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Work, energy stored in solid sphere

  1. Oct 16, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the energy stored in a uniformly charged sphere of charge q, radius R

    2. Relevant equations



    3. The attempt at a solution
    [tex]Ein=\frac{qr}{4\pi\epsilon o R^3}, Eout=\frac{q}{4\pi\epsilon o r^2}... W=\int_{0}^ {R}\int_{0}^{2\pi}\int_{0}^{\pi}[\frac{qr}{4\pi\epsilon oR^3}] ^2sin\theta d\theta d\phi r^2\ dr+ \int_{R}^ {\infty }\int_{0}^{2\pi}\int_{0}^{\pi}[\frac{q}{4\pi\epsilon or^2}] ^2sin\theta d\theta d\phi r^2\ dr= 2\pi\epsilon o(\frac{q}{4\pi\epsilon o})^2(\frac{1}{5R}+\frac{1}{R})=\frac{q^2}{4\pi\epsilon o R}\frac{3}{5}[/tex]

    by the way the work in this case is also like the effort needed to bring the whole solid sphere in from infinity by point charges or also the stored energy?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 17, 2013 #2

    Simon Bridge

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    Is this a uniforms sphere of charge (i.e. as you may find in an insulator), or a spherical shell of charge (i.e. the charges are being placed on a conductor)? It affects the integral.

    But yep - the electrostatic energy stored in a system of charges is the work needed to assemble them from infinity.

    Nice LaTeX ... you can make a newline with a \\ to avoid running off the end of the page;
    you can make subscripts with _{} like this: ##\epsilon_0## and ##E_{out}##.
    trig functions are written \sin \cos etc.
     
  4. Oct 17, 2013 #3
    thanks! this is a solid sphere of charge. I'm curious how you develop the mathematics for this, is it based on a physical intuition or a mathematical result of the electric field equation
     
    Last edited: Oct 17, 2013
  5. Oct 17, 2013 #4

    Simon Bridge

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    OK... you seem to be using:
    $$U=\int_V E^2d\tau + \int_S VEda$$
    ... it's a good idea to explain your process.

    You ended up with: $$U=\frac{1}{2k}k^2q^2\left( \frac{1}{5R}+\frac{1}{R}\right)$$ ... where ##k=1/4\pi\epsilon_0##

    Your next step is to simplify the expression.
    Did you have any other questions?
     
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