Four charges of same magnitude are placed at points $X, Y, Y,$ and $Z$ respectively, as shown in the following figure.

A point is located at $P$, which is $r$ distance away from point $Y$. The system of charges forms an electric quadrupole.

It can be considered that the system of the electric quadrupole has three charges.

Charge $+ q$ placed at point $X$

Charge $-2 q$ placed at point $Y$

Charge $+ q$ placed at point $Z$

$Y P=r$

$PX = r + a$

$PZ =r- a$

Electrostatic potential caused by the system of three charges at point $P$ is given by,

$V =\frac{1}{4 \pi \epsilon_{0}}\left[\frac{q}{ XP }-\frac{2 q}{ YP }+\frac{q}{ ZP }\right]$

$=\frac{1}{4 \pi \epsilon_{0}}\left[\frac{q}{r+a}-\frac{2 q}{r}+\frac{q}{r-a}\right]$

$=\frac{q}{4 \pi \epsilon_{0}}\left[\frac{r(r-a)-2(r+a)(r-a)+r(r+a)}{r(r+a)(r-a)}\right]=\frac{q}{4 \pi \epsilon_{0}}\left[\frac{2 a^{2}}{r\left(r^{2}-a^{2}\right)}\right]$

$=\frac{q}{4 \pi \epsilon_{0}}\left[\frac{r^{2}-r a-2 r^{2}+2 a^{2}+r^{2}+r a}{r\left(r^{2}-a^{2}\right)}\right]$

$=\frac{2 q a^{2}}{4 \pi \epsilon_{0} r^{3}\left(1-\frac{a^{2}}{r^{2}}\right)}$

since $\frac{r}{a}\,>\,>\,1$

$\therefore \frac{a}{r} \,<\, <\, 1$

$\frac{a^{2}}{r^{2}}$ is taken as negligible.

$\therefore V=\frac{2 q a^{2}}{4 \pi \epsilon_{0} r^{3}}$

It can be inferred that potential, $V \propto \frac{1}{r^{3}}$

However, it is known that for a dipole, $V \propto \frac{1}{r^{2}}$ And,

for a monopole, $V \propto \frac{1}{r}$