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For a continuously differentiable vector field $F$ the divergence theorem can be used to give $$(\nabla\cdot F)(a) = \lim_{r\to 0} \frac{3}{4\pi r^3}\int_{|x-a|=r} F \cdot n dA$$ This should mean that for $c<3 $ $$\lim_{r\to 0} \frac{3}{4\pi r^c}\int_{|x-a|=r} F \cdot n dA=0$$ In particular for $c=2$ that $$\lim_{r\to 0} \frac{3}{4\pi r^2}\int_{|x-a|=r} F \cdot n dA=0$$

Although I understand mathematically why divergence is associated with the first equation given above, I don't have a good grasp about why intuitively it is correct to divide by $r^3$ instead of $r^2$. I would have thought that since the flux is a surface integral we should divide by $r^2$ to account for the changing size of the surface area. Can anyone give an intuitive explanation for what I am missing?

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The basic reason for this is that we want the limit to be finite. As $r$ approaches zero, there are two reasons why the flux shrinks.

One is that the flux is proportional to the surface area, which goes like $r^2$. The other is that flux tends to cancel on opposite sides of the volume. If the field is constant, this cancellation is always exact. The change in the field from one side of the volume to the other is proportional to $r$.

Putting these two factors together, the flux shrinks in proportion to $r^3$. To make the limit finite, we need to divide by $r^3$.

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  • $\begingroup$ Great "counting" of the geometric factors involved :-) $\endgroup$ – rajb245 Sep 9 '13 at 4:25
  • $\begingroup$ I don't find it a convincing argument as to why the corrective factor should be $r^3$ as opposed to say $r^{3-\epsilon}$. $\endgroup$ – user782220 Sep 9 '13 at 4:54

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