(a) Suppose a charge distribution \rho_{1}( r ) produces a potential V_{1}( r ), and some other charge distribution \rho_{2}( r ) produces a potential V_{2}( r ). [The two situations may have nothing in common, for all I care—perhaps number 1 is a uniformly charged sphere and number 2 is a parallel-plate capacitor. Please understand that \rho_{1}( r ) and \rho_{2}( r ) are not present at the same time; we are talking about two different problems, one in which only \rho_{1}( r ) is present, and another in which only \rho_{2}( r ) is present.] Prove Green’s reciprocity theorem:^{25}
\int\limits_{all space}^{}{} \rho_{1} V_{2} d \tau=\int\limits_{all space}^{}{} \rho_{2} V_{1} d \tau .
[Hint: Evaluate \int E _{1} \cdot E _{2} d \tau two ways, first writing E _{1}=-\nabla V_{1} and using integration by parts to transfer the derivative to E _{2}, then writing E _{2}=-\nabla V_{2} and transferring the derivative to E _{1}.]
(b) Suppose now that you have two separated conductors (Fig. 3.41). If you charge up conductor a by amount Q (leaving b uncharged), the resulting potential of b is, say, V_{a b} . On the other hand, if you put that same charge Q on conductor b (leaving a uncharged), the potential of a would be V_{b a} . Use Green’s reciprocity theorem to show that V_{a b}=V_{b a} (an astonishing result, since we assumed nothing about the shapes or placement of the conductors).
{ }^{23} There are other ways of getting the delta-function term in the field of a dipole—my own favorite is Prob. 3.49. Note that unless you are right on top of the dipole, Eq. 3.104 is perfectly adequate .
{ }^{24} See C. P. Frahm, Am. J. Phys. 51, 826 (1983). For applications, see D. J. Griffiths, Am. J. Phys. 50,
698 (1982). There are other (perhaps preferable) ways of expressing the contact (delta-function) term in Eq. 3.106; see A. Gsponer, Eur. J. Phys. 28, 267 (2007), J. Franklin, Am. J. Phys. 78, 1225 (2010), and V. Hnizdo, Eur. J. Phys. 32, 287 (2011).
{ }^{25} For interesting commentary, see B. Y.-K. Hu, Am. J. Phys. 69, 1280 (2001).
E _{ dip }( r )=\frac{1}{4 \pi \epsilon_{0}} \frac{1}{r^{3}}[3( p \cdot \hat{ r }) \hat{ r }- p ] (3.104)
E _{\text {dip }}( r )=\frac{1}{4 \pi \epsilon_{0}} \frac{1}{r^{3}}[3( p \cdot \hat{ r }) \hat{ r }- p ]-\frac{1}{3 \epsilon_{0}} p \delta^{3}( r ) (3.106)
