# Induced charge due to charge distribution between two plates

I am trying to work through this problem: given two parallel square conducting plates (side w), one grounded(at x=0) and the other held at +V (at x=L where L < < w), and there is a charge distribution $\rho(\vec x)$ (arbitrary, not given) between them. What is the charge accumulated on each plate, i terms of the moments of the distribution?

My issue comes in trying to find the potential. In general: $$\phi=\frac{1}{4\pi\epsilon_0}\int\frac{\rho(\vec x')}{|x-x'|}d^3x'$$ Which can be expanded in terms of the charge distribution moments $q_{lm}$ $$\phi=\frac{1}{\epsilon_0}\sum_{l,m}\frac{q_{l,m}Y_{lm}}{(2l+1)r^{l+1}}$$

This expression blows up at the origin (i.e the grounded plate), so I am not sure how to account for that. Should I have some kind of sine envelope with this potential, so that it vanishes at x=0? Possibly use a greens function with dirichlet boundary conditions?

• So it is not a capacitor, it is just two conducting plates. There is no capacitance given (eventually want to solve for it). But we need to find the charge on the plates in terms of moments of the charge density – yankeefan11 Mar 13 '16 at 22:13
• First problem is with your geometry, don't use spherical expansions for non-spherical symmetries, since there is no spherical symmetry here. Also, you must have some sort of definition of the charge density, the general case can be quite challenging; even ignoring border effects. Are the plates infinite? More details are certainly needed to solve this problem. – user82635 Mar 13 '16 at 22:21
• So all we have is the charge density rho, the plates are separated by a distance d, and the plates are a square with side w (x>>d), and one is help at ground, the other at V – yankeefan11 Mar 13 '16 at 22:31
• So, to be more clear, is $\rho$ constant? Is the problem from Jackson? – user82635 Mar 13 '16 at 22:33
• $rho$ is arbitrary, could be constant, could not be. This is not from jackson – yankeefan11 Mar 13 '16 at 22:44

Instead of solving Laplace's equation, you need to consider Poisson's equation. The reason for this is that you have free charges available in the enclosed region between two plates. However, you should solve this equation in Cartesian coordinates along with imposing the boundary conditions. For this particular problem, this methodology would be more appropriate than the multipole moment expansion. The singularity you claim for vanishingly small distances will be removed if boundary conditions is applied correctly.

Another way of approaching it would be Green's Function methodology where you need to realize that your boundary conditions are of Dirichlet type.

Take a look at the following documents where particular charge distributions are considered too:

Poisson's Equation in Cartesian Coordinates

Example: Charge filled Parallel Plates

Multipole Moments in Electrostatics

• So I believe (from wikipedia) that the greens function for poissons equation is $$G=[(x-x')^2+(y-y')^2+(z-z')^2]^{-1/2}$$, which means $\phi = \int dV' G(x,y,z,x',y',z')\rho(x',y',z')$ I do not see how this results in something I can find the moments of – yankeefan11 Mar 14 '16 at 5:19
• Or is that G only free space? Seeing as it does not meet any of the BC (which because of dirichlet should be 0 correct?) – yankeefan11 Mar 14 '16 at 5:20
• Why are you insisting on a need to find moments? Do you really need them? Also, is your charge distribution really unknown? – Benjamin Mar 14 '16 at 5:26
• The exact wording is: "A blob of charge of density ρ(x) lies between two square conducting plates of side w that are separated by distance d<<w. One plate is grounded and the other held at potential V. Write expressions for the charge on each plate in terms of moments of the charge density. – yankeefan11 Mar 14 '16 at 5:28
• Added to my response is also another paper (No. 3) which particularly discusses multipole expansion methodology in Cartesian Coordinates. – Benjamin Mar 14 '16 at 5:33