild

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    The volume element is a shell of radius r and thickness dr, which volume is dV=4pi r2dr

    ehild
     
  4. Oct 4, 2011 #3
    then when do I use r^2 sin(theta) dr d(theta) d(phi)?
     
  5. Oct 4, 2011 #4

    ehild

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    You are right now, the volume element is dV=r^2sin(theta)dr d(theta) d(phi) in spherical coordinates, but not r^2dr as you wrote in your first post. Integrate the charge density for the sphere of radius r.

    ehild
     
  6. Oct 4, 2011 #5
    So lets say a question gives you the charge density p=kr where k is a constant, and it wants you to find the electric field inside a solid sphere of radius R.

    Id say that [int] E ds = q/e0

    and i can pull out E, -> E [int] ds = q/e0

    and using what you said, E [int] ds = [int] kr r^2sin(theta)dr d(theta) d(phi) / e0

    I would be left with 4piR^2 E = (4piR^3)/3e0 -> E=R / 3e0 ?
     
  7. Oct 4, 2011 #6

    ehild

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    You have to integrate kr^3 with respect to r.

    ehild
     
  8. Oct 5, 2011 #7
    So would the total enclosed charge be [int 0,R] kr * 4pi r^2 dr = k pi R^4 ?

    and then I would get (k pi R^4) / (4 pi R^2 e0) = E = kR/4e0
     
    Last edited: Oct 5, 2011
  9. Oct 5, 2011 #8

    ehild

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    Yes.

    ehild
     
  10. Oct 5, 2011 #9
    So if i use [int] E ds = k pi R^4 / e0, would I get E = kR^4 / 4r^2 e0 ?
     
    Last edited: Oct 5, 2011
  11. Oct 5, 2011 #10

    ehild

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    It depends what you mean on R and r and [int] E ds.

    ehild
     
  12. Oct 5, 2011 #11
    R is the radius of the sphere, r is your variable radius

    it'd be the surface integral of E ds = k pi R^4 / e0

    and the surface would be 4 pi r^2, divide on both sides leaves you with E = kR^4 / 4r^2 e0
     
  13. Oct 5, 2011 #12

    ehild

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    Gauss' Law states that the surface integral of the electric field is equal to the enclosed charge divided by ε0. If you want to find the electric field at distance r from the centre of the charged sphere of radius R, you calculate the surface integral of E for a sphere of radius r. If r>R it encloses the whole charge, and your formula is rigth. If r≤R the enclosed charge is the charge inside the sphere of radius r. In this case, E=kr^4/(4r^2 ε0)=kr^2(4 ε0).

    ehild
     
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