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Electrostatics, Energy of a uniformly charged sphere

  1. Feb 13, 2007 #1
    From Griffiths, Third edition Intro Electrodynamics

    I've been working on this problem, and I get 2 different answers. I get the same answer for parts b and c, but not for a. I know I'm probably just doing something silly along the way, but I can't find my mistake...

    1. The problem statement, all variables and given/known data
    Find the energy stored in a uniformly charged solid sphere of radius R and charge q. Do it three different ways:
    (a) Use Eq. 2.43
    (b) Use Eq. 2.45
    (c) Use Eq. 2.44

    2. Relevant equations
    Eq. 2.43 [tex]W=\frac{1}{2}\int \rho V d\tau[/tex]
    Eq. 2.45 [tex]W=\frac{\epsilon_{0}}{2}\int_{all space} E^2 d\tau[/tex]
    Eq. 2.44 [tex]W=\frac{\epsilon_{0}}{2}\left(\int_{V} E^2 d\tau + \oint_{S} VE\cdot da\right)[/tex]

    3. The attempt at a solution

    For part (a) we have:
    [tex] E(r)=\left\{\begin{array}{cc}\frac{kq}{r^2}, &\mbox{ if } r > R\\ \frac{kq}{R^3} r, &\mbox{ if } r < R \end{array}\right
    \\ \Rightarrow V(r) = \left\{\begin{array}{cc}\frac{kq}{r}, & r > R\\ \frac{3kq}{2R} - \frac{kq}{R^3} r^2, & r < R \end{array}\right[/tex]

    Since it's uniformly charged, we know that:
    [tex] \rho = \frac{3q}{4\pi R^3} \\ [/tex]

    Then we can evaluate Eq. 2.43 to get:
    [tex] W=2\pi\rho\int_{0}^R V(r) r^2 dr \\
    = \frac{9}{20} \frac{kq^2}{R} [/tex]

    For (b) and (c) however I get:

    [tex] W=\frac{3}{5} \frac{kq^2}{R} [/tex]

    So they are off by a factor of 3/4.

    If you need to see more work to help me find my mistake, let me know and I will post more details.
    Last edited: Feb 13, 2007
  2. jcsd
  3. Feb 14, 2007 #2


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    E is the NEGATIVE gradient of the potential. You have a sign error. Also as written V is not even continuous across r=R, but I assume that is just a typo.
  4. Feb 14, 2007 #3
    Thanks, there's no sign errors, but I forgot a factor of 1/2 in the second term of V for r<R
  5. Feb 14, 2007 #4


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    I guess you're right. The sign error was mine!
  6. Feb 14, 2007 #5
    Yeah, I worked it out with the correct expression for V, and it gives me the same answer now, thanks.
  7. Apr 27, 2009 #6
    NeoDevin, you wouldn't happen to be in professor schick's class would you?
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