Question 10.29: We are now in a position to treat the example in Sect. 8.2.1...

\text { The electric field of } \left.q_{1} \text { at } q_{2} \text { [Eq. 10.75, with } \theta=45^{\circ} \text { and } R =(-v t \hat{ x }+v t \hat{ y })\right] is

E _{1}=\frac{q_{1}}{4 \pi \epsilon_{0}} \frac{1-v^{2} / c^{2}}{\left(1-v^{2} / 2 c^{2}\right)^{3 / 2}} \frac{1}{2 \sqrt{2}(v t)^{2}}(-\hat{ x }+\hat{ y }).

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)

The magnetic field [Eq. 10.76, with\left. v _{1}=-v \hat{ x }\right] is

B _{1}=\frac{1}{c^{2}}\left( v _{1} \times E \right)=-\frac{v}{c^{2}}(\hat{ x } \times E )=-\frac{v}{c^{2}} \frac{q_{1}}{4 \pi \epsilon_{0}} \frac{1-v^{2} / c^{2}}{\left(1-v^{2} / 2 c^{2}\right)^{3 / 2}} \frac{1}{2 \sqrt{2}(v t)^{2}} \hat{ z }.

The force on q_{2} is therefore [Lorentz force law with \left. v _{2}=-v \hat{ y }\right]

F _{2}=q_{2}\left( E _{1}+ v _{2} \times B _{1}\right)=\frac{q_{1} q_{2}}{4 \pi \epsilon_{0}} \frac{1-v^{2} / c^{2}}{\left(1-v^{2} / 2 c^{2}\right)^{3 / 2}} \frac{1}{2 \sqrt{2}(v t)^{2}}\left(-\hat{ x }+\hat{ y }+\frac{v^{2}}{c^{2}} \hat{ x }\right) .

The electric field q_{2} \text { at } q_{1} \text { is reversed, } E _{2}=- E _{1} ; \text { so is the magnetic field } B _{2}=- B _{1} . The electric force is also reversed, but the magnetic force now points in the y direction instead of the x direction. So the force on q_{1} is

F _{1}=\frac{q_{1} q_{2}}{4 \pi \epsilon_{0}} \frac{1-v^{2} / c^{2}}{\left(1-v^{2} / 2 c^{2}\right)^{3 / 2}} \frac{1}{2 \sqrt{2}(v t)^{2}}\left(\hat{ x }-\hat{ y }+\frac{v^{2}}{c^{2}} \hat{ y }\right).

The forces are equal in magnitude, but not opposite in direction. No,