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Imagine this:

You have a sphere of air where you have no charge and around this sphere you have a charge distribution $\rho(r,\theta,\phi)$. (For instance, this could be $\rho(r,\theta,\phi)=e^{-r}$) Now my question is: What is the most general equation that will give me the potential inside the sphere?-You can use that we have azimuthal symmetry. I am just interested in the equation.

Probably this will contain a series with Legendre Polynomials and so on.

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The electric potential $\Phi$ is defined through the following relation:

$$\mathbf{E}=-\nabla \Phi\tag{1}$$

Now consider a vector field $\mathbf{F}$ such that:

$$\nabla.\mathbf{F}=D$$ $$\nabla \times \mathbf{F}=\mathbf{C}$$

According to Helmholtz theorem, if the divergence $D(\mathbf{r})$ and the curl $\mathbf{C(r)}$ are specified and if they both go to zero faster than $\dfrac{1}{ r^2}$ as $r\to \infty$, and if $\mathbf{F(r)}$ goes to zero as $r\to \infty$ then $\mathbf{F}$ is uniquely given by

$$\mathbf{F}=-\nabla U+ \nabla \times \mathbf{W}$$ where

$$U(\mathbf{r})=\frac{1}{4\pi}\int \frac{D(\mathbf{r'})}{|\mathbf{r-r'}|}d^3r'\tag{2}$$ $$\mathbf{W}(\mathbf{r})=\frac{1}{4\pi}\int \frac{\mathbf{C}(\mathbf{r'})} {|\mathbf{r-r'}|}d^3r'\tag{3}$$

For a static electric field, $D=\dfrac{\rho}{\epsilon_0}$ and $\mathbf{C}=0$. So, according to $(1)$ and $(2)$ the electric potential of a charge distribution that goes to zero faster than $\dfrac{1}{r^2}$ as $r\to \infty$ can be calculated as $$\Phi(\mathbf{r})=\frac{1}{4\pi \epsilon_0}\int \frac{\rho(\mathbf{r'})}{|\mathbf{r-r'}|}d^3r'$$ where the integral is over all of space.

$\dfrac{1}{|\mathbf{r-r'}|}$ can be expanded using spherical harmonics to obtain a multipole expansion. So the multipole expansion is valid only under the above conditions too.

If the above condition doesn't hold, you have to use the $(1)$ equation, i.e. you have to find $\mathbf{E}$ first and then perform the integration to find $\Phi$ (as in the case of an infinite uniformly charged wire).

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  • $\begingroup$ Yes, actually I am looking at an exponentially decaying charge distribution(so fast decaying) and now my question would be: What would be the correct expansion(of the denominator in harmonic function) if you are looking at the potential inside a sphere, when the charge distribution is only nonzero outside the sphere? $\endgroup$ – Xin Wang Aug 12 '13 at 20:08
  • $\begingroup$ my question is, whether the expansion $\frac{1}{|r-r'|} = \sum_{l=0}^{\infty}\frac{r'^l}{r^{l+1}}P_l(\cos(\theta))$, where r' is the interior position, where I want to know the potential. Is this correct, so can I integrate $\Phi(r',\theta) = \frac{1}{4 \pi \epsilon_0 } \int_0^{\infty} \int_0^{\pi} \int_0^{2 \pi} \rho(r, \theta) \sum_{l=0}^{\infty}\frac{r'^l}{r^{l+1}}P_l(\cos(\theta)) d \phi d \theta dr$ $\endgroup$ – Xin Wang Aug 13 '13 at 9:38
  • $\begingroup$ sorry forgot the $r^2 \sin(\theta)$ and I think there is something wrong with my $\theta$ dependence, as I integrate over it, this one is somehow lost $\endgroup$ – Xin Wang Aug 13 '13 at 9:46
  • $\begingroup$ Your expansion gives the potential on the $z$ axis. To find the potential everywhere, replace $\theta$ with $\gamma$, which is the angle between $\mathbf{r}$ and $\mathbf{r'}$; i.e. expand in terms of spherical harmonics (here). $\endgroup$ – Mo_ Aug 13 '13 at 9:59
  • $\begingroup$ but we do not have any dependence on $\phi$-dependence. is it not true that in this case the spherical harmonics reduce to the legendre polynomials? $\endgroup$ – Xin Wang Aug 13 '13 at 10:12
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The most general expression for the potential (assuming a static charge distribution, as you use) is:

$$V({\vec r}) = C + \int d^{3}r'\frac{\rho({\vec r'})}{4\pi \epsilon_{0}\left|{\vec r}-{\vec r}'\right|}$$

Where C satisfies ${\vec \nabla}C = 0$, and the integral covers the region where $\rho \neq 0$.

If you doubt this, you can work out that $\nabla^{2} \frac{1}{\left|{\vec r}\right|} = \delta^{3}({\vec r})$, and then it should be pretty obvious that this equation satisfies the differential form of Gauss's law.

EDIT:

I see that you're asking what happens inside a gap inside of a spherically symmetric distribution. In this case, you can use the high school physics version of Gauss's law to show that:

$$\left|{\vec E}\right| = k\frac{Q_{inc}}{r^{2}}$$

Since, inside of your inner gap, the charge enclosed by any gaussian surface is zero, you have $E = 0$ and $V=$ Constant

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    $\begingroup$ no, the first version was okay, but I was wondering how one can find C from your definition and whether there is some taylor expansion (of the denominator) that I could use in this case, since it is hard to integrate this thing in general $\endgroup$ – Xin Wang Aug 9 '13 at 10:15
  • $\begingroup$ @Lipschitz: The potential is arbitrary, and C reflects this arbitrariness. Typically, you specify one point where the potential is zero, and after you've done the integral, you evaluate the potential at this point, and set C such that the potential is zero here. $\endgroup$ – Jerry Schirmer Aug 9 '13 at 15:52
  • $\begingroup$ As far as evaluating the integral, you're probably safer doing a taylor expansion of the density in powers of $r-r'$, which should be nice easy integrals. Most people just resort to numerics as this point. $\endgroup$ – Jerry Schirmer Aug 9 '13 at 15:54
  • $\begingroup$ The first equation you've provided for $V$ is not the most general. @Lipschitz asks for the most general case. $\endgroup$ – Mo_ Aug 12 '13 at 17:43
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    $\begingroup$ @Mostafa: that is the most general solution to Maxwell's equations in the static gauge in the case where $\rho$ is time-independent, which is the only reasonable way of answering the question. It's certainly the most general solution to Poisson's equation. If you want to write your own answer that tackles the time-dependence and a general gauge, have at it. That's way beyond the scope of the question, and Jefemienko's equations aren't very enlightening. Downvoting over that is petty. $\endgroup$ – Jerry Schirmer Aug 12 '13 at 17:52

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