91Ó°ÊÓ

The magnitude of an electric field along the dipole axis due to a dipole is given by $E_{axis}=\frac{2qsr}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{\left(r-{\displaystyle \frac{s}{2}}\right)}^2{\left(r+{\displaystyle \frac{s}{2}}\right)}^2}$. In this formula, the charge is represented by q, the distance between the two charges of dipole is represented by s and the distance of position of third charge from the center of dipole is represented by r.

From the given figure, it is concluded that the charge +q and -q forms a dipole. Let the separation between this dipole bes that is s = x.

Write the distance, r of charge +Q from the center of the dipole by using figure,

$\begin{array}{rcl}r& =& 1.5x+0.5x\\ & =& 2x\end{array}$

The magnitude of the force applied on the charge +Q will be role="math" localid="1661171139412" $F_{Q^{+}}=Q{E}_{axis}$.

Substitute the value of E_axis into $F_{Q^{+}}=Q{E}_{axis}$, this gives,

$E_{axis}=\frac{2qsr}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{\left(r-{\displaystyle \frac{s}{2}}\right)}^2{\left(r+{\displaystyle \frac{s}{2}}\right)}^2}$

Therefore, the above obtained equation is the magnitude of the force on the charge +Q.

When r>> then the term $\frac{s}{2}$will vanish.

Use the approximation r >> s into the obtained equation.

role="math" localid="1661172224854" width="134" height="102">FQ+=2Qqsr4πε0r2r2=2Qqs4πε0r3

From this it is concluded that the given equation is only possible when r >> s but according to the given information, the value of r is not very much greater than s.

Therefore, the electric field on charge +Q is not $F_{Q^{+}}=\begin{array}{rcl}& & \frac{2Qqs}{4{\mathrm{Ï€Î\mu }}_0{r}^3}\end{array}$.