A thin circular ring of radius R has charge Q/2

[hitemup]

Homework Statement

A thin circular ring of radius R has charge Q/2 uniformly distributed on the top half, and -Q/2 on the bottom half.

a) What is the value of the electric potential at a point a distance x along the axis through the center of the circle?
b) What can you say about the electric field E at a distance x along the axis. Let V = 0 at r = infinity.

Homework Equations

Coulomb's Law

The Attempt at a Solution

Since the electric potential is a scalar, the answer for the part a is simply zero.
But there is something confuses me in part b.

https://www.physicsforums.com/threads/electric-field-of-a-semi-circle-ring.799746/#post-5021977

This is what I drew for the flat ring problem while studying Coulomb's Law. As you see, dEsin(theta) was not equal to the y component of dE.

Now I have a solutions manual, and for part b, this is the answer.

(b) We follow Example 21-9 from the textbook. But because the upper and lower halves of the ring are oppositely charged, the parallel components of the fields from diametrically opposite infinitesimal segments of the ring will cancel each other, and the perpendicular components add, in the negative y direction. We know then that E_x = 0 .

The solution continues with dEsin(theta) integrated from 0 to 2*pi*r
How can the perpendicular components be added even though they are not in the same direction? Shouldn't we do what we did for the semi-circle problem to find y component of the field?

 

[TSny]

Maybe the solution manual is saying that the perpendicular components dE_⊥ of two diametrically opposite segments add. That is true. But, you are right that you still need to project dE_⊥ onto the y direction before integrating, as you did in the problem with the semi-circle in the other thread. We would need to see the entire form of the integral used in the solution manual in order to comment on whether it is correct or not.

 

[hitemup]

Maybe the solution manual is saying that the perpendicular components dE_⊥ of two diametrically opposite segments add. That is true. But, you are right that you still need to project dE_⊥ onto the y direction before integrating, as you did in the problem with the semi-circle in the other thread. We would need to see the entire form of the integral used in the solution manual in order to comment on whether it is correct or not.

Thanks for helping me, again... Here is the answer.

 

[TSny]

I think you are right. The solution manual is wrong. They did not project dE_⊥ onto the vertical direction. They should have considered two diametrically opposite elements that are in an arbitrary position rather than at the top and bottom.

If you think of the entire ring as a superposition of two semicircles, then you get the correct answer by simple superposition of your solution for the semicircle in the other thread.

 

[TSny]

Something else that you can check. If you know how to calculate the dipole moment of a charge distribution, then you can easily show that the dipole moment of the entire ring is in the y direction and has a magnitude p_y= (2QR)/π, where Q/2 is the charge on the top semicircle.

The "far field" on the x-axis for a dipole in the y direction should be E_y = -p_y/(4πε_ox³). So, you can check to see if your answer reduces to this for large x.