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Consider the famous work equation due to a continuous charge distribution:

$W=\frac{\varepsilon_{0}}{2}\left ( \int_{volume \space space}\left \| \vec{E} \right \|^{2}.d \tau+\oint_{S}V.\vec{E}.d \vec{a} \right )$

Note:

$V\left(r \right)=\frac{1}{4 \pi \varepsilon _{0}}\frac{q}{r}$

$\vec{E}= \frac{1}{4\pi\varepsilon _{0}}\frac{q}{r^{2}}$

The author of the text Introduction to Electrodynamics claims that

..well the integral of $E^{2}$ can only increase; evidently the surface integral must decrease correspondingly to leave the sum intact. In fact, at large distance from the charge, E goes like $\frac{1}{r^{2}}$ and V like $\frac{1}{r}$ while the surface area grows like $r^{2}$; roughly speaking, then, the surface integral goes down like $\frac{1}{r}$

Could someone explain further on the bold?

Thanks in advance.

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Intuitively speaking, any finite continuous object with charge $Q$ will become increasingly smaller when looked at from larger distances. Furthermore, recall that for a point-like charge the electric field can be written as

$$ E \sim \frac{Q}{r^2} \tag{1} $$

and its electric potential

$$ V \sim \frac{Q}{r} \tag{2} $$

From a more formal point of view, you can make a multipole expansion of the potential

$$ V({\bf r}) = \frac{Q}{4\pi \epsilon_0}\frac{1}{r} + \frac{{\bf p}\cdot \hat{\bf r}}{4\pi\epsilon_0}\frac{1}{r^2} + \cdots \tag{3} $$

And note that the monopole term drops slower than any other term, so at sufficiently large distances the potential can be approximately written as

$$ V({\bf r}) \approx \frac{Q}{4\pi \epsilon_0}\frac{1}{r} \tag{4} $$

and its associated electric field

$$ {\bf E}({\bf r}) \approx \frac{Q}{4\pi\epsilon_0}\frac{\hat{\bf r}}{r^2} \tag{5} $$

Now, for a sufficiently large integrating surface $S$, you can also think of it as a sphere of radius $R$, so the area is approximately equal to $4\pi R^2$, since the potential is spherically symmetric at very large distances then

$$ \oint_S V {\bf E}\cdot {\rm d}{\bf S} \approx \left(\frac{Q}{4\pi\epsilon_0}\right)^2 \oint_S \frac{1}{r^3}\underbrace{dS}_{R^2 {\rm d}\Omega} = \frac{Q^2}{4\pi\epsilon_0^2} \frac{R^2}{R^3} \sim \frac{1}{R} $$

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