# Two dimensional hyperbolic potential has no charge density but a surface charge density?

I am learning about electric potential, and I am confused about what exactly is going on in the following situation:

Say we have a potential $$V =- V_0 \frac{xy}{a^2}$$

This means that $$\nabla^2V=0$$. So there is no volume charge density.

However, the surface charge density is $$\epsilon_0 E$$ which is proportional to $$\sqrt{x^2+y^2}$$ because the electric field is $$\frac{V_0 y}{a^2} \mathbf{\hat x} + \frac{V_0 x}{a^2} \mathbf{\hat y}$$.

How can surface charge density be increasing as one moves further from the origin if volume charge density is non-existent?

What I interpreted this as was if each equipotential is treated as an thin ideal conductor in the x-y plane, there would be electric fields going into and out of the conductor but not through the conductor. The region between surfaces is neutral and has no charge density. However, because there is a field around the conductor, there must be a surface charge density. But if the conductors are infinitesimally thin, shouldn't there be a steady increase in charge density as one goes further from the origin?

• What is this "surface charge" that you mentioned? Also, what are the relevant boundary conditions? Since you've already found $\nabla^2 V = 0$, the field must be coming from charge at the boundaries. Commented Sep 23, 2018 at 6:56