Columbs law in vector form
Let the position vector of charges $\displaystyle q_{1}$ and $\displaystyle \ q_{2}$ be $\displaystyle \vec{r}_{1}$ and $\displaystyle \overrightarrow{r_{2}}$
We denote the force on $\displaystyle q_{1}$ due to $\displaystyle \ q_{2}$ by $\displaystyle \overrightarrow{F_{12}}$ and the force on $\displaystyle \ q_{2}$ due to $\displaystyle q_{1}$ by $\displaystyle \overrightarrow{F_{21}}$
$\displaystyle F_{12} =F_{21} \ $
Here $\displaystyle \vec{r}_{1} +\overrightarrow{r_{12}}$ =$\displaystyle \overrightarrow{r_{2}}$
$ $$ $$\displaystyle \overrightarrow{r_{12}} =\overrightarrow{r_{2}} -\vec{r}_{1}$
Also $\displaystyle \overrightarrow{r_{2}} +\overrightarrow{r_{21}} =\vec{r}_{1}$
or $\displaystyle \overrightarrow{r_{21}} =\vec{r}_{1} -\overrightarrow{r_{2}}$ and $\displaystyle \overrightarrow{r_{12}} =-\overrightarrow{r_{21}}$
The magnitude of vectors $\displaystyle \overrightarrow{r_{12}}$ and $\displaystyle \overrightarrow{r_{21}}$ is denoted by $\displaystyle |\overrightarrow{r_{12}} |$ and $\displaystyle |\overrightarrow{r_{21}} |$ and their unit vectors are given as
$\displaystyle \hat{r}_{12} =\frac{\overrightarrow{r_{12}} \ }{|\overrightarrow{r_{12}} |}$ and $\displaystyle \hat{r}_{21} =\frac{\overrightarrow{r_{21}} \ }{|\overrightarrow{r_{21}} |}$
So the force $\displaystyle \vec{F}_{12}$ and $\displaystyle \vec{F}_{21}$ can be given as
$\displaystyle \vec{F}_{12} =\frac{1}{4\pi \epsilon _{o}} \cdotp \frac{q_{1} q_{2}}{|\overrightarrow{r_{21}} |^{2}}\hat{r}_{21}$ $\displaystyle \&\ $ $\displaystyle \vec{F}_{21} =\frac{1}{4\pi \epsilon _{o}} \cdotp \frac{q_{1} q_{2}}{|\overrightarrow{r_{12}} |^{2}}\hat{r}_{12}$
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