91Ó°ÊÓ

Consider thedipole $\mathrm{p}$ .

Consider the dipole $p$with a distance$r$ from a point charge $q$.

Write the formula of force on $p$.

role="math" localid="1657541332834" $\stackrel{\mathbf{→}}{\mathbf{F}}=(\stackrel{\mathbf{→}}{\mathbf{p}}â‹\dots ∇)\stackrel{\mathbf{→}}{\mathbf{E}}$

Here,role="math" localid="1657541296947" $\stackrel{\mathbf{→}}{\mathbf{p}}$is dipole and $\stackrel{\mathbf{→}}{\mathbf{E}}$is electric field on dipole.

Write the formula offorce on $q$.

$\stackrel{\mathbf{→}}{\mathbf{F}}\mathbf{=}\mathbf{q}\stackrel{\mathbf{→}}{\mathbf{E}}$ …… (1)

Here, $\mathbf{q}$ is point charge and $\stackrel{\mathbf{→}}{\mathbf{E}}$is electric field on dipole.

Determine the force on $p$.

Substitute$\left(px\frac{∂}{∂x}+py\frac{∂}{∂y}+pz\frac{∂}{∂z}\right)$for$(\stackrel{→}{p}â‹\dots ∇)$and$(E_x\widehat{x}+E_y\widehat{y}+E_z\widehat{z})$for$\stackrel{→}{\mathrm{E}}$into equation (1).

$\stackrel{→}{F}=\left(px\frac{∂}{∂x}+py\frac{∂}{∂y}+pz\frac{∂}{∂z}\right)(E_x\widehat{x}+E_y\widehat{y}+E_z\widehat{z})$

Let us take the $x$-component:

$\begin{array}{c}{F}_x=p_x\left(\frac{∂}{∂x}+\frac{∂}{∂y}+\frac{∂}{∂z}\right)F_x\\ =\frac{p_{x}q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{∂}{∂x}+\frac{∂}{∂y}+\frac{∂}{∂z}\right)\frac{x}{r^3}\\ =\frac{p_{x}q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{1}{r^3}+x\frac{∂}{∂x}\frac{1}{r^3}+x\frac{∂}{∂y}\frac{1}{r^3}+x\frac{∂}{∂z}\frac{1}{r^3}\right)\\ =\frac{p_{x}q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{1}{r^3}+\stackrel{→}{x}â‹\dots ∇\frac{1}{r^3}\right)\end{array}$

Solve further as

${\stackrel{→}{F}}_x=\frac{p_{x}q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{1}{r^3}−3\frac{\stackrel{→}{x}â‹\dots \stackrel{→}{r}}{r^5}\right)$

Summing the force components:

$\begin{array}{c}\stackrel{→}{F}=F_x\widehat{x}+F_y\widehat{y}+F_z\widehat{z}\\ =\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^3}[\stackrel{→}{p}−3(\stackrel{→}{p}â‹\dots \widehat{r})\widehat{r}]\end{array}$

Therefore, the value of force on$P$ is$\stackrel{→}{F}=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^3}[\stackrel{→}{p}−3(\stackrel{→}{p}â‹\dots \widehat{r})\widehat{r}]$ .

The charge $q$is subjected to a force on the dipole $p$, that is (reversing the position vector $\stackrel{→}{P}$).

Determine the force on $q$.

Substitute $\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^3}$for $q$and$[3(\stackrel{→}{p}â‹\dots (−\widehat{r}))−\stackrel{→}{p}]$ for$\stackrel{→}{E}$ into equation (1).

$\begin{array}{c}\stackrel{→}{F}=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^3}[3(\stackrel{→}{p}â‹\dots (−\widehat{r}))(−\widehat{r})−\stackrel{→}{p}]\\ =−\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^3}[\stackrel{→}{p}−3(\stackrel{→}{p}â‹\dots \widehat{r})\widehat{r}]\end{array}$

Therefore, the value of force on$q$ is $\stackrel{→}{F}=−\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^3}[\stackrel{→}{p}−3(\stackrel{→}{p}â‹\dots \widehat{r})\widehat{r}]$.