Why does the surface integral of V E·dA vanish in EM energy derivation?

[Swapnil]

The following wikipedia article derives the energy stored in an E-field (under "Energy stored in an electric field"):
http://en.wikipedia.org/wiki/Electrical_energy

I don't quite get how the following term goes to zero in the article?
$$\frac{\epsilon_o}{2}\int V\mathbf{E}\cdot dA$$

 

[jtbell]

First consider what would happen if the V were not in that term. Then for a very large surface that encloses all the charge in the system, the integral would always equal $Q_{total} / \epsilon_0$ (a constant) because of Gauss's Law, no matter how large the surface is. For a spherical surface of radius r, with the charge more or less at the center, the area would increase as $r^2$, but the electric field at the surface would decrease as $1/r^2$, and the two effects would cancel out.

Now put the V inside the integral. It decreases like $1/r$ for very large r, so it forces the integral towards zero as r goes to infinity.

 

[Manchot]

The original volume integral from which the above expression was derived is taken over all space. Therefore, the surface integral above is taken across a surface located "at" infinity. (You can think of it as a sphere whose radius approaches infinity.) Now, for all physical charge distributions, both V and E vanish as r approaches infinity. Therefore, the integrand vanishes at the surface of integration, and the integral is also zero.

To be more precise, you need to be careful about the limiting behavior of the integrand. The integration itself will contribute an r^2 term, meaning that even slowly-decaying integrands like 1/r won't decay fast enough to overcome this. Luckily, V(r) goes like 1/r and E(r) goes like 1/r^2, so the overall behavior is 1/r * 1/r^2 * r^2 = 1/r, which is still a decaying function.

 

[Swapnil]

I got it now.Thanks for the help!