W9_parallel_resonance.eps

112 Week 3: Potential Energy and Potential

Example 3.4.2: Potential of a Dipole at an Arbitrary Point in Space

+q

−q

z

x

+a

−a

r

r

r 1

θ^2

Figure 29: A simple dipole aligned with thez-axis, in a spherical coordinate system.

Find the potential of this dipole at an arbitrary pointP= (r, φ, θ). Because the problem

is manifestlyazimuthally symmetricthe answer cannot depend in any way onφ(the

azimuthal/longitude coordinate), so we might as well label the pointP= (r, θ) in the

plane of the figure, where the answer can be azimuthally rotated byφabout thez-axis

to any other plane without changing the form of the answer.

The potential in this problem is extremely easy to findif you can remember the law of cosines:

r 1 = +

√

r^2 +a^2 − 2 arcos(θ) (160)

r 2 = +

√

r^2 +a^2 + 2arcos(θ) (161)

so that the potential can be read off by inspection:

V(r, θ) = keq

(r^2 +a^2 − 2 arcos(θ))^1 /^2

− keq

(r^2 +a^2 + 2arcos(θ))^1 /^2

(162)

Of course, if youdon’tremember the law of cosines, you should visit the math chapter and learn to
derive it in two or three lines so you don’t ever forget it again, as we willuse it fairly often and you
don’t want this to be an obstacle to your learning!

To find the fieldnow, one can take the gradient of this exact result. However, actuallytaking
gradients is beyond the immediate scope of this course, so just bear in mind that youcan(and if
you are a physics major, almost certainly sooner or laterwill) and otherwise forget it. Doing so isn’t
particularly simple in any event because of the fairly complicated denominators (although it is still
much easier than finding the field directly).

Consider what happens, though, when one looks at the potential at a pointr ≫a, so far
away that the dipole looks like a “point object”. To find the potentialthen, we must use the
binomial expansion to factor out the leading r dependence and to move the complicated stuff from