# Potential of circular electric field

Suppose, in a 2D plane, there is an electric field: $$\vec{E}(x,y)=E_0\begin{pmatrix}-y \\ +x\end{pmatrix}=E_0(-y\vec{e}_x+x\vec{e}_y)$$ $$\vec{E}(r,\theta)=E_0 r \vec{e}_\theta$$ where $r,\theta$ are polar coordinates. This electric field goes around in a circle and closes in itself. What would the potential, defined through $\vec{E}=-\nabla V$, be as a function of $\theta$ at a given radius from the origin?

If we define the $-\infty<\theta<+\infty$, it would be a simple linear function in $\theta$: $V(r,\theta)=-E_0 r \theta$. But if we restrict $0\leq\theta<2\pi$, surely it would become a matter of convention, whether it is a single-valued function (therefore leading to a discontinuity in $\vec{E}$) or a multi-valued function?

I think you're missing something very important here. The electric field you definied , $\vec{E} = E_0(-y,x)$ is not conservative, for $\vec{E}$ to be conservative you need $\oint\vec{E}d\vec{l}$ to be $0$, which is not the case because $\frac{\partial(x)}{\partial{x}} \neq \frac{\partial(-y)}{\partial{y}}$ (Green's theorem). So you cannot assign a scalar potential V to $\vec{E}$, that's probably why you are having problems.
Also note that you would get a similar looking magnetic field $\vec{B}$ when there was a current along the z-axis. Ampère's circuital law states that: $$\oint_C\vec{B}.d\vec{l} = \mu_0I \neq 0$$ So you can't assign a scaler potential to the magnetic field as well.