Explain vector form of Coulomb’s law and its importance. Write some important points for vector form of Coulomb’s law.

Suppose, position vectors of $q_{1}$ and $q_{2}$ are $r_{1}$ and $r_{2}$ respectively as shown in figure (a).

Let, force acting on $q_{1}$ by $q_{2}$ is $\vec{F}_{12}$ and force on $q_{2}$ by $q_{1}$ is $\vec{F}_{21} \cdot$

If $1$and $2$ numbers are given to $q_{1}$ and $q_{2}$, then $\overrightarrow{r_{21}}$ is position vector from 1 to 2 and $\overrightarrow{r_{12}}$ is

position vector from $2$ to $1$ .

By using triangle method for vector addition,

$\overrightarrow{r_{1}}+\overrightarrow{r_{21}}=\overrightarrow{r_{2}}$

$\therefore \overrightarrow{r_{21}}=\overrightarrow{r_{2}}-\overrightarrow{r_{1}}$ and $\overrightarrow{r_{12}}=\overrightarrow{r_{1}}-\overrightarrow{r_{2}}=-\overrightarrow{r_{21}}$

and $\left|\overrightarrow{r_{12}}\right|=r_{12}$ also $\left|\overrightarrow{r_{21}}\right|=r_{21}$

$\therefore \vec{r}_{12}=\frac{r_{12}}{r_{12}}$ and $\hat{r}_{21}=\frac{\overrightarrow{r_{21}}}{r_{21}}$

Force acting on $q_{2}$ by $q_{1}$ '

$\overrightarrow{\mathrm{F}_{21}}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q_{1} q_{2}}{r_{21}^{2}} \cdot \hat{r}_{21}$ and

Force acting on $q_{1}$ by $q_{2}$ '

$\overrightarrow{\mathrm{F}_{12}}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q_{1} q_{2}}{r_{12}^{2}} \cdot \hat{r}_{12}$

but $\hat{r}_{12}=-\hat{r}_{21}$,

$\overrightarrow{\mathrm{F}_{21}}=-\overrightarrow{\mathrm{F}_{12}}$