# Homework Help: Weird charge distribution

1. Apr 14, 2012

### weaver159

1. The problem statement, all variables and given/known data
We have a infinite plate on the yz plane from $x=-d/2$ to $x=d/2$. The plate has a uniform volume charge distribution $ρ_{0}$. Parallel to the z axis at $y=y_{0}$ we have a cylindrical hole with a radius $a$. At the center of the hole (paralle to the z-axis) we have an infinite line distribution $λ_{0}$.
We need to find the field everywhere and the condition that $λ_{0}$, $ρ_{0}$ must satisfy in order to have zero field outside the hole.

2. Relevant equations
Gauss's law and the boundary contitions for $E,D$

3. The attempt at a solution
My first though was offcourse the superpossition principal. I found a problem using it:

The field inside the hole doesn't match the field from an infinite line, as it supposed to.

2. Apr 15, 2012

### rude man

What is the x-coordinate of the hole's central axis? Is it x = 0? Not sure it matters, but I think it does.

3. Apr 15, 2012

### weaver159

Yep. My mistake. It is at x = 0.

4. Apr 15, 2012

### rude man

I'd go with superposition all right.

1. Sheet without the hole and line charge.
2. Right circular cylinder of where the hole is, charge density the negative of the charge density of the sheet.
3. Just the line charge by itself.
4. Add the whole business.
5. Invoke the requirement of zero E field outside the hole.

5. Apr 15, 2012

### weaver159

Thanks for the answer.
That's exactly what I did, but I have a small concern.
The field inside the hole should be $λ_{0}/2πε_{0}r$, because all we have in the hole is the line distribution. That is not result I get when I add the field from the above 3 distinct distributions. Do I miss anything?

*It is not actually a sheet but a plate, but that I think has minimal effect on the methodology.

6. Apr 16, 2012

### rude man

The fied inside the hole is not just due to the line charge. It's also due to the plate's charge distribution. The field in the hole is due to three separate charge distributions, as I outlined.

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