# What is the potential $\phi(x,y)$ inside quadrupole/Paul trap?

Suppose we have a configuration like the figure below.

I'm asked to compute the potential $$\phi (x,y)$$ inside the trap.

I'm attempting this by solving the Laplace equation $$\nabla^2 \phi = 0$$ because there is no charge in the interior. I'm doing this change of variables:

$$u= xy$$ ; $$v= y$$

And I'm impossing this boundary conditions:

$$\phi (u=1,y) = 5$$ and $$\phi (u=-1,y) = -5$$

$$\phi(u,-v)=-\phi(u,v)$$

However, I'm not able to solve the resulting PDE by separation of variables.

Doing some try and error with the boundary conditions, it seems that $$\phi(x,y)=5xy$$ solves the problem. But that's no rigorous at all.

I want to understand a mathematical rigorous way to get this same potential. In general, I'm getting problems trying to solve Laplace equation in non spherical, cylindrical nor plane symmetry.

• What is the resulting PDE, acc. you? – Gert Sep 20 '20 at 17:54