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I'm reading Griffiths's E&M book and he derives the following result for the energy required in assembling a continuous charge distribution $$W=\dfrac{\epsilon_0}{2}\left(\iiint E^2\ dV +\oint V\textbf{E}\cdot d\textbf{A} \right)$$ where $E$ is the electric field, $V$ is the potential and $\textbf{A}$ is the normal area vector. I understood the derivation but he then claims that as we extend our volume integral to all of space, the surface integral vanishes and we are left with $$W=\dfrac{\epsilon_0}{2}\iiint E^2\ dV $$ I really don't understand the reasoning behind this. I searched elsewhere and someone tried to justify this by imagining integrating over a spherical volume. The claim was that since $E$ goes like $1/r^2$, $V$ goes like $1/r$ and $A$ goes like $r^2$ so that the surface integral goes down by $1/r$. But this doesn't make much sense, since the volume integral must also go down by $1/r$? ($E^2$ goes like $1/r^4$ and the volume goes by $r^3$?). Is there a more rigorous or better explanation for the reasoning behind this?

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In the image, I have shows the region over which we're integrating.The orange line is the area over which were doing the surface integral and the blue area is the region over which we're doing the volume integral. Assume that the charge distribution is at the centre of the sphere.

Now, Suppose you make the radius of the sphere infinitely large. When your doing the surface integral, you are always the same distance away from the centre, and hence you can neglect it because as you said, it is proportional to $\frac{1}{r}$.

But now look at the volume integral. Yes, near the outer surface the integral becomes negligible (because it's proportional to $\frac{1}{r}$), but notice that since it is a volume integral, we must also do the integration over the inner volume, which also includes near the surface, which is not negligible anymore. This is why you can ignore the surface integral and not the volume integral.

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