# Homework Help: Ring of Charge, Electric Potential

1. Nov 8, 2012

### fornax

1. The problem statement, all variables and given/known data
A proton is moving along the main axis of a uniformly charged thin ring. The charge density on the ring is 5.0nC/cm and the ring radius is 1.0cm. Initially the proton is 2.0cm (along the axis) from the center of the ring with the velocity towards the center of the ring. What initial speed should the proton have such it will cross the plane of the ring?

2. Relevant equations
ΔU = ΔV*q
ΔKE + ΔU = 0
Vcenter = KQ/R
W = -ΔU

3. The attempt at a solution

50nC/cm = 500nC/m

Vcenter = KQ/R
V = ((9x10^9)*(500x10^-9))/.01^2
V = 4.5x10^7

ok, looking at my paper, I then went on to multiply V*d, and I'm not sure why anymore...
I can't seem to find the correct equation for the V on the axis in my notes. I could use V*q for ΔU then user ΔU to ger ΔKE(1/2mv^2) but that gets me velocity if it accelerated toward the ring... ugh.

If I had V at .02m, the I could get ΔV between the center and there, then get ΔU, assume ΔKE is zero at the center, then get the velcoity at .02m with ΔU = ΔKE. Am I reasoning this correctly? Also, mind sharing that formula? :)

2. Nov 8, 2012

### collinsmark

That looks about right to me so far, excluding the typo in bold and crossed out. (5.0 nC/cm, not 50 nC/cm)
Okay, above in red is 'K'. So far so good.
Above in green is supposed to be the total charge 'Q'. But 500 nC/m is the charge per unit length. You need to multiply that by the length of the ring to get the total charge [Edit: By 'length' here, I am referring to the ring's 'circumference'].
Why are you squaring the distance?

(Your goal is to find the electric potential, not the electric force.)
Nor am I. :uhh:
That would give you the potential energy with respect to infinity.
That's exactly the correct reasoning (if I'm understanding you correctly).

(Just make sure that when you calculate the electric potential at a distance of 0.02 m [from the center of the ring], realize that the distance to the charge involves the Pythagorean theorem. )
The Pythagorean theorem, I'm sure you already know. You already have the equation for electrical potential (with respect to infinity) in your relevant equations.

Last edited: Nov 8, 2012