# Electric potential and conservation of energy

1. Jan 26, 2012

1. The problem statement, all variables and given/known data
An electron is released from rest on the axis of a uniform positively charged ring, 0.174 m from the ring's center. If the linear charge density of the ring is +0.150 nC/ m and the radius of the ring is 0.348 m, how fast will the electron be moving when it reaches the center of the ring?

2. Relevant equations

ΔU=qΔV

q=λL

ΔU=0.5mv^2

3. The attempt at a solution
I've tried to find the electric potential at each point,the charge of the ring using q=λL where L i found by using 2∏r. From that, i took the difference between the electric potentials and plugged it into the first equation and equated that to the change in kinetic energy and solved for v but it's not right. can someone see what i did wrong?

2. Jan 26, 2012

### vela

Staff Emeritus
What formula did you use to calculate the potential? You're probably using the wrong one.

3. Jan 26, 2012

i used v=ke∫dq/r

4. Jan 26, 2012

### vela

Staff Emeritus
It sounds like you have the right approach, but we can't pinpoint your mistake without seeing what you actually did. Post the details of your work.

5. Jan 26, 2012

1.q=λL (L=2∏r)
2.V1=Ke q/√(r^2+x^2)
3.V2=Ke q/√(r)
4.ΔV=V2-V1
5.ΔU=qΔV=0.5mv^2

6. Jan 26, 2012

### vela

Staff Emeritus
In step 5, what value did you use for q?

7. Jan 26, 2012

The one found in part one

8. Jan 26, 2012

### vela

Staff Emeritus
OK, that's your mistake. The potential V depends on the q you found in part 1, but the potential energy ΔU is that of the electron so you need to use the elementary charge e. That is, ΔU = eΔV.

9. Jan 26, 2012