02-03-2003 08:49									#1
	Do my HW part 45 Increase Your PM Length

these threads are farily common so I am sure we are to 45 now

now I want someone to do my homeowkr so I can screw off on apolyton

first thing needed is the book Jackson

so everyone who has that (and is willing to do a problem ro two) chime in

than Iw ill tell you the problem numbers

Jon Miller

	02-03-2003 08:50									#2
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hurry though, soon there will be no new threads to go to and I will go back and take a derivative and be one problem closer to done

Jon Miller

	02-03-2003 08:54									#3
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The answer is Pi/2.

	02-03-2003 08:54									#4
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Or maybe it's e^(i * Pi). I always forget the difference between the two.

	02-03-2003 08:58									#5
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W00t, not 42?

	02-03-2003 08:59									#6
	Remove this text

probably youa re right (to a few constants)

but I really need the solution, not the answer

Jon Miller

	02-03-2003 09:07									#7
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D'oh, I hate having to show my work on assignments. It makes it a lot tougher to bullshit.

	02-03-2003 09:09									#8
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I have Griffiths.

	02-03-2003 09:11									#9
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quote: Originally posted by loinburger D'oh, I hate having to show my work on assignments. It makes it a lot tougher to bullshit.

Yeah, but if you don't you won't get partial credits.

	02-03-2003 09:12									#10
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if you don't you don't get any credit

Jon Miller
the voice of expereince

	02-03-2003 09:31									#11
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quote: Originally posted by Ramo I have Griffiths.

It's like Jackson, minus all the hard parts.

	02-03-2003 09:32									#12
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I don't have Jackson on me. Any problems you're willing to transmit via writing?

	02-03-2003 09:43									#13
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thanks for the offer

I will post any that are getting me (assuming I get back to working on them)

Jon Miller

	02-03-2003 09:45									#14
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I can't guarantee I'll be able to do them. I'm sometimes incredible at EM and sometimes absolutely useless.

	02-03-2003 09:46									#15
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quote: Originally posted by Urban Ranger Yeah, but if you don't you won't get partial credits.

I just scribble something that might be interpreted as a bunch of equations. My handwriting's so atrocious that I'm usually given the benefit of the doubt, since it's not like they can read my writing any other time...

	02-03-2003 09:56									#16
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I get -100% for handwriting fairly often (when I am right damnit)

Jon Miller

	02-03-2003 09:59									#17
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I have only once had an anal bastard reject my work due to handwriting.

On an Algebra III assignment that took me 10 hours, no less.

	02-03-2003 10:09									#18
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that is horrid

I have to agree though, my HW is bad (he only takes off 100% for the problem, he has threatened to just reject my assignments though)

Jon Miller
who is currently spending a minute on HW for every 5 on apolyton

	02-03-2003 11:04									#19
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heyt Frogger (or anyone else)

what is a boss?

I am suppose to work with a boss on a plane conductor

Jon Miller

	02-03-2003 11:20									#20
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Just a bump. I bet you it's a hemispherical bump of radius "a"...

	02-03-2003 11:26									#21
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yah

but in or out?

JHon Miller

	02-03-2003 11:29									#22
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They have to tell you that.

	02-03-2003 11:30									#23
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probably out...

Jon Miller

	02-03-2003 12:31									#24
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hmm

if it is out

and you had a point charge right above this bump on the conductor

where would yuo place image charges (that is what I am thinking to use)?

I was thinking of just usning the plane, and the sphere placements

but that doesn't seem right

Jon Miller

	02-03-2003 12:38									#25
	Increase the size of your Attachments

The question is what? charge density on surface of conductor?

	02-03-2003 12:42									#26
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yah

Jon Miller

	02-03-2003 12:42									#27
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That should work. The "hole" in the conducting plane is okay since the plane image charge method works for finite conducting planes as well...

	02-03-2003 12:42									#28
	Lose 30 kilos (of popups)

no, on the bump

Jon Miller

	02-03-2003 12:44									#29
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ok, hmm, wonder why it is not coming out

thanks

Jon Miller

	02-03-2003 12:46									#30
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Image charges always seemed like hoodoo to me.

Just make sure that the resulting image charge configuration gives you the right boundary condition on the conducting plane (i.e. that V is constant on the surface)...