Homework Help: Work, A Charged Ring, etc

1. Oct 12, 2009

BustedBreaks

A charge of 10 nC is uniformly distributed around a ring of radius 10 cm that has its center at the origin and its axis along the x axis. A point charge of 1 nC is located at x = 1.75 m. Find the work required to move the point charge to the origin. Give your answer in both joules and electron volts.

I'm using this equation, but I'm I keep getting it wrong. $$U=\frac{kqQ}{\sqrt{a^{2}+z^{2}}}$$

a=10 cm
z=1.75 m
q=1 nC
Q=10 nC

2. Oct 12, 2009

RoyalCat

Look at the units of your equation.
It is in Newtons*Meters

What are the units of electrical potential? How would you adjust your equation to compensate, or what do you think it DOES describe that is of use to you?

3. Oct 12, 2009

BustedBreaks

I guess what's confusing me is that it is asking for an answer in Joules. Isn't electrical potential N*m/C? So I should divide by a charge for potential?

4. Oct 12, 2009

RoyalCat

Exactly! What you have there is an expression for the potential energy of a particle of charge $$q$$

$$1 V = 1\frac{J}{C}=1\frac{N\cdot m}{C}$$

I do, however, think that your equation doesn't have anything to do with the potential energy of a particle in the system described? How did you get it?

Try finding the field at that point, and the field at the origin (The center of the ring) and use the fields there to find the electrical potential. Using the potential, you can find the work that the question asks for.

For the two dimensional case:

$$U_{AB}=\int^{A}_{B} E \cdot dr$$

Last edited: Oct 12, 2009
5. Oct 12, 2009

BustedBreaks

My original equation, $$U=\frac{kqQ}{\sqrt{a^{2}+z^{2}}}$$ gives units N*m which is Joules, which is what the question asks for... Why is that wrong?

The equation you gave at the bottom won't give Joules if I Integrate E for a charged ring, $$E=\frac{KQz}{(z^{2}+a^{2})^{\frac{3}{2}}}$$

I would have to multiply that by a charge, maybe q, to get units of work...

Still not working for me.