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Electric field inside a solid sphere

  1. Sep 5, 2012 #1
    1. The problem statement, all variables and given/known data

    We have a uniformly charged solid sphere whose radius is R and whose total charge is q. I'm trying to find the electric field inside a (r<R).

    The correct answer must be:

    [itex]E=\frac{1}{4 \pi \epsilon_0} \frac{q}{R^3} r \hat{r}[/itex]

    How did they get that answer?

    3. The attempt at a solution

    Since the sphere is not a shell, the E is not 0. And it should be found by considering a concentric Gaussian sphere with radius smaller than R. So by using Gauss's law for electric fields in the integral form we obtain:

    [itex]\oint E . da = E \oint da = E (4 \pi r^2) = \frac{q}{\epsilon_0}[/itex]

    [itex]\therefore E = \frac{q}{4 \pi \epsilon_0 r^2}[/itex]

    For E outside the sphere we use r>R, and inside the sphere we use r<R. But why is my answer so different from the correct result?

    Any help is greatly appreciated.
     
  2. jcsd
  3. Sep 5, 2012 #2

    TSny

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    In Gauss' law as stated above, what is the interpretation of q?
     
  4. Sep 5, 2012 #3

    AGNuke

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    In the first answer, q is the charge spread out throughout the solid sphere, while in your attempt, q must be the charge enclosed by the gaussian surface.
     
  5. Sep 6, 2012 #4
    Thank you very much for the hint! The q in Gauss's law is the amount of charge inside the Gaussian surface. The q in the question is the total amount of charge spread across a volume of 4/3 π R3.

    I found the charge per unit volume by q dividing by the volume to get:

    [itex]q_{per \ unit \ volume}=\frac{q}{\frac{4}{3} \pi R^3}[/itex]

    And to find the total charge

    Then I multiplied this by 4/3πr2 to find the total charge in the Gaussian surface of that volume. I got

    [itex]q=\frac{qr^3}{R^3}[/itex]

    Substituting this back into Gauss's equation I got the right answer. I hope this was the correct thinking and approach.
     
    Last edited: Sep 6, 2012
  6. Sep 6, 2012 #5

    TSny

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    Good work :smile:
     
  7. Sep 6, 2012 #6
    Thanks a lot for your help! :)
     
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