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Write some important points for vector form of Coulomb’s law.
Solution
$\overrightarrow{\mathrm{F}_{21}}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q_{1} q_{2}}{r_{21}^{2}} \cdot \hat{r}_{21}$
and equation is true for both positive and negative values of $q_{1}$ and $q_{2}$.
If $q_{1}$ and $q_{2}$ both are positive or both are negative, then $\overrightarrow{F_{21}}$ is in direction of $\overrightarrow{r_{21}}$ which shows repulsion (like charges).
If $q_{1}$ and $q_{2}$ are unlike charges, then $\overrightarrow{F_{21}}$ is along $\hat{r}_{21}\left(=-\hat{r}_{12}\right)$ which shows attraction.
By replacing 1 and 2 in above equations $(1)$, $\overrightarrow{\mathrm{F}_{12}}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q_{1} q_{2}}{r_{12}^{2}} \hat{r}_{12}=-\overrightarrow{\mathrm{F}}_{21}$
Coulomb's law is agrees with Newton's third law.
If two charges are placed in medium, then the force between them becomes $\frac{1}{K}$ times means Coulomb force decreases.
Coulomb forces are central forces means they pass through line connecting centres of charges. Coulomb's law is inverse square law.
According to this law, electric force are of two types : attractive and repulsive.
There is no effect of third charge on electric force between two charges. Thus, Coulomb force is called two body force.
