# Modified Coulomb potential

I'm working through Byron and Fuller's "Mathematics of Classical and Quantum Physics" and came across this problem:

If the electric potential of a point charge were

$\phi(r) = \frac{q}{r^{1-\epsilon}}, \epsilon \ll 1$

instead of $\frac{q}{r}$, many of the results in physics would be altered. Compute $E = -\nabla \phi, \nabla \cdot E, \nabla \times E$ and derive the analog of Gauss's Law for a spherical region of radius R. Examine the limit as $\epsilon \rightarrow 0$ and compare with the usual Coulomb potential.

I worked them out and while the answers are different the physics seems essentially the same, and reduces to the normal answer as $\epsilon \rightarrow 0$. I'm trying to find more information about this problem. Thanks!

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What do you get for the divergence of $E$? What should it read (for $r > 0)$? –  Gerben Oct 22 '11 at 9:48
$q\epsilon(1-\epsilon)r^{\epsilon - 3}$ –  xuanji Oct 22 '11 at 9:55
That seems right. Now turn to Maxwell's equations, in particular $\rho = \nabla E.$ The modification has introduced a non-zero charge distribution outside of the origin $\propto \epsilon r^{-3}$, which isn't coherent with the point charge idea. For the Gauss Law, you'll find a similar issue. –  Gerben Oct 22 '11 at 10:06
Good question! @Gerben: that sounds like it could be an answer. –  David Z Oct 23 '11 at 5:45