(a) \sum_{i, j=1}^{3} \hat{ r }_{i} \hat{ r }_{j} Q_{i j}=\frac{1}{2} \int\left\{3 \sum_{i=1}^{3} \hat{ r }_{i} r_{i}^{\prime} \sum_{j=1}^{3} \hat{ r }_{j} r_{j}^{\prime}-\left(r^{\prime}\right)^{2} \sum_{i, j} \hat{ r }_{i} \hat{ r }_{j} \delta_{i j}\right\} \rho d \tau^{\prime}
But \sum_{i=1}^{3} \hat{ r }_{i} r_{i}^{\prime}=\hat{ r } \cdot r ^{\prime}=r^{\prime} \cos \alpha=\sum_{j=1}^{3} \hat{ r }_{j} r_{j}^{\prime} ; \quad \sum_{i, j} \hat{ r }_{i} \hat{ r }_{j} \delta_{i j}=\sum \hat{ r }_{j} \hat{ r }_{j}=\hat{ r } \cdot \hat{ r }=1 . So
V_{\text {quad }}=\frac{1}{4 \pi \epsilon_{0}} \frac{1}{r^{3}} \int \frac{1}{2}\left(3 r^{\prime 2} \cos ^{2} \alpha-r^{\prime 2}\right) \rho d \tau^{\prime} = the third term in Eq. 3.96.
V( r )=\frac{1}{4 \pi \epsilon_{0}}\left[\frac{1}{r} \int \rho\left( r ^{\prime}\right) d \tau^{\prime}+\frac{1}{r^{2}} \int r^{\prime} \cos \alpha \rho\left( r ^{\prime}\right) d \tau^{\prime}\right.\left.+\frac{1}{r^{3}} \int\left(r^{\prime}\right)^{2}\left(\frac{3}{2} \cos ^{2} \alpha-\frac{1}{2}\right) \rho\left( r ^{\prime}\right) d \tau^{\prime}+\ldots\right] (3.96)
(b) Because x^{2}=y^{2}=(a / 2)^{2} \text { for all four charges, } Q_{x x}=Q_{y y}=\frac{1}{2}\left[3(a / 2)^{2}-(\sqrt{2} a / 2)^{2}\right](q-q-q+q)= 0 .
Because z = 0 for all four charges, Q_{z z}=\frac{1}{2}\left[-(\sqrt{2} a / 2)^{2}\right](q-q-q+q)=0 \text { and } Q_{x z}=Q_{y z}=Q_{z x}=Q_{z y}=0 .
This leaves only
Q_{x y}=Q_{y x}=\frac{3}{2}\left[\left(\frac{a}{2}\right)\left(\frac{a}{2}\right) q+\left(\frac{a}{2}\right)\left(-\frac{a}{2}\right)(-q)+\left(-\frac{a}{2}\right)\left(\frac{a}{2}\right)(-q)+\left(-\frac{a}{2}\right)\left(-\frac{a}{2}\right) q\right]=\frac{3}{2} a^{2} q .
(c)
2 \bar{Q}_{i j}=\int\left[3\left(r_{i}-d_{i}\right)\left(r_{j}-d_{j}\right)-( r – d )^{2} \delta_{i j}\right] \rho d \tau ( I ^{\prime}ll drop the primes, for simplicity.)
=\int\left[3 r_{i} r_{j}-r^{2} \delta_{i j}\right] \rho d \tau-3 d_{i} \int r_{j} \rho d \tau-3 d_{j} \int r_{i} \rho d \tau+3 d_{i} d_{j} \int \rho d \tau+2 d \cdot \int r \rho d \tau \delta_{i j}
-d^{2} \delta_{i j} \int \rho d \tau=Q_{i j}-3\left(d_{i} p_{j}+d_{j} p_{i}\right)+3 d_{i} d_{j} Q+2 \delta_{i j} d \cdot p -d^{2} \delta_{i j} Q .
So if p = 0 and Q = 0 then \bar{Q}_{i j}=Q_{i j} . qed
(d) Eq. 3.95 with n = 3:
V( r )=\frac{1}{4 \pi \epsilon_{0}} \sum_{n=0}^{\infty} \frac{1}{r^{(n+1)}} \int\left(r^{\prime}\right)^{n} P_{n}(\cos \alpha) \rho\left( r ^{\prime}\right) d \tau^{\prime} (3.95)
V_{ oct }=\frac{1}{4 \pi \epsilon_{0}} \frac{1}{r^{4}} \int\left(r^{\prime}\right)^{3} P_{3}(\cos \alpha) \rho d \tau^{\prime} ; \quad P_{3}(\cos \theta)=\frac{1}{2}\left(5 \cos ^{3} \theta-3 \cos \theta\right) .
V_{ oct }=\frac{1}{4 \pi \epsilon_{0}} \frac{1}{r^{4}} \sum_{i, j, k} \hat{ r }_{i} \hat{ r }_{j} \hat{ r }_{k} Q_{i j k} .
Define the “octopole moment” as
Q_{i j k} \equiv \frac{1}{2} \int\left[5 r_{i}^{\prime} r_{j}^{\prime} r_{k}^{\prime}-\left(r^{\prime}\right)^{2}\left(r_{i}^{\prime} \delta_{j k}+r_{j}^{\prime} \delta_{i k}+r_{k}^{\prime} \delta_{i j}\right)\right] \rho\left( r ^{\prime}\right) d \tau^{\prime} .
