Question

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).

Answer 1

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,

$V=\frac{1}{4\pi {\in }_0}[\frac{q}{XP}-\frac{2q}{YP}+\frac{q}{ZP}]$

$=\frac{1}{4\pi {\in }_0}[\frac{1}{r+a}-\frac{2q}{r}+\frac{q}{r-a}]$

$=\frac{q}{4\pi {\in }_0}\left[\frac{r(r-a)-2(r+a)(r-a)+r(r+a)}{r(r+a)(r-a)}\right]$

$=\frac{q}{4\pi {\in }_0}\left[\frac{r^2-ra-2r^2+2a^2+r^2+ra}{r(r^2-a^2)}\right]$

$=\frac{q}{4\pi {\in }_0}\left[\frac{2a^2}{r(r^2-a^2)}\right]$

$=\frac{2q{a}^2}{4\pi {\in }_0{r}^3(1-\frac{a^2}{r^2})}$

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

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

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

$\therefore V=\frac{2q{a}^2}{4\pi {\in }_0{r}^3}$

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