Finding V and E above a uniformly charged ring

[maherelharake]

Homework Statement

We are given a ring of charge in the xy plane with a line density of λ(φ)=λ₀cos(φ). Here φ is measured as a rotation from the +x axis.

First, calculate the electric potential along the z axis. Then, calculate the electric field along the z axis.
Hint: The problem may become simpler if you examine it in Cartesian coordinates.

Homework Equations

The Attempt at a Solution

I have a few ideas on this problem, but I am not sure if they are correct. For the first part, I am trying to use the equation: V=k Integral(lamda*dl/r). I tried to plug the given value of lamda into this equation for V, and I tried to integrate from 0 to 2Pi. Would the value of 'r' be sqrt(x²+y²)? Also, could you replace dl with dz since it is asking for the value along the z axis? Am I even thinking correctly up this point? Thanks.

 

[fzero]

What does dl correspond to physically in that formula? Remember it's related to what you're supposed to integrate over. What does the distance r correspond to? Think about the concept before you try to write a formula down.

 

[maherelharake]

I believe that dl represents small segments of the line of charge. Also, 'r' represents the distance from the charged surface to the point in question.

 

[fzero]

So if the ring is in the xy-plane can you agree that dl is independent of z? If the ring has radius R, can you relate dl to the angle φ? Can you write a formula for r in terms of z and R?

 

[maherelharake]

Yes I agree that is independent of z. Also, wouldn't dφR=dl? I'm not sure on how to write For how to write a formula for r in terms of z and R, would it be r^2=R^2+z^2.

 

[fzero]

Yes I agree that is independent of z. Also, wouldn't dφR=dl?

This is correct. An easy way to check is to integrate dl around the ring and compare it to the circumference.

I'm not sure on how to write For how to write a formula for r in terms of z and R, would it be r^2=R^2+z^2.

This is correct too. You can verify it by drawing a sketch of the xz or yz plane.

You should have enough information to use the integral formula to find the potential.

 

[maherelharake]

Ok I pulled all constants in front of the integral sign and got a final value of V=(2Pi*k*lamda*R/sqrt(R^2+z^2.)) Does that look ok?

 

[fzero]

Ok I pulled all constants in front of the integral sign and got a final value of V=(2Pi*k*lamda*R/sqrt(R^2+z^2.)) Does that look ok?

Looks good.

 

[maherelharake]

Great. Now for the E value, I want to use the formula E=k Integral(lamda*dl/r^2). However, I know that E is a vector quantity, and I also know that it is only going to have a nonzero 'z' component. So would the integral be E=k*integral(lamda*R*dφ*z/(R^2+z^2))?

 

[fzero]

Great. Now for the E value, I want to use the formula E=k Integral(lamda*dl/r^2). However, I know that E is a vector quantity, and I also know that it is only going to have a nonzero 'z' component. So would the integral be E=k*integral(lamda*R*dφ*z/(R^2+z^2))?

Almost, you'll need another factor of r in the denominator if you check the dimensions. You could also obtain E directly from V if you've covered that in class.

 

[maherelharake]

I'm not sure what the other factor in the denominator would be. I thought I just had to square the 'r' term. and also, I know that E=gradV. To use this, should I convert R into sqrt(x^2+y^2) so I can use that formula?

 

[fzero]

I'm not sure what the other factor in the denominator would be. I thought I just had to square the 'r' term. and also, I know that E=gradV. To use this, should I convert R into sqrt(x^2+y^2) so I can use that formula?

Think about where you got the factor of z from. If you did it the right way, it came in as a factor z/r. As for the 2nd question, no, R is a constant, you don't rewrite it. Note that if you did, E_x and E_y would be nonzero.

 

[maherelharake]

Hmm ok. So I should just take the negative gradient and I am going to have to take the negative gradient of k*2Pi*lamda*R/sqrt(R^2+z^2). I basically hold everything constant and differentiate with respect to z?

 

[fzero]

Hmm ok. So I should just take the negative gradient and I am going to have to take the negative gradient of k*2Pi*lamda*R/sqrt(R^2+z^2). I basically hold everything constant and differentiate with respect to z?

Yes, everything else apart from z is constant, assuming that you mean \lambda_0 and not \lambda.

 

[maherelharake]

Wait for the potential part again, shouldn't I have transformed Lamda into Lamda₀cosPhi? Wouldn't that alter my answer since I integrated over Phi?

 

[fzero]

Wait for the potential part again, shouldn't I have transformed Lamda into Lamda₀cosPhi? Wouldn't that alter my answer since I integrated over Phi?

I thought that you did and just forgot to put the subscript, I wasn't tracking numerical factors.

 

[maherelharake]

Sorry my fault. Ok I got that part sorted out. Now back to finding E. I tried to differentiate that equation with respect to 'z' and got this...

E=k*2Pi*R*Lamda₀*z/(z^2+R^2)^3/2

 

[maherelharake]

I actually looked at this problem again, and got both values equal to 0. I got V=0 because the integral was from the values of phi=0 to phi=2Pi, and the integrand was a cos term. When you integrate, it becomes sin, and gives a value of 0.