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After studying the definition (& derivation) of the potential to an electric field and the Poisson equation I'm currently wondering whether the following is possible:

  • Can one give an example of a physical setup where the Poission equation fails to provide the electric field?

I have tried to come up with an example but failed. What I thought of was, that considering a valid solution of the Poisson equation, the electric field derived from the solution would have to be differentiable (since the second derivative of the potential appears in the differential equation). If one were to find a setup with a discontinuous electric field, would the Poisson equation fail?

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Any potential not stisfying either of the electrostatic uniqueness theorems, which are a consequence of the form of Poisson's equation, will fail to satisfy electrostatic condition in the given configuration –  Torsten Hĕrculĕ Cärlemän Dec 27 '13 at 13:59
    
Thanks for your comment. I'm not quite sure, whether I understood it correctly. So what you are implying is that if the Poisson equation doesn't provide a solution for a certain configuration there will be no possibility to derive the electric field from there? (kind of logical). Could you elaborate a little more? –  R.G. Dec 27 '13 at 14:16
    
Poisson's equation states that the Laplacian of the electrostatic potential must vanish subject to the uniqueness theorems. If the theorems themselves fail to be correct, then arguing backwards, there would not be a potential(hence a field) satisfying the equation. Hence, if you can find a potential violating the uniqueness conditions, then you have the one you seek. –  Torsten Hĕrculĕ Cärlemän Dec 27 '13 at 15:04
    
Also read @xuanji's answer carefully. –  Torsten Hĕrculĕ Cärlemän Dec 27 '13 at 15:06
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Electric field induced in the secondary winding of a transformer can not be described as a gradient of a potential. –  Maxim Umansky Dec 27 '13 at 15:49
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1 Answer

up vote 3 down vote accepted

Yes, it is possible, but not in the way you suggest.

It's totally possible for Poisson's equation with non-continuous electric field; the charge density (and hence the second derivative of potential) could be non-finite. This is less esoteric than it seems; the electric field of a point charge is discontinuous at the location of the point charge itself.

There's another way to show that Poisson's equation is not enough, though. Consider an electric field which is constant, ie $\mathbf{E} = \mathbf{E_0}$ for some constant $\mathbf{E_0}$. We can deduce that the charge density is $0$ everywhere, but this is independent of $\mathbf{E_0}$! So if we are given that $\rho=0$ to start with and nothing else, there is no unique solution for the electric field.

This shows that in addition to Poisson's equation, we must have boundary conditions. In most cases it is implicitly assumed that the boundary conditions are that $E \to 0$ at large distances.

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Thanks for the answer! –  R.G. Dec 27 '13 at 19:58
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