Griffiths' : Electrostatic Energy

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Discussion Overview

The discussion revolves around the derivation of the expression for electrostatic energy of a continuous charge distribution, specifically the formula W = \frac{\epsilon_o}{2}\int (\vec{E})^2d\tau. Participants explore the implications of integrating over a volume and the behavior of electric field and potential at large distances from the charge.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the derivation of the electrostatic energy expression and questions whether the volume integral behaves similarly to the surface integral as the volume of integration increases.
  • Another participant clarifies that the variable r in the volume integral changes from 0 to the radius of the bounding sphere, implying that small r values are always present in the volume integral.
  • A third participant notes that the approach taken by Griffiths is standard practice in integrating by parts, emphasizing the importance of eliminating boundary terms based on physical arguments.
  • A later reply introduces an alternative method for deriving the expression by assembling the charge distribution adiabatically, suggesting that this is a common approach found in introductory texts.
  • The same participant mentions that a more advanced method involves Poynting's theorem, referencing higher-level texts like Jackson.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the behavior of the volume integral compared to the surface integral, and there are multiple perspectives on the derivation methods presented.

Contextual Notes

The discussion includes assumptions about the behavior of electric fields and potentials at large distances, as well as the implications of integrating over different regions of space. There are unresolved questions regarding the specific limits and conditions under which the integrals are evaluated.

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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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