PHYS208 Fundamentals of Physics II

Electric Field on Axis of Split Ring

This approach is the hardest for visualizing vectors!!!

[split ring dipole]


Solution:

Step 1: Understand the geometry

This configuration is similar to the ring of charge example presented in class. However, the right half of the ring is now oppositely charged.

Step 2: Span the charge distribution

Since the charge is on a ring, the angle variable theta is a convenient choice. The entire charge distribution can then be spanned by varying theta from 0 to 2 pi, changing the sign of charge half-way around.

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).

phi is also introduced now to get ready for consideration of components of dE. Note that phi is constant once the point of evaluation is chosen; it is does not vary as the angle theta is varied as the charge distribution is spanned.

[split ring dipole]

Step 3: Evaluate the contribution from the infinitesimal charge element

Begin by focusing on the left half of the ring. The infinitesimal contribution dE will point directly away from the charge infinitesimal, along the line connecting it and the point of evaluation.

[split ring dipole]

Step 4: Exploit symmetry as appropriate

The symmetry is exploited by considering the four conjugate elements shown.

[split ring dipole]

The vertical and axial components of dE from one charge infinitesimal will always be balanced by the contribution from its conjugate.

[split ring dipole]

Thus, only the horizontal x-components of the four dE need be integrated. All four quadrants of charge contribute the same horizontal component, so we need only integrate from 0 to pi / 2 and multiply the answer by 4.

[split ring dipole]

It takes some maneuvering to visualize exactly what component of dE is needed, but the right perspective shows that it is proportional to sin(theta) and sin(phi).

[split ring dipole]

[split ring dipole]

Step 5: Set up the integral

[split ring dipole]

Step 6: Solve the integral

[split ring dipole]

Step 7: The result!

[split ring dipole]


"http://www.physics.udel.edu/~watson/phys208/quiz3extra2.html"
Last updated Oct. 14, 1997.
Copyright George Watson, Univ. of Delaware, 1997.