## Is every electric field possible?

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FrancovS
Posts: 59
Joined: Fri Apr 16, 2010 1:47 am UTC

### Is every electric field possible?

This is a problem about electrostatics I've been trying to solve. I believe it has been solved before, but I couldn't find anything about it. So here we go: I want to know if any vector field with zero curl can be an electric field, in the sense that it can be generated by a charge distribution with coulomb's law.

To avoid dealing with electric fields that come from charges which are infinitely far and non-volumetric charge distributions, I stated the problem in the following way: assume a vector field E defined inside an sphere S. If div E = 0 and curl E = 0, can I find a charge arrangement outside the sphere that will have E as the resultant electric field? I will try to state it in a more rigorous way: can I always find a function [imath]\rho(x, y, z)[/imath] that satisfies the identity

$\forall{(x_0, y_0,z_0)\in S}\left(\int_{-\infty }^{\infty }\int_{-\infty }^{\infty}\int_{-\infty}^{\infty} \frac{\rho (x,y,z)\cdot (x_0-x, y_0-y, z_0-z)}{|(x_0-x, y_0-y, z_0-z)|^{3}}dxdydz = \vec{\mathbf{E}}(x_0,y_0,z_0)\right)$no matter how E is defined, as long as div(E) = 0 and curl E = 0?Did I do it right?

I am using the fact that div E = 0 because it seems to simplify the problem; if there is always a solution when div E = 0, I can use superposition to find the solutions when div E !=0 by dividing the field into a "source only" field and a "source free" field.

imatrendytotebag
Posts: 152
Joined: Thu Nov 29, 2007 1:16 am UTC

### Re: Is every electric field possible?

Here's the basic idea (to your general problem): If E is a curl-less vector field, then p = e0*div E is the charge distribution you're looking for, where e0 is the electric constant. Why? Well if E' is the electric field generated by the charge distribution p, gauss's law says div E' = p/e0, hence div E' = div E. Also, curl E' = curl E = 0. This would hopefully be enough to say that E' = E but it's not quite. We do, however, have the following result: There does not exist a nonzero vector field with 0 curl and 0 div which also goes to 0 at infinity. This means that if E and E' both go to 0 at r = infinity, then E = E'.

On the other hand, the vector field Ex = yz, Ey = xz, Ez = xy cannot be created by any charge distribution (since it has 0 curl and div but is not identically 0).
Hey baby, I'm proving love at nth sight by induction and you're my base case.

FrancovS
Posts: 59
Joined: Fri Apr 16, 2010 1:47 am UTC

### Re: Is every electric field possible?

Well, I kinda mis-explained the problem. I wouldn't even have gone so far: I could simply argue that an constant field Ex = 1 (x component) has zero divergence, so no charge originates it, so it can't exist. My goal with this problem is to know which field patterns can be generated in practice. Since the charges have to be somewhere, I added an "I don't care anymore" boundary from which I can't specify the electric field.