(a) For the configuration in Prob. 4.5, calculate the force on P _{2} due to P _{1}, and the force on P _{1} due to P _{2}. Are the answers consistent with Newton’s third law?
(b) Find the total torque on P _{2} with respect to the center of P _{1}, and compare it withthe torque on P _{1} about that same point. [Hint: combine your answer to (a) withthe result of Prob. 4.5.]
(a) Eq. 4.5 \Rightarrow F _{2}=\left( p _{2} \cdot \nabla \right) E _{1}=p_{2} \frac{\partial}{\partial y}\left( E _{1}\right) ;
F =( p \cdot \nabla ) E (4.5)
Eq. 3.103 \Rightarrow E _{1}=\frac{p_{1}}{4 \pi \epsilon_{0} r^{3}} \hat{ \theta }=-\frac{p_{1}}{4 \pi \epsilon_{0} y^{3}} \hat{ z } . Therefore
E _{ dip }(r, \theta)=\frac{p}{4 \pi \epsilon_{0} r^{3}}(2 \cos \theta \hat{ r }+\sin \theta \hat{ \theta }) (3.103)
F _{2}=-\frac{p_{1} p_{2}}{4 \pi \epsilon_{0}}\left[\frac{d}{d y}\left(\frac{1}{y^{3}}\right)\right] \hat{ z }=\frac{3 p_{1} p_{2}}{4 \pi \epsilon_{0} y^{4}} \hat{ z }, \text { or } F _{2}=\frac{3 p_{1} p_{2}}{4 \pi \epsilon_{0} r^{4}} \hat{ z } (upward).
To calculate F _{1}, put p _{2} at the origin, pointing in the z direction; then p _{1} is at -r \hat{ z }, \text { and it points in the }-\hat{ y } \text { direction. So } F _{1}=\left( p _{1} \cdot \nabla \right) E _{2}=-\left.p_{1} \frac{\partial E _{2}}{\partial y}\right|_{x=y=0, z=-r} ; we need E _{2} as a function of x, y, and z.
From Eq. 3.104: E _{2}=\frac{1}{4 \pi \epsilon_{0}} \frac{1}{r^{3}}\left[\frac{3\left( p _{2} \cdot r \right) r }{r^{2}}- p _{2}\right], \text { where } r =x \hat{ x }+y \hat{ y }+z \hat{ z }, p _{2}=p_{2} \hat{ z }, \text { and hence } p _{2} \cdot r =p_{2} z .
E _{ dip }( r )=\frac{1}{4 \pi \epsilon_{0}} \frac{1}{r^{3}}[3( p \cdot \hat{ r }) \hat{ r }- p ] (3.104)
E _{2}=\frac{p_{2}}{4 \pi \epsilon_{0}}\left[\frac{3 z(x \hat{ x }+y \hat{ y }+z \hat{ z })-\left(x^{2}+y^{2}+z^{2}\right) \hat{ z }}{\left(x^{2}+y^{2}+z^{2}\right)^{5 / 2}}\right]=\frac{p_{2}}{4 \pi \epsilon_{0}}\left[\frac{3 x z \hat{ x }+3 y z \hat{ y }-\left(x^{2}+y^{2}-2 z^{2}\right) \hat{ z }}{\left(x^{2}+y^{2}+z^{2}\right)^{5 / 2}}\right]
\left.\frac{\partial E _{2}}{\partial y}\right|_{(0,0)}=\frac{p_{2}}{4 \pi \epsilon_{0}} \frac{3 z}{r^{5}} \hat{ y } ; \quad F _{1}=-p_{1}\left(\frac{p_{2}}{4 \pi \epsilon_{0}} \frac{-3 r}{r^{5}} \hat{ y }\right)=\frac{3 p_{1} p_{2}}{4 \pi \epsilon_{0} r^{4}} \hat{ y } .
But \hat{ y } in these coordinates corresponds to -\hat{ z } in the original system, so these results are consistent with Newton’s third law: F _{1}=- F _{2} .
(b) From the remark following Eq. 4.5, N _{2}=\left( p _{2} \times E _{1}\right)+\left( r \times F _{2}\right) . The first term was calculated in Prob. 4.5; the second we get from (a), using r =r \hat{ y } ;
p _{2} \times E _{1}=\frac{p_{1} p_{2}}{4 \pi \epsilon_{0} r^{3}}(-\hat{ x }) ; \quad r \times F _{2}=(r \hat{ y }) \times\left(\frac{3 p_{1} p_{2}}{4 \pi \epsilon_{0} r^{4}} \hat{ z }\right)=\frac{3 p_{1} p_{2}}{4 \pi \epsilon_{0} r^{3}} \hat{ x } ; \text { so } \quad N _{2}=\frac{2 p_{1} p_{2}}{4 \pi \epsilon_{0} r^{3}} \hat{ x }.
This is equal and opposite to the torque on p _{1} due to p _{2}, with respect to the center of p _{1} (see Prob. 4.5).
