(a) The total torque is twice the torque on +q; we might as well calculate it at time t = 0. First we need the electric field at +q, due to “q when it was at the retarded point P (Eq. 10.72).
E ( r , t)=\frac{q}{4 \pi \epsilon_{0}} \frac{ᴫ}{( ᴫ \cdot u )^{3}}\left[\left(c^{2}-v^{2}\right) u + ᴫ \times( u \times a )\right] (10.72)<span
class=”fontstyle0″>From the figure, \hat{ ᴫ}=\cos \theta \hat{ x }-\sin \theta \hat{ y }, \quad ᴫ=2 R \cos \theta . The velocity of –q (at P) was v =-\omega R(\sin 2 \theta \hat{ x }+\cos 2 \theta \hat{ y }) , and its acceleration was a =\omega^{2} R(\cos 2 \theta \hat{ x }-\sin 2 \theta \hat{ y }) . Quantities we will need in Eq. 10.72 are:
u =c \hat{ ᴫ }- v =(c \cos \theta+\omega R \sin 2 \theta) \hat{ x }-(c \sin \theta-\omega R \cos 2 \theta) \hat{ y },
ᴫ\cdot u =2 R \cos \theta(c+\omega R \sin \theta), ᴫ \cdot a =2(\omega R \cos \theta)^{2}.
E =\frac{-q}{4 \pi \epsilon_{0}} \frac{2 R \cos \theta}{(2 R \cos \theta)^{3}(c+\omega R \sin \theta)^{3}}\left\{\left[c^{2}-(\omega R)^{2}+2(\omega R \cos \theta)^{2}\right][(c \cos \theta+\omega R \sin 2 \theta) \hat{ x }-(c \sin \theta-\omega R \cos 2 \theta) \hat{ y }]\right.
\left.-2 R \cos \theta(c+\omega R \sin \theta) \omega^{2} R(\cos 2 \theta \hat{ x }-\sin 2 \theta \hat{ y })\right\}
The total torque (about the origin) is
N =2(R \hat{ x }) \times(q E )=\frac{-2 q^{2} R}{4 \pi \epsilon_{0}} \frac{\hat{ z }}{(2 R \cos \theta)^{2}(c+\omega R \sin \theta)^{3}}\left\{-\left[c^{2}-(\omega R)^{2}+2(\omega R \cos \theta)^{2}\right](c \sin \theta-\omega R \cos 2 \theta)\right.
\left.+2(\omega R)^{2} \cos \theta(c+\omega R \sin \theta) \sin 2 \theta\right\}
=-\frac{q^{2}}{4 \pi \epsilon_{0}} \frac{\hat{ z }}{2 R \cos ^{2} \theta(c+\omega R \sin \theta)^{3}}\left[-c^{3} \sin \theta+c^{2} \omega R\left(2 \cos ^{2} \theta-1\right)+c(\omega R)^{2}\left(2 \cos ^{2} \theta+1\right) \sin \theta+(\omega R)^{3}\right]
=-\frac{q^{2}}{4 \pi \epsilon_{0}} \frac{1}{2 R \cos ^{2} \theta(1+\beta \sin \theta)^{3}}\left[-\sin \theta+\beta\left(2 \cos ^{2} \theta-1\right)+\beta^{2}\left(2 \cos ^{2} \theta+1\right) \sin \theta+\beta^{3}\right] \hat{ z },
where \beta \equiv \omega R / c . [Since E and ᴫ both lie in the xy plane, B =(1 / c) \hat{ᴫ} \times E is along the z direction, v × B is radial, and hence the magnetic contribution to the torque is zero.]
The angle θ is determined by the retarded time condition, ᴫ=-c t_{r} \text { (note that } t_{r} is negative, here), and 2θ is the angle through which the dipole rotates in time -t_{r}, \text { so } 2 R \cos \theta=-c t_{r}=c(2 \theta / \omega), \text { or } \theta=\beta \cos \theta .
We can use this to eliminate the trig functions:
N =-\frac{q^{2}}{4 \pi \epsilon_{0}} \frac{\beta}{2 R \theta^{2}\left(1+\sqrt{\beta^{2}-\theta^{2}}\right)^{3}}\left[\left(\beta^{4}-\beta^{2}+2 \theta^{2}\right)+\sqrt{\beta^{2}-\theta^{2}}\left(\beta^{2}+2 \theta^{2}-1\right)\right] \hat{ z }.
Meanwhile, expanding in powers of θ:
\beta=\theta \sec \theta=\theta+\frac{1}{2} \theta^{3}+\frac{5}{24} \theta^{5}+\frac{61}{720} \theta^{7}+\ldots
This can be “solved” (for \theta \text { as a function of } \beta ) by reverting the series:
\theta=\beta-\frac{1}{2} \beta^{3}+\frac{13}{24} \beta^{5}-\frac{541}{720} \beta^{7}+\ldots
Then
\theta^{2}=\beta^{2}\left(1-\beta^{2}+\frac{4}{3} \beta^{4}+\ldots\right), \sqrt{\beta^{2}-\theta^{2}}=\beta^{2}\left(1-\frac{2}{3} \beta^{2}+\frac{4}{5} \beta^{4}+\ldots\right),
\frac{1}{\theta^{2}}=\frac{1}{\beta^{2}}\left(1+\beta^{2}+\ldots\right), \frac{1}{\left(1+\sqrt{\beta^{2}-\theta^{2}}\right)^{3}}=1-3 \beta^{2}+\ldots,
\left[\left(\beta^{4}-\beta^{2}+2 \theta^{2}\right)+\sqrt{\beta^{2}-\theta^{2}}\left(\beta^{2}+2 \theta^{2}-1\right)\right]=\frac{8}{3} \beta^{4}\left(1-\frac{4}{5} \beta^{2}+\ldots\right).
To leading order in \beta, then,
N =-\frac{q^{2}}{4 \pi \epsilon_{0}} \frac{\beta}{2 R} \frac{1}{\beta^{2}} \frac{8 \beta^{4}}{3} \hat{ z }=-\frac{q^{2}}{4 \pi \epsilon_{0}} \frac{4 \beta^{3}}{3 R} \hat{ z }.
(b) The radiation reaction force on +q is (Eq. 11.80) F =\frac{\mu_{0} q^{2}}{6 \pi c} \dot{ a } . In this case \dot{ a }=-\omega^{2} v =-\omega^{3} R \hat{ y } , so the net torque (counting both ends) is
F _{ rad }=\frac{\mu_{0} q^{2}}{6 \pi c} \dot{ a } (11.80)
N =-2 R \frac{\mu_{0} q^{2}}{6 \pi c} \omega^{3} R \hat{ z }=-\frac{q^{2}}{4 \pi \epsilon_{0}} \frac{4 \beta^{3}}{3 R} \hat{ z }.
Adding this to the interaction torque from (a), the total is
N =-\frac{q^{2}}{4 \pi \epsilon_{0}} \frac{8 \beta^{3}}{3 R} \hat{ z }=-\frac{\mu_{0} p^{2} \omega^{3}}{6 \pi c} \hat{ z }.
\text { (c) } \ddot{ p }=2 q \ddot{ r }=2 q\left(-\omega^{2}\right) r =-\omega^{2} p , so Eq. 11.60 says the power radiated is P=\frac{\mu_{0}}{6 \pi c} \omega^{4} p^{2}
. The power associated with the torque in (b) is N \omega=-\frac{\mu_{0} p^{2} \omega^{3}}{6 \pi c} \omega , so they are in agreement.
