Modified Coulomb potential [closed]

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!

closed as off-topic by sammy gerbil, user36790, CuriousOne, ACuriousMind♦, honeste_vivereJul 22 '16 at 11:52

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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
• I'm not sure what the question is here. You say you worked out the problem, so it's not how to solve it. You're "trying to find more information", but what exactly do you want to know? – ACuriousMind Jul 22 '16 at 11:08