PHGN462: Advanced E&M
Group Quiz 7 – November 9, 2009
NAMES: Key
z
P
ρ
p
r
s/2
x
The electric potential for a single oscillating dipole located at the origin is given by
V (~r, t) = −
p~ · r̂ sin[ω(t − r/c)]
,
4π0 −
λ
r
where −
λ = c/ω is the reduced wavelength. As described in problem 11.11, an electric quadrupole can be thought of as two
oppositely oriented oscillating electric dipoles. Shown in the figure is one of the electric dipoles, p~ = pẑ, which, at arbitrary
time, t, is shifted from the origin to ~r 0 (t) = + 2s cos ωt ẑ. You are to evaluate some of the expressions needed to calculate
the retarded electric potential of this dipole in the radiation zone (r s) in the long wavelength limit ( −
λ s). Define the
relative coordinate, ρ
~ = ~r − ~r 0 (the same as script-r in the text).
s
a. Show that |~
ρ| ' r(1 − 2r
ẑ · r̂ cos ωt) to first order in rs .
q
q
p
2
s2
2
Solution: |~
ρ| = (~r − 2s cos ωtẑ)2 = r2 − 2 2s cos ωtẑ · ~r + s4 cos2 ωt = r 1 − rs ẑ ·~ˆr + 4r
2 cos ωt. Keeping only first order
s
s
in r , gives: r(1 − 2r ẑ · r̂ cos ωt)
b. Find p~ · ρ̂ = pẑ · ρ̂ to first order in rs .
Solution: Using the result from part a, we have: p~ · ρ̂ = pẑ ·
s
(1 + (ẑ · r̂)2 ) cos ωt]
r̂) ' p[ẑ · r̂ − 2r
c. Find sin[ω(t − |~
ρ|/c)] to first order in
(~
r −ẑ 2s cos ωt)
|~
ρ|
=p
ẑ·~
r −ẑ cot ẑ 2s cos ωt
s
r(1− 2r
ẑ·r̂ cos ωt)
s
s
' p(ẑ · r̂ − 2r
cos ωt)(1+ 2r
cos ωtẑ ·
s
−.
λ
s
ẑ·r̂ cos ωtR ))/c] = sin(α+β), where α = ω(t−r/c) = ωtr
Solution: Using part a, we have sin ω(t−|~
ρ(tR )|/c) ' sin[ω(t−r(1− 2r
s
(note the difference betwee tR and tr .) and β = 2 λ− ẑ · r̂ cos ωt (using ω/c = 1/ −
λ). Note that β is of order sλ− . Using the angle
addition formula for the sine-function, we have: sin(α + β) = sin α cos β + cos α sin β. For small β, cos β ' 1 and sin β ' β.
Thus, sin ω(t − |~
ρ(tR )|/c) ' sin ωtr + 2sλ− ẑ · r̂ cos2 ωtr .
1