Question

Question 2.22: Figure 2.34 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).

Class 12 - Physics - Electrostatic Potential And Capacitance Electrostatic Potential And Capacitance Page-89

Answer 1

see figure,
so, total potential at P = potential at P due to A + potential at P due to B + potential at P due to C + potential at P due to D
e.g., $V_P=V_{PA}+V_{PB}+V_{PC}+V_{PD}$
$\bf{now},V_{PA}=\frac{1}{4\pi\epsilon_0}\frac{q}{r+a}\\\\V_{PB}=\frac{1}{4\pi\epsilon_0}\frac{-q}{r}\\\\V_{PC}=\frac{1}{4\pi\epsilon_0}\frac{-q}{r}\\\\V_{PD}=\frac{1}{4\pi\epsilon_0}\frac{+q}{r-a}$

$so,V_P=\frac{1}{4\pi\epsilon_0}\left[\begin{array}{c}\frac{+q}{r+a}-\frac{q}{r}-\frac{q}{r}+\frac{q}{r-a}\end{array}\right]\\\\=\frac{q}{4\pi\epsilon_0}\left[\begin{array}{c}\frac{r(r-a)-2(r^2-a^2)+r(r+a)}{r(r^2-a^2)}\end{array}\right]\\\\V_P=\frac{q.2a^2}{4\pi\epsilon_0r(r^2-a^2)}=\frac{pa}{4\pi\epsilon_0r(r^2-a^2)}$
for r/a >> 1 or, r >> a
then, $V_\approx\frac{pa}{4\pi\epsilon_0r^3},or,V_p\propto\frac{1}{r^3}$

However , electric potential at any point on the kids of electric dipole is $V=\frac{1}{4\pi\epsilon_0}\frac{p}{r^2}\:or,V\propto\frac{1}{r^2}$ and due to point charge is given by $V=\frac{1}{4\pi\epsilon_0}\frac{q}{r}\:\:or,V\propto\frac{1}{r}$