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Electric field inside a sphere

  1. Oct 6, 2015 #1
    1. The problem statement, all variables and given/known data
    A sphere with radius R has a spherically symmetric charge density that varies as 1/r. What is the electric field outside and inside the sphere?

    2. Relevant equations
    E=kQ/r^2, ε=permitivity of free space, Q=total charge, ρ=charge density, dτ=infinitesimal volume

    3. The attempt at a solution
    For the case (outside), due to the concept that we can treat the sphere as a point charge, E = (1/4*pi*e)(Q/r^2).

    For the case (inside), by using a Gaussian surface, we have E (4πr^2) = Q/ε. By evaluating Q = ∫ ρ dτ. I got E = (R^2)/(2εr^2).

    Is this correct? Or am I missing something?
     
    Last edited: Oct 6, 2015
  2. jcsd
  3. Oct 6, 2015 #2

    BvU

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    I don't see them matching up at r = R

    Your inside case solution description isn't detailed enough for me to tell you where you derail.
    And your problem description has holes in it. Is Q the total charge on the sphere? Is it a given ? What does that mean for the enclosed charge when r < R ?
     
  4. Oct 6, 2015 #3
    Sorry, there was a typo, it should be E(4πr^2) = Q/ε and Q is the total charge. What I've done for the inside is, given ∫E⋅da = Q/ε, we have E(4πr^2) = Q/ε. But Q=∫ρ dτ and ρ=1/r. So Q=∫ dτ/r. Also, dτ= r^2 sinθ dr dθ d∅ is the change of the volume in spherical coordinates. So, Q=∫ r sinθ dr dθ d∅. 0≤r≤R, 0≤θ≤π, 0≤∅≤ 2π. By integration, I got Q = (4πR^2)/2. Thus, I got E = (R^2)/ 2εr^2.
     
  5. Oct 6, 2015 #4

    BvU

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    Aha! Check your integration bounds for the contained charge when r < R ! Contained charge as a function of r is not a constant ! You confuse yourself by using the symbol Q both for the total charge and also for contained charge as a function of r...
     
  6. Oct 6, 2015 #5
    Oh! So to rephrase, E(4 π r'^2) = Qenc/ε with Qenc as the enclosed charge in the gaussian surface. Qenc= ∫ r sinθ dr dθ d∅. 0≤r≤r', 0≤θ≤π, 0≤∅≤ 2π. Qenc = (4 π r'^2)/2. But this will cancel the r'. E(4 π r'^2) = (4 π r'^2)/2ε.
     
  7. Oct 6, 2015 #6

    BvU

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    So nice it doesn't diverge for r to 0, and that now it matches at r = R !
     
  8. Oct 6, 2015 #7
    I've done everything over and over, If I integrate, the two r' really cancel, is there something wrong here? I'm confused.
     
  9. Oct 6, 2015 #8

    BvU

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    I think it's the right answer.
     
  10. Oct 6, 2015 #9
    What do you mean right? But that will mean the electric field is not varying as we go from r=0 to r=r'. How is that?
     
  11. Oct 6, 2015 #10

    BvU

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    It means area and charge contained grow with the same power of r . In another example (number 4 here) charge grows with r3 and area with r2, so E grows with r1
     
  12. Oct 6, 2015 #11
    Thank you so much!!! That is kinda subtle.
     
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