Griffiths' : Electrostatic Energy

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SUMMARY

The electrostatic energy of a continuous charge distribution is expressed as W = \frac{\epsilon_o}{2}\int (\vec{E})^2d\tau. This expression is derived by considering the limit of integration volume and the behavior of electric field \vec{E} and potential V at large distances from the charge. The surface integral approaches \frac{1}{r} while the volume integral increases due to the positive integrand, confirming Griffiths' assertion that the boundary term can be neglected. Alternative derivations include assembling the charge distribution adiabatically and applying Poynting's Theorem as discussed in advanced texts like Jackson.

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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.

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

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

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

For large ditances from the charge V goes like \frac{1}{r} while \vec{E} goes like \frac{1}{r^2}, and the area increases as r^2. Therefore the surface integral is roughly \frac{1}{r}. 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 \frac{1}{r} as we increase of the volume of integration?
 
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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.
 
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!
 
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: )
 
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.

Regards,
Reilly Atkinson
 

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