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Let us consider a $1D$ real function $V(x)$. When is this a classical electrostatic potential?

My take on the problem:

  1. $V(x)$ must be differentiable everywhere. In fact, we should be able to differentiate it $n$ times.
  2. $V(x)$ should vanish at $\pm \infty$.

I think these are necessary and sufficient conditions. Is this right? How do I deal with discrete charge distributions, where the potential is not differentiable at the points where the charges are present?

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  • $\begingroup$ Can you strictly define what is a classical electrostatic potential? $\endgroup$
    – Heath
    Oct 15 '20 at 7:55
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    $\begingroup$ @AndreasMastronikolis a physical potential that obeys poisson's equation. $\endgroup$
    – Soham
    Oct 15 '20 at 9:36
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I think that most conditions can be seen from the Poisson equation: $$ \frac{d^2 V(x)}{dx^2} = -\rho(x). $$ Thus, it should be differentiable everywhere except a few singular points. One often bypasses this issue by using generalized functions (delta-function and Heaviside step function).

However, there is no requirement that the potential vanishes in infinity. For example, the potential corresponding to a constant electric field of magnitude $E$ is $$ V(x) = - Ex $$

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    $\begingroup$ Aren't you missing a negative sign in Poisson's Equation? (oh, and an $\epsilon_0$...) $\endgroup$
    – Philip
    Oct 15 '20 at 7:50
  • $\begingroup$ @Philip Thanks. Fixed. $\endgroup$ Oct 15 '20 at 7:56
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    $\begingroup$ Obviously, E cannot be constant over an infinite range. $\endgroup$
    – my2cts
    Oct 15 '20 at 8:19
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    $\begingroup$ @my2cts I think the question is about the mathematical constraints. But physically - yes, E cannot be constant over an infinite range. $\endgroup$ Oct 15 '20 at 8:34

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