Question 3.57: A stationary electric dipole p = p zˆ is situated at the ori...

Symmetry suggests that the plane of the orbit is perpendicular to the z axis, and since we need a centripetal force, pointing in toward the axis, the orbit must lie at the bottom of the field loops (Fig. 3.37a), where the z component of the field is zero. Referring to Eq. 3.104,

E _{ dip }( r )=\frac{1}{4 \pi \epsilon_{0}} \frac{1}{r^{3}}[3( p \cdot \hat{ r }) \hat{ r }- p ]                              (3.104)

E \cdot \hat{ z }=0 \Rightarrow 3( p \cdot \hat{ r })(\hat{ r } \cdot \hat{ z })- p \cdot \hat{ z }=0, \text { or } 3 \cos ^{2} \theta-1=0 . \quad \text { So } \cos ^{2} \theta=1 / 3, \cos \theta=-1 / \sqrt{3}, \sin \theta=\sqrt{2 / 3}, z / s=\tan \theta \Rightarrow z=-\sqrt{2} s . The field at the orbit is (Eq. 3.103)

E _{ dip }(r, \theta)=\frac{p}{4 \pi \epsilon_{0} r^{3}}(2 \cos \theta \hat{ r }+\sin \theta \hat{ \theta })                   (3.103)

=\frac{p}{4 \pi \epsilon_{0} r^{3}} \sqrt{\frac{2}{3}}[-\sqrt{3}(\cos \phi \hat{ x }+\sin \phi \hat{ y })]=-\frac{p}{4 \pi \epsilon_{0} r^{3}} \sqrt{2} \hat{ s }=-\frac{p}{3 \sqrt{3} \pi \epsilon_{0} s^{3}} \hat{ s } .

The centripetal force is

F=q E=-\frac{q p}{3 \sqrt{3} \pi \epsilon_{0} s^{3}}=-\frac{m v^{2}}{s} \quad \Rightarrow \quad v^{2}=\frac{q p}{3 \sqrt{3} \pi \epsilon_{0} m s^{2}} \quad \Rightarrow \quad v=\frac{1}{s} \sqrt{\frac{q p}{3 \sqrt{3} \pi \epsilon_{0} m}} .

The angular momentum is

L=s m v=\sqrt{\frac{q p m}{3 \sqrt{3} \pi \epsilon_{0}}} ,

the same for all orbits, regardless of their radius (!), and the energy is

W=\frac{1}{2} m v^{2}+q V=\frac{1}{2} \frac{q p}{3 \sqrt{3} \pi \epsilon_{0} s^{2}}+\frac{q}{4 \pi \epsilon_{0}} \frac{p \cos \theta}{r^{2}}=\frac{q p}{6 \sqrt{3} \pi \epsilon_{0} s^{2}}-\frac{q p}{4 \pi \epsilon_{0} \sqrt{3}(3 / 2) s^{2}}=0 .