What is the Electric Potential of a Uniformly Charged Ring?

AI Thread Summary
The electric potential of a uniformly charged ring is calculated using the formula V = (2kλπa)/(a² + d²)^(1/2), where λ is the charge density and d is the distance from the center along the axis. For a particle displaced slightly from the center, its speed after being repulsed can be derived from energy conservation principles, leading to the equation v = 2πλ(2ak/m)^(1/2). There was a discussion about the correct interpretation of the variable d, which should be treated as zero for initial calculations. The final expressions for both potential and speed were confirmed with some simplifications. The calculations and reasoning provided were validated by participants in the discussion.
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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.
 

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Cantspel said:
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.
 
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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
 
Cantspel said:
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.
 
haruspex said:
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.

haruspex said:
I think there may be a factor of 2 error.

I see it now the (2k)1/2 is actually a (8k)1/2
 
Cantspel said:
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.
 
TSny said:
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.
 
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?
 
Cantspel said:
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?
Cantspel said:
displaced ever so slightly from the ring's center
 
  • #10
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.
 
  • #11
Cantspel said:
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.
 
  • #12
So does this mean the answer is just v = πλa(8k/ma)1/2
 
  • #13
Cantspel said:
So does this mean the answer is just v = πλa(8k/ma)1/2
Yes, but you can simplify a bit more.
 
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  • #14
v = 2πλ(2ak/m)1/2
 
  • #15
Cantspel said:
v = 2πλ(2ak/m)1/2
Looks good.
 
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  • #16
Thanks for all the help!
 
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