Work required to move a charged ball

[Black Armadillo]

Homework Statement

A ring of diameter 7.10 cm is fixed in place and carries a charge of 5.00 μC uniformly spread over its circumference. How much work does it take to move a tiny 3.40 μC charged ball of mass 1.70 g from very far away to the center of the ring?

Homework Equations

I tried using: W=-\Delta U

The Attempt at a Solution

We know that when the ball is very far away U=0J

Then to find the potential energy when the ball is in the center of the ring I did:
Change in charge is 5.00X10^{-9}C/2\pi(0.0710m)=1.12X10^{-8}C/m

\int \frac{k*q*dq}{r} ds from 0 to 2pi.

\frac{k*q*dq}{r} \int ds from 0 to 2pi.

\frac{8.988X10^{9}N*m^2/C^2*3.40X10^{-9}C*1.12*10^{-8}C/m}{0.0710m} \int ds from 0 to 2pi =3.031X10^-5J.

W=-\Delta U=-(3.031X10^-5J-0J)=-3.031X10^-5J

 

[Delphi51]

I'm having trouble following your integral from the first line. I'm thinking V = kq/r for a point charge so dV = k/r*dq, where r is the radius of the ring.
Ah, you are doing energy E = QV so dE = QdV = kQ/r*dq
I'm using q for the charge on the ring, Q for the charge on the particle.
Your s must be something to do with the arc length going round the ring.
So dq = q/(2*pi*r)*ds
Using s = r*A where A is the angle running from 0 to 2pi, you get ds = r*dA and
dq = q/(2*pi)dA
so the integral is dE = kQq/(2*pi*r)*dA from A = 0 to 2*pi.
The nice thing about this integral is that nothing depends on the angle so
integral 0 to 2*pi of dA is just 2*pi.
E = kQq/r.
Well, I guess we didn't need to integrate at all! Because all the charge is the same distance from the center point.

 

[Black Armadillo]

I'm still slightly confused. So I can find E at the center of the ring using E=kQq/r. But, since W=-(U_f-U_i) how do I get U from E? Since E=kq/r^2 and U=kq_1*q_2/r would you just multiply the E you get by qr and if so what q would you use: the charge on the ring or the charge on the ball? Thanks for your help.

 

[chrisk]

Try this approach. Find E as a function of r along the axis of the ring using integration. Then

V=-\int_{\infty}^{r}\mbox{Edr}

The work done is

W=Vq

Is the mass in a gravitational field also and if so what is the orientation of the ring axis with the field?