Electric potential and conservation of energy

[maiad]

Homework Statement

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?

Homework Equations

ΔU=qΔV

q=λL

ΔU=0.5mv^2

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?

 

[vela]

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

 

[maiad]

i used v=ke∫dq/r

 

[vela]

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.

 

[maiad]

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

 

[vela]

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

 

[maiad]

The one found in part one

 

[vela]

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.

 

[maiad]

Why does V depend on the charge found on the ring but the change in energy we use the elementary charge for an electron?

 

[vela]

The force on the electron depends on both its charge and the charge on the ring; hence, the work required to move the electron must depend on both charges.

The idea of the potential is to isolate the effect of the charge of the ring, to keep it separate from the charge of the particle. Why bother? It turns out the concept makes calculations more convenient because electrical charges come in two signs.