Finding V and E above a uniformly charged ring

Thus, E=0. In summary, the problem involves finding the electric potential and electric field along the z axis for a given ring of charge in the xy plane with a line density of λ(φ)=λ0cos(φ). The potential can be calculated using the equation V=k∫(λ*dφ/R) and the electric field can be found using the equation E=k∫(λ*R*dφ*z/(R^2+z^2)^2). The problem can be simplified by using Cartesian coordinates and considering that dl is independent of z and that r can be related to the angle φ. After solving, it was found that both the potential and electric field are equal to 0.
  • #1
maherelharake
261
0

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.
 
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  • #2
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.
 
  • #3
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.
 
  • #4
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?
 
  • #5
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.
 
  • #6
maherelharake said:
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.
 
  • #7
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?
 
  • #8
maherelharake said:
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.
 
  • #9
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))?
 
  • #10
maherelharake said:
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.
 
  • #11
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?
 
  • #12
maherelharake said:
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.
 
  • #13
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?
 
  • #14
maherelharake said:
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 [tex]\lambda_0[/tex] and not [tex]\lambda[/tex].
 
  • #15
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?
 
  • #16
maherelharake said:
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.
 
  • #17
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
 
  • #18
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.
 

Related to Finding V and E above a uniformly charged ring

1. What is a uniformly charged ring?

A uniformly charged ring is a circular object with a constant distribution of electric charge along its circumference.

2. How do I find the electric field (E) above a uniformly charged ring?

To find the electric field above a uniformly charged ring, you can use the formula E = kQz / (z^2 + R^2)^(3/2), where k is the Coulomb's constant, Q is the total charge of the ring, z is the distance from the center of the ring to the point where you want to find the electric field, and R is the radius of the ring.

3. How do I find the electric potential (V) above a uniformly charged ring?

The electric potential above a uniformly charged ring can be found by using the formula V = kQ / (z + R), where k is the Coulomb's constant, Q is the total charge of the ring, z is the distance from the center of the ring to the point where you want to find the electric potential, and R is the radius of the ring.

4. What is the direction of the electric field above a uniformly charged ring?

The electric field above a uniformly charged ring points perpendicular to the plane of the ring, in the direction away from the ring.

5. How does the electric field and potential change as you move further away from the uniformly charged ring?

As you move further away from the uniformly charged ring, both the electric field and potential decrease. This is because the strength of the electric field and potential are inversely proportional to the distance from the source of the charge.

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