Do my HW part 45

[Jon Miller]

I think you need something else to cancel the effects of the charge that sets the hemisphere out (on the plane)

Jon Miller

[KrazyHorse]

I don't think so...

Hmm..

Is the point charge right on the bump, or some distance above?

[Jon Miller]

distance above

Jon Miller

[KrazyHorse]

I've got the answer.

Took a bit of skull scratching. Hold while I type it out...

[KrazyHorse]

Oops. Maybe not.

[Jon Miller]

I am thinking of calling it a night anyway

I did a proof and surfed apolyton while thinking about that problem

it should be easier than this

Jon Miller

[Jon Miller]

8 problems, damn

Jon Miller

[Jon Miller]

(what are you talking this semester?)

Jon Miller

[Ramo]

I think I know the solution for the bump alone (put a charge of rq/r' a distance r²/r' away from the bottom of the bump where q is the original charge, r is the radius of the bump, and r' is the distance of the charge to the plane), but I can't figure out how to handle the non-bumped part...

[Jon Miller]

yah

that is my problem

you can't fit the two solutions together (plane and bump)

Jon Miller

[Jon Miller]

I acutally haven't been working on it intensly (mostly posting on apolyton as well as working on a couple of other problems)

going home now, will return in 40 minutes
(if anyone has looked at this farther)

Jon Miller

[KrazyHorse]

I'm thinking it'll end up being 3 image charges:

1 like standard plane (-q at distance d on the other side of the plane) and two equal but opposite charges q' and -q' mirroring each other about the centre of the bump at a distance b. The solution that way looks nasty, though. But it's better than my last solution, which involved an infinite number of image charges...

[Jon Miller]

hmm

yah, a 3rd charge sounds right (to cancel)

I will work it out as I bike back home

Jon Miller

[Jon Miller]

which will be in a bit (it looks like the streets are rivers out therE)

Jon Miller

[Jon Miller]

it's Jackson, nasty is ok

Jon Miller