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## Homework Statement

The [itex]z=0[/itex] plane is covered with the uniform surface charge density, [itex]\rho_s=10^{-9}[/itex], while at the same time the point charge [itex]q=10^{-9}[/itex] is located at (0,0,1). Where can an electron be place so if released, it will not move.

## Homework Equations

[itex]\mathbf{E}(\mathbf{r})=\frac{1}{4\pi\varepsilon_0}\sum_{n=1}^N\frac{q_n(\mathbf{r}-\mathbf{r}_n)}{\left|\mathbf{r}-\mathbf{r}_n\right|^3}[/itex]

[itex]\mathbf{F}=q\mathbf{E}[/itex]

## The Attempt at a Solution

I assume that the forces will net to zero,

[itex]\mathbf{F}=e\mathbf{E}_{\rho_s}+e\mathbf{E}_q=0[/itex]

[itex]\therefore\mathbf{E}_{\rho_s}=-\mathbf{E}_q[/itex]

I know that, due to symmetry, the surface charge only acts in the z-direction. Since the point charge is given to be at (0,0,1), I expect the solution to be of the form (0,0,m). I tried using the electric field of the surface charge to be

[itex]\mathbf{E}_{\rho_s}=\frac{\rho_s}{2\varepsilon_0}[/itex]

but I am not sure that that is acceptable since there is a charge in this problem and that answer is for just the plane alone.

Not sure where to start really, other than the force equations, but after that I'm stuck as to how to go about solving this problem, any and all help is appreciated!