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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?

Gavin S. Yancey
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  • 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); https://brilliant.org/discussions/thread/vector-fields-and-potential-functions/ – Prasad Mani Oct 19 '16 at 07:20
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    Possible duplicates: http://physics.stackexchange.com/q/279344/2451 , http://physics.stackexchange.com/q/269402/2451 Related: http://physics.stackexchange.com/q/1324/2451 , http://physics.stackexchange.com/q/248101/2451 , http://physics.stackexchange.com/q/133363/2451 and links therein. – Qmechanic Oct 19 '16 at 08:45
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    Well, no. It is not differentiable wherever a point charge is. – valerio Oct 19 '16 at 09:36

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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".

tparker
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    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! ;) – valerio Oct 19 '16 at 09:39
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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.

Lelouch
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  • 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 – Prasad Mani Oct 19 '16 at 08:04