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This is somehow the inverse of the Poisson problem, which i could reformulate as follows:

  • Given a volume in the empty space, bounded by a bidimensional surface, and an arbitrary electric field inside the volume, determine a charge distribution on the surface which generates the field.

I don't know how to approach this problem, and I can't find any book or paper that addresses it. Is it trivial? Is it ill-posed?

I would be happy to have just some hints at answers for the spherical case

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  • $\begingroup$ Can't you simple use $\rho(r)=\epsilon_0\nabla \cdot \mathbf{E}$? $\endgroup$ Commented Mar 1, 2021 at 13:18
  • $\begingroup$ How do I enforce that the charge must lie on the surface? On a side note, if both a volume and a charge distribution can generate the same field, does this imply that the problem of finding E given the charge distribution is ill-posed? $\endgroup$
    – ygh
    Commented Mar 1, 2021 at 16:54

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You can just solve Poisson's law or Gauss'law (its integral version) depending on which is the easiest as you have the complete description of the field alredy.

I think the best starting point is actually Poisson's:

$$\nabla\cdot\vec{E}={\rho\over\epsilon_0}$$

then given a field $$\vec{E}(x, y, z)$$ you get $$\partial_x E_x+\partial_yE_y+\partial_zE_z = {\rho(x, y, z)\over\epsilon_0}$$

Whether that is easy to solve or not depends on the details.

If your surface is spherical it might be convenient to write everything is spherical coordinates.

Let's do an easy example: $$E(r)={1\over 4\pi \epsilon_0 }{Q\over r^2} \hat{r}$$ if $r>R$ and $0$ otherwise, which is the field of a uniform spherical charge distribution over a sphere of radius $R$.

Then the divergence of the field is (in spherical coordinates)

$$\nabla\cdot\vec{E} = {Q\over 4\pi\epsilon_0} \nabla\cdot {\hat{r}\over r^2}$$ if $r>R$ and 0 otherwise. The other terms of the divergence (in $\theta$ and $\phi$, the polar angles) vanish as they are constant - the field only varies along $\hat{r}$.

So we get, using Gauss' law $$\rho(r)=0$$ if $r<R$ and otherwise

$$\rho(r)=\epsilon_0 \nabla\cdot\vec{E}= {Q\over 4\pi} \nabla\cdot {\hat{r}\over r^2}$$

so we only need to compute $$\nabla\cdot {\hat{r}\over r^2}$$ in $[R, \infty]$.

Now this requires a bit of calculus which I have to hand-wave [I am happy if somebody solves it more generally]

In general, $$\nabla\cdot {\hat{r}\over r^2} = 4\pi\delta(r)$$ [you can see why here] for example. However in our case we don't have a discontinuity not in the origin but in $r=R$ and I am pretty sure the result in our case would be that the divergence is $0$ everywhere and infinite in $r=R$ so

$$\nabla\cdot {\hat{r}\over r^2} = \delta(r-R)/r^2$$ meaning that $$\rho(r)= {Q\over 4\pi r^2}\delta(r-R)$$ which is a spherical distribution of total charge $Q$ on a shell centered at the origin of radius $R$.

The main problem to deal with is the behavior of the divergence on the surface, where the charge distribution is discontinous. But in principle

$$\rho(r)=\epsilon_0 \nabla\cdot\vec{E}$$

is the solution you are looking for if you already know the field $\vec{E}$ everywhere. It's just a matter of doing derivatives.

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  • $\begingroup$ How do I enforce, a priori, that the charge must lie on the surface? Also, you gave a field outside the volume, but what interests me is a field inside the volume, because I don't know how to apply Gauss's law in this case $\endgroup$
    – ygh
    Commented Mar 1, 2021 at 17:00
  • $\begingroup$ you don't enforce the charge density. The divergence tell you where there is charge, that's how you get the delta function. It is enough to have the field close to the surface to find the div. However, if you don't have the field everywhere/near the surface, then the only way I can think of is to write an integral equation for the electric field of an unknown charge distribution on a surface, something like $E(\vec{r})={1\over 4\pi \epsilon_0} \int_S\rho dS {\vec{r}-\vec{r'}\over |r-r'|^3}$ where $r'$ is on the surface and $r$ in the volume and hope to solve it for known values of E.. $\endgroup$
    – JalfredP
    Commented Mar 2, 2021 at 8:17

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