Is electric potential always differentiable?
If so, why?
If it isn't always, then what properties of a charge-distribution are required to make it differentiable?
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$\begingroup$ They have to be twice differentiable! Because $\triangledown {V}$ must exist and $\triangledown^2{V}$ must also exist (for a conservative electric field of course); brilliant.org/discussions/thread/… $\endgroup$– Prasad ManiOct 19, 2016 at 7:20
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4$\begingroup$ Possible duplicates: physics.stackexchange.com/q/279344/2451 , physics.stackexchange.com/q/269402/2451 Related: physics.stackexchange.com/q/1324/2451 , physics.stackexchange.com/q/248101/2451 , physics.stackexchange.com/q/133363/2451 and links therein. $\endgroup$– Qmechanic ♦Oct 19, 2016 at 8:45
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1$\begingroup$ Well, no. It is not differentiable wherever a point charge is. $\endgroup$– valerioOct 19, 2016 at 9:36
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2 Answers
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One typically starts with the physically measurable electric field $\vec{E}$ and then defines the electric potential $\phi$ such that $\vec{E}$ is its derivative, so if $\phi$ weren't differentiable then it wouldn't be a very useful concept...
One exception is that it can fail to be differentiable at points where the electric field itself is not well-defined, e.g. exactly at the location of a point charge, line charge, or sheet of charge. But it will certainly be differentiable "almost everywhere".
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1$\begingroup$ From a physical point of view, this is certainly right. On the other hand, from a mathematical point of view, I think that we can build distributions of point charges which are not almost everywhere differentiable...Well, anyway, we are physicists, so who cares! ;) $\endgroup$– valerioOct 19, 2016 at 9:39
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The electrostatic potential might not always be differentiable. A simple theoretical example of the following is a point charge in empty space. V is differentiable everywhere except at the point where the charge is placed. Here, the function blows up and therefore becomes non-differentiable. To generalise even further, we know that $\vec E = -\nabla V$ for any conservative electric field $\vec E$. E itself may not be defined at all points in space, for various charge distributions.
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$\begingroup$ Of course it is not differentiable smack dab in the middle of the charge. I guess thats not what he meant to ask. He is asking about the general behavior of the potential which from laplace/poisson equation has to be twice differentiable $\endgroup$ Oct 19, 2016 at 8:04