Question

1) In class, we worked out the multipole expansion for the potential of a system of two charges q1 and q2, distances di and d2 from the origin. For r » di, d2, the first three terms of the expansion looked like: 4πε0 I also claimed that if we generalized things to systems with lots of charge, the multipole potential out to the quadrupole term would look like V(X) = where q is the monopole moment, p is the dipole moment vector, and Q is the quadrupole moment tensor. (Note to Griffiths readers: He uses the same form, but reserves it for one of the end-of-chapter problems instead of the main text) I don't consider it entirely obvious that equation (1) follows from the (2), so let's grind it out and get a little practice with the machinery. Start with (2) and the same system of two point charges q1 and q2 from above. Use the definitions of q,p, and 0 to show that you can get (1) directly from (2), though we originally derived (1) via other means

Answer 1

V(s)^vector = 1/4pi epsilon_0(q/r + r^^middot p^vector/r^2 + r^^middot Q^vector middot r^^/r^3) Q^vector rightarrow quadrapole moment tensor q = q_1 + q_2(total change of system) r^^middot p^vector = Vertical-bar r^^ Vertical-bar Vertical-bar p^vector Vertical-bar cos theta = p cos theta = p cos theta p = 2^bar r_i q_i = d_1q_1 + d_2 q_2 q_ij = Tensor of 2nd order for which off diagonal elements are zero Q_ij = [Q_11 0 0 0 Q_22 0 0 0 Q_33] Q_11 = q_1[3d_1^2 - d_1^2] + q_2[3d_2^2 - d_2^2] = q_1 2d_1^2 + q_2 2d_2^2 Q_22 = q_1[0 - d_1^2] + q_2 [0 - d_2^2] = -q_1 d_1^2 - q_2 d_2^2 Q_33 = -q_1 d_1^2 - q_2 d_2^2 Q_ij = [2 (q_1 d_1^2 + q_2 d_2^2) 0 0 0 = q_1 d_1^2 + q_2 d_2^2 0 0 0 -q_1d_1^2 - q_2 d_2^2] r^^= cos theta i^^+ sin theta j^^ r^^middot theta^^middot r^^ [cos theta sin theta] [2(q_1d_1^2 + q_2 d_2^2) 0 0 0 -q_1d_1^2 -q_2d_2^2 0