Potential and total charge on plane

[Yoni V]

Homework Statement

An infinite plane in z=0 is held in potential 0, except a square sheet -2a<x,y<2a which is held in potential V.
Above it in z=d there is a grounded plane. Find:
a) the potential in 0<z<d?
b) the total induced charge on the z=0 plane.

Homework Equations

Green's function for a plane + Green's Theorem for the potential; method of images etc.

The Attempt at a Solution

a) First, I put an image square sheet on z=2d, then solved by superposition, calculating the potential for the z=0 plane and then adding a shifted result.
For the z=0 plane, ##G = \frac{1}{|r-r'|}+\frac{1}{|r-r''|}## where ##r''=(x',y',-z')##, and due to Dirichle b.c. we get $$\phi_1(r)=-\frac{1}{4\pi}\iint\phi(r')\frac{\partial G}{\partial n'}dS=\frac{Vz}{2\pi}\iint_{-2a}^{2a}\frac{dx'dy'}{((x-x')^2+(y-y')^2+z^2)^{3/2}}$$ where the integrals are from -2a to 2a. This integral is a real pain, so I left it aside for now. I'm not sure, but we might be allowed to leave it in integral form. But still, is there a decent way to calculate it? Finally, $$\phi_2=\frac{V(2d-z)}{2\pi}\iint_{-2a}^{2a}\frac{dx'dy'}{((x-x')^2+(y-y')^2+(2d-z)^2)^{3/2}}$$ and ##\phi=\phi_1+\phi_2##. Thinking more about this integral, I thought maybe I could solve it using separation of variables, which I haven't tried yet, but I don't think it would make my life any easier in getting a closed form which doesn't contain series expansions or integrals.

b) Here I'm kinda stuck... The induced charge can be attributed to the image square sheet. But now I find myself not knowing how to calculate total charge on a sheet given only its size and potential V. And also, I'm not sure if this is correct, since this is not a closed surface. If it were, I'd say from Gauss' Law we can replace the surface charge distribution with point charges inside (or outside depending on the original arrangement) the surface. Here, my equivalent argument is placing yet another image sheet in z=-2d with potential V to cancel out the potential on z=0 due to the sheet on z=2d.

Summing up, any advice on the following would be greatly appreciated:
1. Is there a more elegant way to solve (a), or a way to solve its integral?
2. Is my argument in (b) about the induced charge of the plane correct?
3. How can I calculate the charge given this potential?

Thanks!

 

[anathema]

1. There are a few different ways you could approach solving the integral in (a). One option would be to use a computer program or calculator to numerically evaluate the integral. Another option would be to use the method of separation of variables, as you mentioned. You could also try using a change of variables or making a substitution to simplify the integral. It may take some trial and error to find the most efficient method for this particular integral.

2. Your argument in (b) seems reasonable. By placing an image sheet in z=-2d with potential V, you are effectively creating a mirror image of the original sheet in z=0. This will cancel out the potential on z=0 due to the sheet on z=2d, resulting in a net potential of 0 on the z=0 plane. This means that there is no induced charge on the z=0 plane, as all of the charges are canceled out by the presence of the image sheet.

3. To calculate the charge on the original sheet, you could use the method of images again. By placing a charge of -V on the image sheet in z=-2d, you are effectively creating a mirror image of the original sheet in z=0 with opposite charge. This charge will induce a charge on the original sheet, which you can calculate using Gauss' Law. You can then use this induced charge to calculate the total charge on the original sheet.