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 $r\text{for}r/a>>1$, and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).

Answer 1

Let the four charges are placed at points $A,B,\text{and}C$ respectively.

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

The charges $+q$ is places at $A$.

The charges $-2q$ is places at $B$.

The charges $+q$ is places at $C$.

Then $AB=BC=a$
$PB=r$
$PA=r+a$
$PC=r-a$
Therefore, the electrostatic potential due to the system of three charges is

$V=\frac{1}{4\pi {\epsilon }_0}(\frac{q}{r+a}+\frac{-2q}{r}+\frac{q}{r-a})$

$V=\frac{q}{4\pi {\epsilon }_0}(\frac{1}{r+a}+\frac{2}{r}+\frac{1}{r-a})$

$V=\frac{q}{4\pi {\epsilon }_0}\left(\frac{r(r-a)-2(r+a)(r-a)+r(r+a)}{r(r+a)(r-a)}\right)$

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

$V=\frac{q}{4\pi {\epsilon }_0}\left(\frac{2a^2}{r^3(1-{\left(\frac{a}{r}\right)}^2)}\right)$

Since, $\frac{r}{a}>>1\text{then}\frac{a}{r}<<1$
Therefore,

$V=\frac{q}{4\pi {\epsilon }_0}\left(\frac{2a^2}{r^3}\right)$

Hence, the dependence of potential on $r$ can be given as,

$V\propto \frac{1}{r^3}$
However, for a dipole, $V\propto \frac{1}{r^2}$
and for a monopole, $V\propto \frac{1}{r}$


Final Answer: $V\propto \frac{1}{r^2}$

for a dipole, $V\propto \frac{1}{r^2}$

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