# Find 2 charge densities that gives the same E field

1. Mar 4, 2013

### fluidistic

1. The problem statement, all variables and given/known data
I'm asked to find 2 different charge densities $\rho (\vec x )$ (without any symmetry, discarding the method of images) that produce the same $\vec E (\vec x )$ field.

2. Relevant equations
$\Phi ( \vec x ) = \int _{\Omega } \frac{\rho (\vec x )}{|\vec x - \vec x '|}d^3 x'$.

3. The attempt at a solution
Since $\vec E =- \vec \nabla \Phi$ and E must be the same for both charge distributions, this means that their potential must differ by a constant. In other words, if $\rho _3 (\vec x )=\rho _1 (\vec x )-\rho _2 (\vec x )$ where $\rho _1 (\vec x )$ and $\rho _2 (\vec x )$ are the 2 charge distributions I'm asked to find, then the potential resulting from $\rho _3 (\vec x )$ must be a constant that does not depend on the position (equipotential).
But I've thought about that, I don't know of any charge distribution that gives a constant potential throughout the whole space.
I've really ran out of ideas. Not even an infinite charged non conducting sheet would do the job.
I'd appreciate if someone could throw some idea(s). Thanks.

2. Mar 4, 2013

### Staff: Mentor

Here are some ways to avoid the problem:
- use a space different from R^3
- or simply restrict the analysis to some part of R^3
- did they say "the same electric field in the same coordinate system?
- treat positive and negative charges as independent charge densities
- add a uniform charge density in the whole space, and talk about symmetry and the limited age of the universe to justify this argument
- use current densities instead of charge densities

3. Mar 4, 2013

### fluidistic

Thanks for the help mfb.
As well as the space has still 3 dimensions I'm ok with considering a smaller than R^3 space.
They don't mention the coordinate system, though I'd be surprised if the E field is the same in spherical coordinates but different in Cartesian coordinates!
About adding a uniform charge density, hmm, that may qualify as long as they don't consider this as "symmetric" I guess.
And about the current densities I've no idea how this could help nor if I can really use it since I'm dealing with electrostatics.

4. Mar 4, 2013

### Staff: Mentor

Spheres of variable radius, if you consider the E-field outside only.
I thought of translations and similar things. You have to express the charge density in a different coordinate system than your E-field in that case. This is not a solution to the original problem ;).
That is the trick. A circular current in an uncharged wire (for example) has no electric field, but a different current than empty space.

5. Mar 4, 2013

### fluidistic

So Omega =R^3 - sphere(s)? :)
Ah ok!
Oh, I did not know this had no E field. But then, how do I write up $\rho (\vec x)$ in such a case? With the Heaviside function and modeling a torus? And then I say that the charges move, so "magically" the E field is worth 0?

6. Mar 4, 2013

### Staff: Mentor

That is possible.

The current flow does not work with $\rho(x)$ - to get zero electric field, you need the same amount of positive and negative charges, so $\rho(x)=0$ for the uncharged, conducting ring.

7. Mar 4, 2013

I see.