Find the electrostatic potential due to a hoop of charge

[wam_mi]

Homework Statement

(Part 1) A hoop of charge of radius k lies in the y, z plane, centred on the x axis, so that
it occupies the points (0, y, z) with y^2 + z^2 = k^2. If the (linear) charge density
in the hoop is j, calculate the electrostatic potential Fi at all points on the
x axis, and show that, far from the hoop on the x axis, Fi (x, 0, 0) = (2*pi*j)/|x|.
Explain briefly why this result was only to be expected.

(Part 2) More generally, what do you expect the leading behaviour of Fi(r) to be, far
from the hoop? (You do not need to give any detailed calculations.) Use your
answer to deduce the limiting forms of the equipotential surfaces and field lines,
again far from the hoop.

Homework Equations

The Attempt at a Solution

(Part 1)

Total Charge Q = Integrating j * dl with interval [0 , 2*pi]
= 2*pi*k*j

By Symmetry, Fi (x) = 2*pi*k*j / |x| (Is this wrong?)

And now I'm stuck for the rest of the question.

Please help! Thank you!

 

[chrisk]

Integrating the charge elements and using symmetry is involved for solving this problem.

$$E = E_x =\int\mbox{dEcos}\theta\mbox{d}\theta$$

where

$$dE =\frac{Kd\rho}{r^2}$$

The sine components of E cancel due to symmetry, and

$$cos\theta = \frac{x}{\sqrt{k^2+x^2}}$$

The charge element is

$$d\rho = jkd\phi \ \mbox{and}\ d\phi \ \mbox{ is integrated from 0 to} \ 2\pi$$

Once E is found, the potential is found from integrating E from infinity to x along the x axis.