Figure 2.34 shows a charge array known as an electric quadrupole.

Solution:

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 −2q placed at point Y

Charge +q placed at point Z

XY = YZ = a

YP = r

PX = r + a

PZ = r − a

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

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

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

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

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

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

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, $V \propto \frac{1}{r}$