\text { (a) } F _{12}(t)=\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{(v t)^{2}} \hat{ z } .
(b) From Eq. 10.75, with \theta=180^{\circ}, R=v t, \text { and } \hat{ R }=-\hat{ z }:
F _{21}(t)=-\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}\left(1-v^{2} / c^{2}\right)}{(v t)^{2}} \hat{ z } .
E ( r , t)=\frac{q}{4 \pi \epsilon_{0}} \frac{1-v^{2} / c^{2}}{\left(1-v^{2} \sin ^{2} \theta / c^{2}\right)^{3 / 2}} \frac{\hat{ R }}{R^{2}} (10.75)
No, Newton’s third law does not hold: F _{12} \neq F _{21} because of the extra factor \left(1-v^{2} / c^{2}\right) .
(c) From Eq. 8.28,
p =\mu_{0} \epsilon_{0} \int_{ \nu } S d \tau (8.28)
p =\epsilon_{0} \int( E \times B ) d \tau . \text { Here } E = E _{1}+ E _{2} . Here E = E _{1}+ E _{2}, \text { whereas } B = B _{2}, \text { so } E \times B =\left( E _{1} \times B _{2}\right)+\left( E _{2} \times B _{2}\right) \text {. }.
But the latter, when integrated over all space, is independent of time. We want only the time-dependent part:
p (t)=\epsilon_{0} \int\left( E _{1} \times B _{2}\right) d \tau . \text { Now } E _{1}=\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1}}{r^{2}} \hat{ r }, \text { while, from Eq. } 10.76, B _{2}=\frac{1}{c^{2}}\left( v \times E _{2}\right) , and (Eq. 10.75)
B =\frac{1}{c}(\hat{ᴫ} \times E )=\frac{1}{c^{2}}( v \times E ) (10.76)
E _{2}=\frac{q_{2}}{4 \pi \epsilon_{0}} \frac{\left(1-v^{2} / c^{2}\right)}{\left(1-v^{2} \sin ^{2} \theta^{\prime} / c^{2}\right)^{3 / 2}} \frac{\hat{ R }}{R^{2}} . \operatorname{But} R = r – v t ; R^{2}=r^{2}+v^{2} t^{2}-2 r v t \cos \theta ; \sin \theta^{\prime}=\frac{r \sin \theta}{R} . So
E _{ 2 }=\frac{q_{2}}{4 \pi \epsilon_{0}} \frac{\left(1-v^{2} / c^{2}\right)}{\left[1-(v r \sin \theta / R c)^{2}\right]^{3 / 2}} \frac{( r – v t)}{R^{3}} . Finally, noting that v \times( r – v t)= v \times r =v r \sin \theta \hat{\phi} , we get
B _{2}=\frac{q_{2}\left(1-v^{2} / c^{2}\right)}{4 \pi \epsilon_{0} c^{2}} \frac{v r \sin \theta}{\left[R^{2}-(v r \sin \theta / c)^{2}\right]^{3 / 2}} \hat{\phi} . \text { So } p (t)=\epsilon_{0} \frac{q_{1}}{4 \pi \epsilon_{0}} \frac{q_{2}\left(1-v^{2} / c^{2}\right) v}{4 \pi \epsilon_{0} c^{2}} \int \frac{1}{r^{2}} \frac{r \sin \theta(\hat{ r } \times \hat{ \phi })}{\left[R^{2}-(v r \sin \theta / c)^{2}\right]^{3 / 2}}.
\operatorname{But} \hat{ r } \times \hat{ \phi }=-\hat{ \theta }=-(\cos \theta \cos \phi \hat{ x }+\cos \theta \sin \phi \hat{ y }-\sin \theta \hat{ z }) , and the x and y components integrate to zero, so:
p (t)=\frac{q_{1} q_{2} v\left(1-v^{2} / c^{2}\right) \hat{ z }}{(4 \pi c)^{2} \epsilon_{0}} \int \frac{\sin ^{2} \theta}{r\left[r^{2}+(v t)^{2}-2 r v t \cos \theta-(v r \sin \theta / c)^{2}\right]^{3 / 2}} r^{2} \sin \theta d r d \theta d \phi
=\frac{q_{1} q_{2} v\left(1-v^{2} / c^{2}\right) \hat{ z }}{8 \pi c^{2} \epsilon_{0}} \int \frac{r \sin ^{3} \theta}{\left[r^{2}+(v t)^{2}-2 r v t \cos \theta-(v r \sin \theta / c)^{2}\right]^{3 / 2}} d r d \theta .
I’ll do the r integral first. According to the CRC Tables,
\int_{0}^{\infty} \frac{x}{\left(a+b x+c x^{2}\right)^{3 / 2}} d x=-\left.\frac{2(b x+2 a)}{\left(4 a c-b^{2}\right) \sqrt{a+b x+c x^{2}}}\right|_{0} ^{\infty}=-\frac{2}{4 a c-b^{2}}\left[\frac{b}{\sqrt{c}}-\frac{2 a}{\sqrt{a}}\right]
=-\frac{2}{\sqrt{c}\left(4 a c-b^{2}\right)}(b-2 \sqrt{a c})=\frac{2}{\sqrt{c}} \frac{(2 \sqrt{a c}-b)}{(2 \sqrt{a c}-b)(2 \sqrt{a c}+b)}=\frac{2}{\sqrt{c}}(2 \sqrt{a c}+b)^{-1} .
\text { In this case } x=r, a=(v t)^{2}, b=-2 v t \cos \theta, \text { and } c=1-(v / c)^{2} \sin ^{2} \theta . \text { So the } r \text { integral } is
\frac{2}{\sqrt{1-(v / c)^{2} \sin ^{2} \theta}\left[2 v t \sqrt{1-(v / c)^{2} \sin ^{2} \theta}-2 v t \cos \theta\right]}=\frac{1}{v t \sqrt{1-(v / c)^{2} \sin ^{2} \theta}\left[\sqrt{1-(v / c)^{2} \sin ^{2} \theta}-\cos \theta\right]}
=\frac{\left[\sqrt{1-(v / c)^{2} \sin ^{2} \theta}+\cos \theta\right]}{v t \sqrt{1-(v / c)^{2} \sin ^{2} \theta}\left[1-(v / c)^{2} \sin ^{2} \theta-\cos ^{2} \theta\right]}=\frac{1}{v t \sin ^{2} \theta\left(1-v^{2} / c^{2}\right)}\left[1+\frac{\cos \theta}{\sqrt{1-(v / c)^{2} \sin ^{2} \theta}}\right] .
So
p (t)=\frac{q_{1} q_{2} v\left(1-v^{2} / c^{2}\right) \hat{ z }}{8 \pi c^{2} \epsilon_{0}} \frac{1}{v t\left(1-v^{2} / c^{2}\right)} \int_{0}^{\pi} \frac{1}{\sin ^{2} \theta}\left[1+\frac{\cos \theta}{\sqrt{1-(v / c)^{2} \sin ^{2} \theta}}\right] \sin ^{3} \theta d \theta
=\frac{q_{1} q_{2} \hat{ z }}{8 \pi c^{2} \epsilon_{0} t}\left\{\int_{0}^{\pi} \sin \theta d \theta+\frac{c}{v} \int_{0}^{\pi} \frac{\cos \theta \sin \theta}{\sqrt{(c / v)^{2}-\sin ^{2} \theta}} d \theta\right\} .
\text { But } \int_{0}^{\pi} \sin \theta d \theta=2 \text {. In the second integral let } u \equiv \cos \theta, \text { so } d u=-\sin \theta d \theta :
\int_{0}^{\pi} \frac{\cos \theta \sin \theta}{\sqrt{(c / v)^{2}-\sin ^{2} \theta}} d \theta=\int_{-1}^{1} \frac{u}{\sqrt{(c / v)^{2}-1+u^{2}}} d u=0 (the integrand is odd, and the interval is even).
Conclusion: p (t)=\frac{\mu_{0} q_{1} q_{2}}{4 \pi t} \hat{ z } (plus a term constant in time).
(d)
F _{12}+ F _{21}=\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{v^{2} t^{2}} \hat{ z }-\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}\left(1-v^{2} / c^{2}\right)}{v^{2} t^{2}} \hat{ z }=\frac{q_{1} q_{2}}{4 \pi \epsilon_{0} v^{2} t^{2}}\left(1-1+\frac{v^{2}}{c^{2}}\right) \hat{ z }=\frac{q_{1} q_{2}}{4 \pi \epsilon_{0} c^{2} t^{2}} \hat{ z }=\frac{\mu_{0} q_{1} q_{2}}{4 \pi t^{2}} \hat{ z }.
-\frac{d p }{d t}=\frac{\mu_{0} q_{1} q_{2}}{4 \pi t^{2}} \hat{ z }= F _{12}+ F _{21} . qed
\text { Since } q_{1} \text { is at rest, and } q_{2} is moving at constant velocity, there must be another force \left( F _{\text {mech }}\right) acting on them, to balance F _{12}+ F _{21} ; what we have found is that F _{\text {mech }}=d p _{ em } / d t , which means that the impulse imparted to the system by the external force ends up as momentum in the fields. [For further discussion of this problem see J. J. G. Scanio, Am. J. Phys. 43, 258 (1975).]
