# Is it possible to have a spherically symmetric constant Electric Field $\vec{E} = E_0 \hat{r}$?

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.

• Yes, for that you'll need some charge density $\rho$ everywhere - but spherical symmetry makes it is easy to find out $\rho(r)$. Commented Nov 21, 2018 at 4:04

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

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

• 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? Commented Nov 21, 2018 at 4:24
• Yes, that's right. Commented Nov 21, 2018 at 4:29