91Ó°ÊÓ

$\mathbf{\left|}{\stackrel{\mathbf{→}}{\mathbf{E}}}_{\mathbf{axis}}\mathbf{\right|}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}}\mathbf{.}\frac{\mathbf{2}\mathbf{qs}}{{\mathbf{r}}^{\mathbf{3}}}$

Where q is the one charge of the dipole, s is the distance between two charges of dipole, r is the distance of the point where we need to find the electric field and$\mathbf{r}\mathbf{>}\mathbf{>}\mathbf{s}$.

$\stackrel{\mathbf{→}}{\mathbf{F}}\mathbf{=}\mathbf{Q}\stackrel{\mathbf{→}}{\mathbf{E}}$

It is given that $\mathrm{d}≫\mathrm{s}$.

Therefore, we can use the formula of the electric field produced by dipole given as,

$\left|{\stackrel{→}{\mathrm{E}}}_{\mathrm{axis}}\right|=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}.\frac{2\mathrm{qs}}{{\mathrm{d}}^3}$

We know that,

If $\stackrel{→}{\mathrm{E}}$ is electric field at some location, then the electric force $\stackrel{→}{\mathrm{F}}$acting on any charge Qwe place there is given as,

$\stackrel{→}{\mathrm{F}}=\mathrm{Q}\stackrel{→}{\mathrm{E}}$

The forces acting on the charge $+\mathrm{Q}$ can be given as,

role="math" localid="1661318584471" $\begin{array}{c}\stackrel{→}{{\mathrm{F}}_{\mathrm{v}}}=\mathrm{Q}\stackrel{→}{\mathrm{E}}\\ \stackrel{→}{{\mathrm{F}}_{\mathrm{v}}}=\mathrm{Q}\left[\frac{1}{4{\mathrm{Ï€Î\mu }}_0}.\frac{2\mathrm{qs}}{{\mathrm{d}}^3}\right]\\ {\stackrel{→}{\mathrm{F}}}_{\mathrm{v}}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}.\frac{2\mathrm{qsQ}}{{\mathrm{d}}^3}\end{array}$

Now, the forces acting on the charge $−9\mathrm{Q}$ can be given as,

$\begin{array}{l}{\stackrel{→}{\mathrm{F}}}_{\mathrm{new}}=−9\mathrm{Q}\stackrel{→}{\mathrm{E}}\\ {\stackrel{→}{\mathrm{F}}}_{\mathrm{new}}=−9\mathrm{Q}\left[\frac{1}{4{\mathrm{Ï€Î\mu }}_0}.\frac{2\mathrm{qs}}{{\mathrm{d}}^3}\right]\\ {\stackrel{→}{\mathrm{F}}}_{\mathrm{new}}=−\frac{1}{4{\mathrm{Ï€Î\mu }}_0}.\frac{2\mathrm{qs}\left(9\mathrm{Q}\right)}{{\mathrm{d}}^3}\end{array}$

Therefore,

$\begin{array}{l}\frac{{\stackrel{→}{\mathrm{F}}}_{\mathrm{new}}}{{\mathrm{F}}_{\mathrm{v}}}=\frac{\frac{1}{4{\mathrm{Ï€Î\mu }}_0}.\frac{2\mathrm{qsQ}}{{\mathrm{d}}^3}}{−\frac{1}{4{\mathrm{Ï€Î\mu }}_0}.\frac{2\mathrm{qs}\left(9\mathrm{Q}\right)}{{\mathrm{d}}^3}}\\ \frac{{\stackrel{→}{\mathrm{F}}}_{\mathrm{new}}}{{\mathrm{F}}_{\mathrm{v}}}=−9\end{array}$

This implies that, the magnitude of force at that point will change by a factor of 9.

$\left|\frac{{\stackrel{→}{\mathrm{F}}}_{\mathrm{new}}}{{\mathrm{F}}_{\mathrm{v}}}\right|=9$

The force on the charge $+\mathrm{Q}$is

${\stackrel{→}{\mathrm{F}}}_{\mathrm{v}}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}.\frac{2\mathrm{qsQ}}{{\mathrm{d}}^3}$

The force on charge $−9\mathrm{Q}$ is

${\stackrel{→}{\mathrm{F}}}_{\mathrm{new}}=−\frac{1}{4{\mathrm{Ï€Î\mu }}_0}.\frac{2\mathrm{qs}\left(9\mathrm{Q}\right)}{{\mathrm{d}}^3}$

From above calculation we can conclude that the direction of forces is apposite on the new charge.