Potential and charge on a 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 thez=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 Dirichlet b.c. we get $$\phi_1(r)=-\frac{1}{4\pi}\iint\phi(r')\frac{\partial G}{\partial n'}dS=\frac{Vz}{2\pi}\iint_{-2a}^{2a}\frac{dx'dy'}{((x-x')^2+(y-y')^2+z^2)^{3/2}}$$ 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!

 

[paralleltransport]

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 thez=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 Dirichlet b.c. we get $$\phi_1(r)=-\frac{1}{4\pi}\iint\phi(r')\frac{\partial G}{\partial n'}dS=\frac{Vz}{2\pi}\iint_{-2a}^{2a}\frac{dx'dy'}{((x-x')^2+(y-y')^2+z^2)^{3/2}}$$ 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!

Hi Yoni,

In order to solve this problem elegantly without integrating green all over the place, I think a starting point would be the method of images. Method of images states that if a particular charge configuration creates a potential that satisfies laplace's equation within some boundary condition, then that's the unique solution within those B.C. So our trick is to create some charge configuration that would create the following things:

1. A grounded plane around z = d: This would happen if the charge distribution was symmetric (+q mirrors to -q) by a z -> d-z transformation.
2. Discontinuous voltage of value V across transition from inside to outside the square. Discontinuous voltage means somehow infinite E field @ those locations, so I assume it would look like a line of charge forming some square.

This is just a guess to get you started, I'm not confident (yet) the problem can be solved elegantly with method of images, and I'm too hungry to finish the solution, but hopefully this helps.