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Griffiths' : Electrostatic Energy

  1. May 28, 2006 #1
    Griffiths : Electrostatic Energy

    I'm having a little difficulty in understanding how one arrives at the following expression for electrostatic energy of a continuous charge distribution.

    [tex]W = \frac{\epsilon_o}{2}\int (\vec{E})^2d\tau[/tex]

    This result is obtained when the volume of integration is increased without limit in

    [tex]W = \frac{\epsilon_o}{2} ( \int (\vec{E})^2d\tau + \oint V \vec{E} \cdot d\vec{a}) [/tex]

    For large ditances from the charge [itex]V[/itex] goes like [itex]\frac{1}{r}[/itex] while [itex]\vec{E}[/itex] goes like [itex]\frac{1}{r^2}[/itex], and the area increases as [itex]r^2[/itex]. Therefore the surface integral is roughly [itex]\frac{1}{r}[/itex]. Griffiths says that the volume integral must increase owing to the positive integrand. But can't we apply similar logic as above and say that volume integral also rougly goes like [itex]\frac{1}{r}[/itex] as we increase of the volume of integration?
    Last edited: May 28, 2006
  2. jcsd
  3. May 28, 2006 #2

    Meir Achuz

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    The r in the surface integral is a constant, the radius of a large sphere.
    The r in the volume integral varies from 0 to the radius of the bounding sphere, so there is always small r for the volume integral.
  4. Jun 5, 2006 #3
    What Griffiths does here is standard practice when integrating by parts, always look to eliminate the boundary term by physical argument, i.e. the potential due to some charge distribution vanishes at infinity. Like Meir said, the surface integral is evaluated only at the boudary of the system, which conveniently for us is user definable!
  5. Jun 6, 2006 #4
    Thank you, Meir Achuz and jarvis. It understand it a bit better now. :smile:

    (I actually forgot that I had posted this question, and hence the late reply :blushing: )
  6. Jun 8, 2006 #5


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    Another way to get this expression, which I always thought was the standard elementary derivation, is to assemble the distribution adiabatically, charge element by element. As the distribution is assembled, bringing in the incoming charge requires the assembler to work against the charge already in place. See any freshman text.

    The fancy way, good for static or dynamic fields comes from Poynting's Thrm, which you can find in more advanced texts -- Jackson, for example.

    Reilly Atkinson
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