The figure shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole,obtain the dependence of potential on $r$ for $r / a >> 1,$ and contrast your results with that due to an electric dipole,and an electric monopole (i.e.,a single charge).

(N/A) The system consists of four charges: $+q$ at $x = -a$,$-q$ at $x = 0$,$-q$ at $x = 0$,and $+q$ at $x = a$. This is equivalent to a charge $+q$ at $x = -a$,a charge $-2q$ at $x = 0$,and a charge $+q$ at $x = a$.
Let point $P$ be at a distance $r$ from the origin (where the $-2q$ charge is located) along the axis.
The distances of the charges from $P$ are:
For $+q$ at $x = -a$: $d_1 = r + a$
For $-2q$ at $x = 0$: $d_2 = r$
For $+q$ at $x = a$: $d_3 = r - a$
The total electrostatic potential $V$ at point $P$ is:
$V = \frac{1}{4 \pi \epsilon_{0}} \left[ \frac{q}{r + a} - \frac{2q}{r} + \frac{q}{r - a} \right]$
$V = \frac{q}{4 \pi \epsilon_{0}} \left[ \frac{r(r - a) - 2(r^2 - a^2) + r(r + a)}{r(r^2 - a^2)} \right]$
$V = \frac{q}{4 \pi \epsilon_{0}} \left[ \frac{r^2 - ra - 2r^2 + 2a^2 + r^2 + ra}{r(r^2 - a^2)} \right]$
$V = \frac{q}{4 \pi \epsilon_{0}} \left[ \frac{2a^2}{r(r^2 - a^2)} \right]$
For $r >> a$,we can approximate $r^2 - a^2 \approx r^2$:
$V \approx \frac{2qa^2}{4 \pi \epsilon_{0} r^3}$
Thus,for an electric quadrupole,$V \propto \frac{1}{r^3}$.
Comparing this with other systems:
For an electric monopole (single charge),$V \propto \frac{1}{r}$.
For an electric dipole,$V \propto \frac{1}{r^2}$.