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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?

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# Griffiths' : Electrostatic Energy

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