Quiz 3 -- Charge Distributions

Find the electric field at point P, the center of the split-ring of charge, left side positive and the right side negative.

Solution:

Step 1: Understand the geometry

Step 2: Span the charge distribution

Assuming a uniform charge distribution about the ring, the linear charge density lambda will be Q/(pi a), where we are assigning the variable a to be the radius of the ring. In terms of the charge density, the infinitesimal charge element will be dq = lambda ds = lambda a d(theta).

Step 3: Evaluate the contribution from the infinitesimal charge element

Begin by focusing on the left half of the ring.

The inifinitesimal contribution dE will point directly away from the charge infinitesimal, along the line connecting it and the point of evaluation. The magnitude of dE will be dq/a² = lambda d(theta) / a².

Step 4: Exploit symmetry as appropriate

There is a conjugate to the charge infinitesimal considered above, so that the contribution to dE from the pair will point directly to the right. The vertical component of dE from one charge infinitesimal will always be balanced by the contribution of its conjugate pair. Thus, only the horizontal component of dE need be integrated.

The contribution from any charge infinitesimal will have the same magnitude as the one considered above. However its horizontal component does vary as the charge configuation is spanned. So the integration over dE reduces to that over dE_x, which is the cosine of theta times the magnitude of dE.

Turning to the right hand side of the ring, a quick sketch indicates that the overall contribution to E at the point of evaluation is exactly the same as the left hand side! So we proceed to integrate over the left hand side and then double for the finalanswer.

Step 5: Set up the integral

Step 6: Solve the integral

Step 7: The result!

Unfortunately no limiting behavior of the charge distribution far away may be checked since we evaluated E at a special point, directly in the center. If you would care to rework this configuration at an arbitrary point on its axis, you would have two limits to consider. One is the special case that we just evaluated. Far away, the result for a point dipole should be obtained.

Special bonus: (Fall'97 only!) As a reward for reading my solution of Quiz 3 to the end, I am offering to add an extra 5 points to the score of your first midterm exam if you submit your own correct solution by the end of the first exam to the following question (may be written out ahead of time):

Good luck!   --   Answer!