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Energy Density of a Sphere

  1. Mar 12, 2006 #1
    Let's say that the electric field near the surface of a sphere or radius R is E. I need to show that the energy density D is


    I started by noting that [itex]U=\frac{1}{2}Q\Delta V=\frac{kQ^{2}}{2R}[/itex]. The volume of the sphere is [itex]\frac{4}{3}\pi R^{3}[/itex]. This is how I found the energy density:

    [tex]D=\frac{U}{\textrm{Vol}}=\frac{\left(\frac{kQ^{2}}{2R}\right)}{\left(\frac{4}{3}\pi R^{3}\right)}=\frac{3\epsilon_{0}E^{2}}{2}.[/tex]

    Where is that extra 3 coming from? I can't really see where I went wrong (unless I used volume instead of surface area, in which case my answer would still be different).

  2. jcsd
  3. Mar 12, 2006 #2

    Physics Monkey

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    I assume we're talking about a spherical shell here. The trouble is that none of the energy is stored inside the shell since the electric field is zero there. You have to integrate the energy density over all of space to get to total energy of the system. So what I would suggest is to integrate [tex] \frac{\epsilon_0 E^2}{2} [/tex] over all space and see what you get.

    Hope this helps.
  4. Mar 12, 2006 #3
    I understand that its from all of space, but the formula you told me to integrate is the one I am trying to derive. I mean, all of the usual calculations I have done regarding a spherical shell involve

    [tex]\int\limits_{\infty}^{R}\textrm{some function here},[/tex]

    but each time I do that it comes out different (only by a constant though, as I have done above).
  5. Mar 12, 2006 #4

    Physics Monkey

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    You can't derive a formula for the energy density by considering a special case. You need to go to the basic equations of electrostatics to really derive the energy density formula. If you want to "guess" the formula, try picking a geometry (like a capacitor) where the field is uniform and the energy is all bound up in a finite region. It seemed to me that the problem was asking you to verify the fact that [tex] \frac{\epsilon_0 E^2}{2} [/tex] is the energy density. Perhaps I have misunderstood.
    Last edited: Mar 12, 2006
  6. Mar 12, 2006 #5


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    You might rather think of it this way ...
    epsilon E^2 has units of Energy/Volume ... maybe it is the Energy density.
    so integrate from R to infinity the function epsilon E(r) 4 pi r^2 dr ...
    You do NOT obtain the PE ... off by a factor of 2 ... but that's easy to fix.
  7. Mar 12, 2006 #6
    OK, I may have done something here that is incorrect, but I did get my result:

    [tex]D=\frac{U}{A\cdot r}=\frac{U}{4\pi r^{3}}=\frac{\left(\frac{kQ^{2}}{2r}\right)}{4\pi r^{3}}=\frac{\epsilon_{0}E^{2}}{2}.[/tex]

    Does this work? Thanks.
  8. Mar 12, 2006 #7


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    I would call that nonsense, with a geometry error.

    You're trying to SHOW that some function can be treated as a PE density.
    What does THAT mean?
    It means that the integral of that function is equal to the total PE.

    (how do you know if some function is the charge density? integrate!)

    You know that E(r) = kQ/r^2 , and can presumably square that.
    You know the Area of a sphere (even if you screw up the Volume ;-/
    just DO it.
  9. Mar 12, 2006 #8

    [tex]U=\int\limits_{R}^{\infty}D\left(r\right)\,dV=\int\limits_{R}^{\infty}\left(\frac{\epsilon_{0}k^{2}Q^{2}}{2r^{4}}\right)\left(4\pi r^{2}\,dr\right)=\frac{kQ^{2}}{2R}.[/tex]

    I understand, thank you.
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