A dipole $\mathbf{p}$ is a distance$\mathbf{r}$ from a point charge $\mathbf{q}$, and oriented so that$\mathbf{p}$ makes an angle $\mathbf{\theta }$ with the vector$\mathbf{r}$ from$\mathbf{q}$ to $\mathbf{p}$.

(a) What is the force on $\mathbf{p}$?

(b) What is the force on $\mathbf{q}$?

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" $\overrightarrow{\mathbf{F}}=(\overrightarrow{\mathbf{p}}\cdot \nabla )\overrightarrow{\mathbf{E}}$

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

Write the formula offorce on $q$.

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

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

Determine the force on $p$.

Substitute$\left(px\frac{\partial }{\partial x}+py\frac{\partial }{\partial y}+pz\frac{\partial }{\partial z}\right)$for$(\overrightarrow{p}\cdot \nabla )$and$(E_x\widehat{x}+E_y\widehat{y}+E_z\widehat{z})$for$\overrightarrow{\mathrm{E}}$into equation (1).

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

Let us take the $x$-component:

$\begin{array}{c}{F}_x=p_x\left(\frac{\partial }{\partial x}+\frac{\partial }{\partial y}+\frac{\partial }{\partial z}\right)F_x\\ =\frac{p_{x}q}{4\pi {\epsilon }_0}\left(\frac{\partial }{\partial x}+\frac{\partial }{\partial y}+\frac{\partial }{\partial z}\right)\frac{x}{r^3}\\ =\frac{p_{x}q}{4\pi {\epsilon }_0}\left(\frac{1}{r^3}+x\frac{\partial }{\partial x}\frac{1}{r^3}+x\frac{\partial }{\partial y}\frac{1}{r^3}+x\frac{\partial }{\partial z}\frac{1}{r^3}\right)\\ =\frac{p_{x}q}{4\pi {\epsilon }_0}\left(\frac{1}{r^3}+\overrightarrow{x}\cdot \nabla \frac{1}{r^3}\right)\end{array}$

Solve further as

${\overrightarrow{F}}_x=\frac{p_{x}q}{4\pi {\epsilon }_0}\left(\frac{1}{r^3}-3\frac{\overrightarrow{x}\cdot \overrightarrow{r}}{r^5}\right)$

Summing the force components:

$\begin{array}{c}\overrightarrow{F}=F_x\widehat{x}+F_y\widehat{y}+F_z\widehat{z}\\ =\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^3}[\overrightarrow{p}-3(\overrightarrow{p}\cdot \widehat{r})\widehat{r}]\end{array}$

Therefore, the value of force on$P$ is$\overrightarrow{F}=\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^3}[\overrightarrow{p}-3(\overrightarrow{p}\cdot \widehat{r})\widehat{r}]$ .

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

Determine the force on $q$.

Substitute $\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^3}$for $q$and$[3(\overrightarrow{p}\cdot (-\widehat{r}))-\overrightarrow{p}]$ for$\overrightarrow{E}$ into equation (1).

$\begin{array}{c}\overrightarrow{F}=\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^3}[3(\overrightarrow{p}\cdot (-\widehat{r}))(-\widehat{r})-\overrightarrow{p}]\\ =-\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^3}[\overrightarrow{p}-3(\overrightarrow{p}\cdot \widehat{r})\widehat{r}]\end{array}$

Therefore, the value of force on$q$ is $\overrightarrow{F}=-\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^3}[\overrightarrow{p}-3(\overrightarrow{p}\cdot \widehat{r})\widehat{r}]$.