91Ó°ÊÓ

An electric quadrupole: Two dipoles with dipole moments that are the same in magnitude but opposite in direction.

The given quadruple moment is given as:$Q=2q{d}^2$

The electric quadruple is composed of two dipoles, each with a dipole moment of magnitude,. The dipole moments point in the opposite directions and produce fields in the opposite directions along the quadruple axis. Considering the point P on the axis, a distance z to the right of the quadruple center takes a rightward pointing field to be positive.

The electric field of an electric dipole along the dipole axis by the left dipole,

$E=\frac{1}{2\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{qd}{{(z-d/2)}^3}$ (i)

The electric field of an electric dipole along the dipole axis by the right dipole,

$E=\frac{-1}{2\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{qd}{{(z+d/2)}^3}$ (ii)

Using the binomial expansions of the denominator of equations (i) and (ii), we can get,

${(z–d/2)}^{–3}≈z^{–3}+3z^{–4}(–d/2)$

${(z+d/2)}^{–3}≈z^{–3}–3z^{–4}(d/2)$

Now, the net electric field from both the dipole points is given by substituting the above values in equations (i) and (ii) as follows:

$\begin{array}{c}E=\frac{qd}{2\mathrm{Ï€}{\mathrm{Î\mu }}_o}[\frac{1}{z^3}+\frac{3d}{2z^4}−\frac{1}{z^3}+\frac{3d}{2z^4}]\\ =\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{6q{d}^2}{z^4}\\ =\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{3Q}{z^4}\text{(}âˆ\mu Q=2q{d}^2\text{)}\end{array}$

Hence, the value of the electric dipole is $\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{3Q}{z^4}$for the given quadruple moment.