I went to the tutoring center at my school. Twice. Finally was able to see the professor in the late afternoon. He checked the problem; it's got the wrong solution attached. Thank you everyone, you helped a lot; I was doing it right, -20.104 is the answer, but the site was wrong. It's a relief...
I'm sure I have the units correct. It wants nC/m2; I'm given 6 nC/m on the inner cylinder.
This makes me wonder if, perhaps, I've been calculating sigma on the inner cylinder, rather than the inner surface of the outer cylinder. Still, for the latter I'd need the charge of the inner cylinder...
I went to my school's tutoring center, and they said everything looked correct, as well.
They helped me see more clearly that length doesn't matter. Equating (sigma)(area) = (lambda)(length) removes the length and leaves (sigma) = (lambda)/2(pi)r^2
-20.103 is incorrect, though; the...
All right... I did notice that and was a bit confused as to how to use that information. I suppose that means there's a way to get the charge Q given (lambda) (where (lambda) = Q/L ). To me it seems that Q = 6 nC, if I'm to take an arbitrary L of 1 m, so Qin = -6 nC. But when I do that...
I thought so, too, so I tried L = 1 m.
(sigma) = -6 nC / 2(pi)rL m^2 = -3 / (pi)x0.0475x1 m^2 = -20.1038 nC/m^2
However, that's apparently incorrect.
Unless I'm missing something...
Homework Statement
The figure shows a portion of an infinitely long, concentric cable in cross section. The inner conductor carries a charge of 6 nC/m and the outer conductor is uncharged.
(part 5 of 6)
What is the surface charge density inside the hollow cylinder?
Answer in units...