Charge Distribution on Metal Sheet

[Chestermiller]

Yes. Hats off to Chestermiller.

Thanks TSny. You did a beautiful job of illustrating and articulating what I was trying to say. I might also add that, initially, I was unable to figure out how to do this, and it wasn't until I received the hints from yourself and Tiny Tim that I was able to put this analysis together. So, in reality, this was really a team effort.

Chet

 

[TSny]

My only concern is applying Gauss' Law. For that, it had to be known that the electric field tends to zero faster then 1/r² (r is the distance from the charge in the quarter space).

I think that for the limit of infinite plates, you can argue that the field in the first quadrant for the left figure is the same as in the middle figure where the other three charges have been removed. That's because for infinite plates, the region of the first quadrant alone is a well-defined boundary valued problem. The solution of the field for the first quadrant can then be obtained by the method of images and would be the same field as produced in the first quadrant by the four point charges alone of the figure on the right. This is a quadrupole field that will fall off rapidly with distance.

But, maybe I'm overlooking something.

I've spent about ten times as much time trying to think through this problem without integration as it took to just do the integration. But it's fun.

 

[Saitama]

I am still going through all the replies in this thread, thanks a lot everyone! :)

That was an information overload for me this time. [/size]

 

[tiny-tim]

Thanks TSny. You did a beautiful job of illustrating and articulating what I was trying to say. I might also add that, initially, I was unable to figure out how to do this, and it wasn't until I received the hints from yourself and Tiny Tim …

yes, thanks TSny!

for the record, i had no idea until i saw TSny's hint at post #26!

 

[rude man]

Talk about a late entry! But, I'd go with:

1. Consider the two plates as one.
2. Image the charge below the plate at z = -d.
3. Compute the E field all along the plate based on the removal of the plate and its substitution by the (negative) image charge.
4. Use σ = ε₀E to get σ(x,y).
5. Integrate σ to get the charge distribution along the plate, then divide it up per the location of Q.

Bottom line, I would base my approach on the image technique. I don't know to what extent this was done by all the other posters.