Velocity of charge orbiting infinite line of negative charg

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
3 replies · 2K views
space-time
Messages
218
Reaction score
4
Sorry I couldn't finish the title. I ran out of space. Anyway, here's the question:

A uniformly charged, infinitely long line of negative charge has a linear charge density of -λ and is located on the z axis. A small positively charged particle that has a mass m and a charge q is in circular orbit of radius R in the xy plane centered on the line of charge. (Use the following as necessary: k, q, m, R, and λ.)

(a) Derive an expression for the speed of the particle.

(b) Obtain an expression for the period of the particle's orbit.

Relevant Equations:

E = λ/(2πrε0) for an infinite line of charge. In the case of this problem I used -λ instead of just λ.

F = Eq = mv2/r

T = (2π)/ω = (2π)/(v/r) = (2πr)/v
I tried using the first equation I listed above in order to derive E. This led to:

E = -λ/(2πRε0)
F = Eq = mv2/R , so this leads to:

F = -λq/(2πRε0) = mv2/R

solving algebraically for v yields:

v = sqrt(-λq/(2πmε0)) = sqrt(-λkq / (2m) )

Here I thought I had derived v, but webassign said that this was wrong. I tried taking out the negative sign since having an imaginary velocity makes no sense, but it was still wrong. I can't do (b) until I solve (a). Please help.
 
Physics news on Phys.org
you expression for centripetal acceleratio should have a minus sign, since it points inwards.
remember: vectors.

note: $$\frac{1}{2\pi\epsilon_0} = 2k$$
 
  • Like
Likes   Reactions: space-time
Simon Bridge said:
you expression for centripetal acceleratio should have a minus sign, since it points inwards.
remember: vectors.

note: $$\frac{1}{2\pi\epsilon_0} = 2k$$
Thanks! I solved it now.