(a) F 12 ( t ) = 1 4 π ϵ 0 q 1 q 2 ( v t ) 2 z ^ \text { (a) } F _{12}(t)=\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{(v t)^{2}} \hat{ z } (a) F 1 2 ( t ) = 4 π ϵ 0 1 ( v t ) 2 q 1 q 2 z ^
(b) From Eq. 10.75, with θ = 18 0 ∘ , R = v t , and R ^ = − z ^ \theta=180^{\circ}, R=v t, \text { and } \hat{ R }=-\hat{ z } θ = 1 8 0 ∘ , R = v t , and R ^ = − z ^
F 21 ( t ) = − 1 4 π ϵ 0 q 1 q 2 ( 1 − v 2 / c 2 ) ( v t ) 2 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 } F 2 1 ( t ) = − 4 π ϵ 0 1 ( v t ) 2 q 1 q 2 ( 1 − v 2 / c 2 ) z ^
E ( r , t ) = q 4 π ϵ 0 1 − v 2 / c 2 ( 1 − v 2 sin 2 θ / c 2 ) 3 / 2 R ^ R 2 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}} E ( r , t ) = 4 π ϵ 0 q ( 1 − v 2 s i n 2 θ / c 2 ) 3 / 2 1 − v 2 / c 2 R 2 R ^ (10.75)
No, Newton’s third law does not hold: F 12 ≠ F 21 F _{12} \neq F _{21} F 1 2 = F 2 1 because of the extra factor ( 1 − v 2 / c 2 ) \left(1-v^{2} / c^{2}\right) ( 1 − v 2 / c 2 )
(c) From Eq. 8.28,
p = μ 0 ϵ 0 ∫ ν S d τ p =\mu_{0} \epsilon_{0} \int_{ \nu } S d \tau p = μ 0 ϵ 0 ∫ ν S d τ (8.28)
p = ϵ 0 ∫ ( E × B ) d τ . Here E = E 1 + E 2 p =\epsilon_{0} \int( E \times B ) d \tau . \text { Here } E = E _{1}+ E _{2} p = ϵ 0 ∫ ( E × B ) d τ . Here E = E 1 + E 2 . Here E = E 1 + E 2 , whereas B = B 2 , so E × B = ( E 1 × B 2 ) + ( E 2 × B 2 ) . 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 {. } E = E 1 + E 2 , whereas B = B 2 , so E × B = ( E 1 × B 2 ) + ( E 2 × B 2 ) .
But the latter, when integrated over all space, is independent of time. We want only the time-dependent part:
p ( t ) = ϵ 0 ∫ ( E 1 × B 2 ) d τ . Now E 1 = 1 4 π ϵ 0 q 1 r 2 r ^ , while, from Eq. 10.76 , B 2 = 1 c 2 ( v × E 2 ) 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) p ( t ) = ϵ 0 ∫ ( E 1 × B 2 ) d τ . Now E 1 = 4 π ϵ 0 1 r 2 q 1 r ^ , while, from Eq. 1 0 . 7 6 , B 2 = c 2 1 ( v × E 2 ) , and (Eq. 10.75)
B = 1 c ( ᴫ ^ × E ) = 1 c 2 ( v × E ) B =\frac{1}{c}(\hat{ᴫ} \times E )=\frac{1}{c^{2}}( v \times E ) B = c 1 ( ᴫ ^ × E ) = c 2 1 ( v × E ) (10.76)
E 2 = q 2 4 π ϵ 0 ( 1 − v 2 / c 2 ) ( 1 − v 2 sin 2 θ ′ / c 2 ) 3 / 2 R ^ R 2 . But R = r – v t ; R 2 = r 2 + v 2 t 2 − 2 r v t cos θ ; sin θ ′ = r sin θ R 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} E 2 = 4 π ϵ 0 q 2 ( 1 − v 2 s i n 2 θ ′ / c 2 ) 3 / 2 ( 1 − v 2 / c 2 ) R 2 R ^ . B u t R = r – v t ; R 2 = r 2 + v 2 t 2 − 2 r v t cos θ ; sin θ ′ = R r s i n θ . So
E 2 = q 2 4 π ϵ 0 ( 1 − v 2 / c 2 ) [ 1 − ( v r sin θ / R c ) 2 ] 3 / 2 ( r – v t ) R 3 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}} E 2 = 4 π ϵ 0 q 2 [ 1 − ( v r s i n θ / R c ) 2 ] 3 / 2 ( 1 − v 2 / c 2 ) R 3 ( r – v t ) . Finally, noting that v × ( r – v t ) = v × r = v r sin θ ϕ ^ v \times( r – v t)= v \times r =v r \sin \theta \hat{\phi} v × ( r – v t ) = v × r = v r sin θ ϕ ^ , we get
B 2 = q 2 ( 1 − v 2 / c 2 ) 4 π ϵ 0 c 2 v r sin θ [ R 2 − ( v r sin θ / c ) 2 ] 3 / 2 ϕ ^ . So p ( t ) = ϵ 0 q 1 4 π ϵ 0 q 2 ( 1 − v 2 / c 2 ) v 4 π ϵ 0 c 2 ∫ 1 r 2 r sin θ ( r ^ × ϕ ^ ) [ R 2 − ( v r sin θ / c ) 2 ] 3 / 2 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}} B 2 = 4 π ϵ 0 c 2 q 2 ( 1 − v 2 / c 2 ) [ R 2 − ( v r s i n θ / c ) 2 ] 3 / 2 v r s i n θ ϕ ^ . So p ( t ) = ϵ 0 4 π ϵ 0 q 1 4 π ϵ 0 c 2 q 2 ( 1 − v 2 / c 2 ) v ∫ r 2 1 [ R 2 − ( v r s i n θ / c ) 2 ] 3 / 2 r s i n θ ( r ^ × ϕ ^ )
But r ^ × ϕ ^ = − θ ^ = − ( cos θ cos ϕ x ^ + cos θ sin ϕ y ^ − sin θ z ^ ) \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 }) B u t r ^ × ϕ ^ = − θ ^ = − ( cos θ cos ϕ x ^ + cos θ sin ϕ y ^ − sin θ z ^ ) , and the x and y components integrate to zero, so:
p ( t ) = q 1 q 2 v ( 1 − v 2 / c 2 ) z ^ ( 4 π c ) 2 ϵ 0 ∫ sin 2 θ r [ r 2 + ( v t ) 2 − 2 r v t cos θ − ( v r sin θ / c ) 2 ] 3 / 2 r 2 sin θ d r d θ d ϕ 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 p ( t ) = ( 4 π c ) 2 ϵ 0 q 1 q 2 v ( 1 − v 2 / c 2 ) z ^ ∫ r [ r 2 + ( v t ) 2 − 2 r v t c o s θ − ( v r s i n θ / c ) 2 ] 3 / 2 s i n 2 θ r 2 sin θ d r d θ d ϕ
= q 1 q 2 v ( 1 − v 2 / c 2 ) z ^ 8 π c 2 ϵ 0 ∫ r sin 3 θ [ r 2 + ( v t ) 2 − 2 r v t cos θ − ( v r sin θ / c ) 2 ] 3 / 2 d r d θ =\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 = 8 π c 2 ϵ 0 q 1 q 2 v ( 1 − v 2 / c 2 ) z ^ ∫ [ r 2 + ( v t ) 2 − 2 r v t c o s θ − ( v r s i n θ / c ) 2 ] 3 / 2 r s i n 3 θ d r d θ
I’ll do the r integral first. According to the CRC Tables,
∫ 0 ∞ x ( a + b x + c x 2 ) 3 / 2 d x = − 2 ( b x + 2 a ) ( 4 a c − b 2 ) a + b x + c x 2 ∣ 0 ∞ = − 2 4 a c − b 2 [ b c − 2 a a ] \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] ∫ 0 ∞ ( a + b x + c x 2 ) 3 / 2 x d x = − ( 4 a c − b 2 ) a + b x + c x 2 2 ( b x + 2 a ) ∣ ∣ ∣ ∣ 0 ∞ = − 4 a c − b 2 2 [ c b − a 2 a ]
= − 2 c ( 4 a c − b 2 ) ( b − 2 a c ) = 2 c ( 2 a c − b ) ( 2 a c − b ) ( 2 a c + b ) = 2 c ( 2 a c + b ) − 1 =-\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} = − c ( 4 a c − b 2 ) 2 ( b − 2 a c ) = c 2 ( 2 a c − b ) ( 2 a c + b ) ( 2 a c − b ) = c 2 ( 2 a c + b ) − 1
In this case x = r , a = ( v t ) 2 , b = − 2 v t cos θ , and c = 1 − ( v / c ) 2 sin 2 θ . So the r integral \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 } In this case x = r , a = ( v t ) 2 , b = − 2 v t cos θ , and c = 1 − ( v / c ) 2 sin 2 θ . So the r integral is
2 1 − ( v / c ) 2 sin 2 θ [ 2 v t 1 − ( v / c ) 2 sin 2 θ − 2 v t cos θ ] = 1 v t 1 − ( v / c ) 2 sin 2 θ [ 1 − ( v / c ) 2 sin 2 θ − cos θ ] \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]} 1 − ( v / c ) 2 s i n 2 θ [ 2 v t 1 − ( v / c ) 2 s i n 2 θ − 2 v t c o s θ ] 2 = v t 1 − ( v / c ) 2 s i n 2 θ [ 1 − ( v / c ) 2 s i n 2 θ − c o s θ ] 1
= [ 1 − ( v / c ) 2 sin 2 θ + cos θ ] v t 1 − ( v / c ) 2 sin 2 θ [ 1 − ( v / c ) 2 sin 2 θ − cos 2 θ ] = 1 v t sin 2 θ ( 1 − v 2 / c 2 ) [ 1 + cos θ 1 − ( v / c ) 2 sin 2 θ ] =\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] = v t 1 − ( v / c ) 2 s i n 2 θ [ 1 − ( v / c ) 2 s i n 2 θ − c o s 2 θ ] [ 1 − ( v / c ) 2 s i n 2 θ + c o s θ ] = v t s i n 2 θ ( 1 − v 2 / c 2 ) 1 [ 1 + 1 − ( v / c ) 2 s i n 2 θ c o s θ ]
So
p ( t ) = q 1 q 2 v ( 1 − v 2 / c 2 ) z ^ 8 π c 2 ϵ 0 1 v t ( 1 − v 2 / c 2 ) ∫ 0 π 1 sin 2 θ [ 1 + cos θ 1 − ( v / c ) 2 sin 2 θ ] sin 3 θ d θ 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 p ( t ) = 8 π c 2 ϵ 0 q 1 q 2 v ( 1 − v 2 / c 2 ) z ^ v t ( 1 − v 2 / c 2 ) 1 ∫ 0 π s i n 2 θ 1 [ 1 + 1 − ( v / c ) 2 s i n 2 θ c o s θ ] sin 3 θ d θ
= q 1 q 2 z ^ 8 π c 2 ϵ 0 t { ∫ 0 π sin θ d θ + c v ∫ 0 π cos θ sin θ ( c / v ) 2 − sin 2 θ d θ } =\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\} = 8 π c 2 ϵ 0 t q 1 q 2 z ^ { ∫ 0 π sin θ d θ + v c ∫ 0 π ( c / v ) 2 − s i n 2 θ c o s θ s i n θ d θ }
But ∫ 0 π sin θ d θ = 2 . In the second integral let u ≡ cos θ , so d u = − sin θ d θ \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 But ∫ 0 π sin θ d θ = 2 . In the second integral let u ≡ cos θ , so d u = − sin θ d θ
∫ 0 π cos θ sin θ ( c / v ) 2 − sin 2 θ d θ = ∫ − 1 1 u ( c / v ) 2 − 1 + u 2 d u = 0 \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 ∫ 0 π ( c / v ) 2 − s i n 2 θ c o s θ s i n θ d θ = ∫ − 1 1 ( c / v ) 2 − 1 + u 2 u d u = 0 (the integrand is odd, and the interval is even).
Conclusion: p ( t ) = μ 0 q 1 q 2 4 π t z ^ p (t)=\frac{\mu_{0} q_{1} q_{2}}{4 \pi t} \hat{ z } p ( t ) = 4 π t μ 0 q 1 q 2 z ^ (plus a term constant in time).
(d)
F 12 + F 21 = 1 4 π ϵ 0 q 1 q 2 v 2 t 2 z ^ − 1 4 π ϵ 0 q 1 q 2 ( 1 − v 2 / c 2 ) v 2 t 2 z ^ = q 1 q 2 4 π ϵ 0 v 2 t 2 ( 1 − 1 + v 2 c 2 ) z ^ = q 1 q 2 4 π ϵ 0 c 2 t 2 z ^ = μ 0 q 1 q 2 4 π t 2 z ^ 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 } F 1 2 + F 2 1 = 4 π ϵ 0 1 v 2 t 2 q 1 q 2 z ^ − 4 π ϵ 0 1 v 2 t 2 q 1 q 2 ( 1 − v 2 / c 2 ) z ^ = 4 π ϵ 0 v 2 t 2 q 1 q 2 ( 1 − 1 + c 2 v 2 ) z ^ = 4 π ϵ 0 c 2 t 2 q 1 q 2 z ^ = 4 π t 2 μ 0 q 1 q 2 z ^
− d p d t = μ 0 q 1 q 2 4 π t 2 z ^ = F 12 + F 21 -\frac{d p }{d t}=\frac{\mu_{0} q_{1} q_{2}}{4 \pi t^{2}} \hat{ z }= F _{12}+ F _{21} − d t d p = 4 π t 2 μ 0 q 1 q 2 z ^ = F 1 2 + F 2 1 . qed
Since q 1 is at rest, and q 2 \text { Since } q_{1} \text { is at rest, and } q_{2} Since q 1 is at rest, and q 2 is moving at constant velocity, there must be another force ( F mech ) \left( F _{\text {mech }}\right) ( F mech ) acting on them, to balance F 12 + F 21 F _{12}+ F _{21} F 1 2 + F 2 1 ; what we have found is that F mech = d p e m / d t F _{\text {mech }}=d p _{ em } / d t F mech = d p e m / 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).]
