# Electrostatics problem using Green's function

I am trying to solve the following: $$\vec \nabla ^2\Phi=\delta(x-x_0)\delta (y-y_0)$$ where $\Phi$ is a scalar function of the form $\Phi(x,y)$. Boundary conditions are $\Phi(0,y)=\Phi(x,0)=0$. The exercise is to find the potential in the area $x>0,y>0$ that arises due to two infinite electric grounded planes lying perpendicular one the other on the $x-z$ and the $y-z$ planes, with the following charge distribution: $$\rho(\vec r)=\frac{q}{l} \frac{\text{Coulomb}}{\text{meter}}\delta(x-x_0)\delta(y-y_0)$$ (A charged wire parallel to the z axis)

How do I find $\Phi$?

Green's function in 2-d is $\frac{1}{2\pi} \ln r$, but is it a 2-d problem or 3-d?

Once choosing the correct function, how is the solution carried out? Should I solve for $$\vec \nabla ^2G(x,y,x_0,y_0)=0$$or for

$$\vec \nabla ^2G(x,y,x_0,y_0)=\frac{\rho(x,y)}{\epsilon_0}?$$ Any hint would be helpful.

• Is $\Phi$ a function $\Phi(x,y)$ or is it $\Phi(x,y,z)$? In the first case you have a point charge, in the second case you have a line charge. The situation is different between the two. – Ian Aug 10 '16 at 15:44
• It has no dependence on $z$ because of the geometry of the later described setup. – E Be Aug 10 '16 at 15:45
• I understand that, but that comes from the solution, it's not intrinsic. Intrinsically $\Phi$ depends on all your spatial variables, of which there are three. – Ian Aug 10 '16 at 16:11
• I agree with you. – E Be Aug 10 '16 at 16:15
• In that case, the relevant Green's function is 3D, i.e. the solution to $\nabla^2 \Phi = \delta(x) \delta(y) \delta(z)$. – Ian Aug 10 '16 at 16:22

If you have an infinite wire, you should try to use cilindrical coordinates insted of cartesian so the solution is $1/\rho$ where $\rho^2=x^2+y^2$. Then you should think an appropiated cutoff, like multiplying by an appropiated Heaviside function or stuff like that. The book of Jackson-Classical Electrodynamics has a lot of examples and a good mathematical and physical approach