What is the Electric Potential of a Uniformly Charged Ring?

[Cantspel]

Homework Statement

A ring of radius a is made from a charge wire with a uniform charge density λ.

a) Calculate the electric potential due to the ring as a function of distance from its center along the axis of the ring passing through the center, perpendicular to its plane

b) If a particle of mass m with a charge identical to that of the ring displaced ever so slightly from the ring's center what speed will the particle eventually attain after being repulsed by the ring.

Homework Equations

V = k∫(1/r)dq[/B]

The Attempt at a Solution

[/B]
So for a I drew the ring here is my best redraw of it in ms paint below. After drawing it i used dq = λ dL than i plug that into V = k∫(1/r)dq and I plug (a² + d²)^1/2 for r. getting me V = k∫(λdL)/(a² + d²)^1/2 getting me
V = (kλL)/(a² + d²)^1/2 since L = 2πa I can replace it in the fuction. So I get V = (2kλπa)/(a² + d²)^1/2

I have to clue weather or not this is right and have no idea how to begin part b.

 

[haruspex]

I have to clue weather or not this is right and have no idea how to begin part b.

You have part a correct.
For part b, think about energy.

 

[Cantspel]

So since V = U/Q I can set it as (2kλπa)/(a² + d²)^1/2 = U/Q since the Q is the same as the one the ring has I can replace Q as λL since L = 2πa, Q = λ2πa So U = (4kπ²λ²a²)/(a² + d²)^1/2 and U = KE = 1/2mv², so v² = (2kπ²λ²a²)/(m(a² + d²)^1/2)
then I take the square of both sides and I get v = (πλa(2k)^1/2)/(m(a² + d²)^1/2)^1/2

 

[haruspex]

So since V = U/Q I can set it as (2kλπa)/(a² + d²)^1/2 = U/Q since the Q is the same as the one the ring has I can replace Q as λL since L = 2πa, Q = λ2πa So U = (4kπ²λ²a²)/(a² + d²)^1/2 and U = KE = 1/2mv², so v² = (2kπ²λ²a²)/(m(a² + d²)^1/2)
then I take the square of both sides and I get v = (πλa(2k)^1/2)/(m(a² + d²)^1/2)^1/2

Where d is...?
I think there may be a factor of 2 error.

 

[Cantspel]

Where d is...?

Can you please elaborate on what you mean? I don't know if you are saying that is the speed at d or if you are asking where d is at.

I think there may be a factor of 2 error.

I see it now the (2k)^1/2 is actually a (8k)^1/2

 

[TSny]

Can you please elaborate on what you mean? I don't know if you are saying that is the speed at d or if you are asking where d is at.

haruspex is pointing out that you never defined what the symbol d represents.

I see it now the (2k)^1/2 is actually a (8k)^1/2

Yes.

 

[haruspex]

haruspex is pointing out that you never defined what the symbol d represents.

Right, but more importantly that d is not given as part of the question, so should not appear in the answer.

 

[Cantspel]

Since d can't be in the solution should I be looking to relate d to a or am I just going about this the wrong way?

 

[haruspex]

Since d can't be in the solution should I be looking to relate d to a or am I just going about this the wrong way?

displaced ever so slightly from the ring's center

 

[Cantspel]

I understand the problem is saying that the particle is starting a distance ds, but i still don't understand how that helps me solve the problem. Is ds somehow related to d cause I just don't see it.

 

[haruspex]

I understand the problem is saying that the particle is starting a distance ds, but i still don't understand how that helps me solve the problem. Is ds somehow related to d cause I just don't see it.

It means that you can treat d as zero initially.

 

[Cantspel]

So does this mean the answer is just v = πλa(8k/ma)^1/2

 

[haruspex]

So does this mean the answer is just v = πλa(8k/ma)^1/2

Yes, but you can simplify a bit more.

 

[Cantspel]

v = 2πλ(2ak/m)^1/2

 

[haruspex]

v = 2πλ(2ak/m)^1/2

Looks good.

 

[Cantspel]

Thanks for all the help!