Question 10.31: A particle of charge q is traveling at constant speed υ alon...

S =\frac{1}{\mu_{0}}( E \times B ) ; B =\frac{1}{c^{2}}( v \times E ) (Eq. 10.76).

B =\frac{1}{c}(\hat{ᴫ} \times E )=\frac{1}{c^{2}}( v \times E )                                            (10.76)

\text { So } S =\frac{1}{\mu_{0} c^{2}}[ E \times( v \times E )]=\epsilon_{0}\left[E^{2} v -( v \cdot E ) E \right] .

The power crossing the plane is P = ∫ S · da,

\text { and } d a =2 \pi r d r \hat{ x } (see diagram). So

P=\epsilon_{0} \int\left(E^{2} v-E_{x}^{2} v\right) 2 \pi r d r ; E_{x}=E \cos \theta, \text { so } E^{2}-E_{x}^{2}=E^{2} \sin ^{2} \theta .

=2 \pi \epsilon_{0} v \int E^{2} \sin ^{2} \theta r d r . From Eq. 10.75,

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)

E =\frac{q}{4 \pi \epsilon_{0}} \frac{1}{\gamma^{2}} \frac{\hat{ R }}{R^{2}\left[1-(v / c)^{2} \sin ^{2} \theta\right]^{3 / 2}} \text { where } \gamma \equiv \frac{1}{\sqrt{1-v^{2} / c^{2}}}.

=2 \pi \epsilon_{0} v\left(\frac{q}{4 \pi \epsilon_{0}}\right)^{2} \frac{1}{\gamma^{2}} \int_{0}^{\infty} \frac{r \sin ^{2} \theta}{R^{4}\left[1-(v / c)^{2} \sin ^{2} \theta\right]^{3}} d r . \quad \text { Now } r=a \tan \theta \Rightarrow d r=a \frac{1}{\cos ^{2} \theta} d \theta ; \frac{1}{R}=\frac{\cos \theta}{a}.

=\frac{v}{2 \gamma^{4}} \frac{q^{2}}{4 \pi \epsilon_{0}} \frac{1}{a^{2}} \int_{0}^{\pi / 2} \frac{\sin ^{3} \theta \cos \theta}{\left[1-(v / c)^{2} \sin ^{2} \theta\right]^{3}} d \theta . \quad \text { Let } u \equiv \sin ^{2} \theta, \text { so } d u=2 \sin \theta \cos \theta d \theta .

=\frac{v q^{2}}{16 \pi \epsilon_{0} a^{2} \gamma^{4}} \int_{0}^{1} \frac{u}{\left[1-(v / c)^{2} u\right]^{3}} d u=\frac{v q^{2}}{16 \pi \epsilon_{0} a^{2} \gamma^{4}}\left(\frac{\gamma^{4}}{2}\right)=\frac{v q^{2}}{32 \pi \epsilon_{0} a^{2}} .