Two long metal plates of length L>>H (their height) intersect each other at right angles. Their cross section is a cross with each line of length H. This configuration is held at a potential V=Vo and the total charge up to a distance d (d<<H) from the center of the cross is given as q. Use separation of variables in 2D to find the near field potential for distances abs(x), abs(y) less than/equal to d from the center of the cross. How is the charge distributed from the center up to this distance d?
In class we solved this problem for one 'quadrant' of the open space of the cross:
V = -[Q/(εoL^2)]xy+Vo
By that I mean the rest of the cross was not there, just 2 conducting planes connected at 90 degrees.
The Attempt at a Solution
I was thinking I could use principle of super position here, combined with the fact that if I set the line where the 2 metal plates as my 'z' axis, that my signs of x and y will alternate as I go around the cross. This gave me a result of 4Vo when I added them all together. I'm not entirely sure that is right. Also my professor mentioned something about the sign of the electric field changing from 'quadrant' to 'quadrant'... this also would seem to suggest I have made an error in my thinking.
Any suggestions or tips?