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I've been studying the Laplace equation for zero charge distribution:

$$\nabla^2 V = 0$$

and I know that this is the general solution to the Laplace equation for an azimuthally symmetric distribution in spherical coordinates:

$$V(r, \theta) = \sum_{\ell=0} (A_\ell r^\ell + B_\ell r^{-(\ell + 1)} ) P_\ell (\cos(\theta))$$

So I was thinking, let's say that I also have symmetry in the $\theta$ angle. So, this must mean that $\ell = 0$ so that the $\cos(\theta)$ terms disappear, so the potential is now:

$$V(r) = A + \frac{B}{r}$$

Now let me think about a spherically symmetric and constant electric field $\vec{E} = E_0 \hat{r}$. This must arise from the potential $V' = - E_0 r + V_0$ since $\vec{\nabla} r = \hat{r}$. But this doesn't match the form of the solution $V(r)$ from the Laplace Eq. sol'n. Does this imply that I cannot have a spherically symmetric constant electric field without charge at the center? I may just be overthinking this, so let me know. Thank you.

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    $\begingroup$ Yes, for that you'll need some charge density $\rho$ everywhere - but spherical symmetry makes it is easy to find out $\rho(r)$. $\endgroup$ Nov 21 '18 at 4:04
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That field requires a particular nonzero charge density throughout all of space. You can compute it from one of Maxwell's equations,

$$\vec{\nabla}\cdot\vec{E}=4\pi\rho.$$

In spherical coordinates, the divergence is

$$\vec{\nabla}\cdot\vec{E}=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2 E_r)+\text{angular terms}$$

So the charge density required to produce the field $\vec{E}=E_0\hat{r}$, where $E_r=E_0$, is

$$\rho=\frac{1}{4\pi}\frac{1}{r^2}\frac{\partial}{\partial r}(r^2 E_0)=\frac{E_0}{2\pi r}.$$

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  • $\begingroup$ That totally makes sense. So then I guess it's just impossible to have an empty cavity with a spherically symmetric constant electric field? You need a charge distribution in there right? $\endgroup$ Nov 21 '18 at 4:24
  • $\begingroup$ Yes, that's right. $\endgroup$
    – G. Smith
    Nov 21 '18 at 4:29

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