Coulomb's Law in Vector Form :
Question: Write Coulomb's law in vector form. Show that Coulomb's law agrees with Newton's third law of motion ?
Consider two point charges $q_1$ and $q_2$ of same polarity separated by a distance $r$ (Figure 16). Coulomb's force acting on $q_1$ due to $q_2$ is given by,
$\vec{F}_{12} = k \frac{q_1 q_2}{r^2} \hat{r}_{21} \quad \dots (i)$
where $\hat{r}_{21}$ is the unit vector directed from $q_2$ to $q_1$.
Thus, Coulomb's force acting on point charge $q_1$ due to point charge $q_2$ is directed along the line joining $q_1$ and $q_2$ from charge $q_2$ to charge $q_1$.
Similarly, force acting on $q_2$ due to $q_1$ is given by
$\vec{F}_{21} = k \frac{q_1 q_2}{r^2} \hat{r}_{12} \quad \dots (ii)$
where $\hat{r}_{12}$ is the unit vector directed from $q_1$ to $q_2$.
Thus, Coulomb's force acting on point charge $q_2$ due to point charge $q_1$ is directed along the line joining $q_1$ and $q_2$ from charge $q_1$ to charge $q_2$.
Eqns. (i) and (ii) give Coulomb's law in vector form.
But
$\hat{r}_{12} = -\hat{r}_{21}$
so eqn. (ii) can be written as
$ \vec{F}_{21} = -k \frac{q_1 q_2}{r^2} \hat{r}_{21} \quad \dots (iii)$
From eqns. (i) and (iii), we have
$ \vec{F}_{12} = -\vec{F}_{21} \quad \dots (iv)$
Thus, the forces exerted by two point charges on each other are equal in magnitude and opposite in direction. In other words, Coulomb's law agrees with Newton's third law of motion.
Coulomb's Force between Charges in terms of their Position Vectors :
Question: Derive an expression for Coulomb's force between charges in terms of their position vectors?
Consider two point charges $q_1$ and $q_2$ lying in free space. Let $\vec{r}_1$ and $\vec{r}_2$ be their position vectors respectively
Coulomb's force acting on $q_2$ due to $q_1$ is given by,
$ \vec{F}_{21} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{|\vec{AB}|^2} \hat{r}_{12} \quad \dots (i)$
But $\vec{AB} = \vec{r}_2 - \vec{r}_1 = \vec{r}_{12}$
Substituting the above result in equation (i), we get
$\vec{F}_{21} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{|\vec{r}_{12}|^2} \hat{r}_{12} \quad \dots (ii)$
But $|\vec{r}_{12}| = |\vec{r}_2 - \vec{r}_1|$ and $(\vec{r}_2 - \vec{r}_1) = |\vec{r}_2 - \vec{r}_1| \hat{r}_{12}$ or $\hat{r}_{12} = (\vec{r}_2 - \vec{r}_1) / |\vec{r}_2 - \vec{r}_1|$
$\therefore \vec{F}_{21} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2 (\vec{r}_2 - \vec{r}_1)}{|\vec{r}_2 - \vec{r}_1|^3} \quad \dots (iii)$
Similarly, Coulomb's force acting on $q_1$ due to $q_2$ is given by,
$\vec{F}_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2 (\vec{r}_1 - \vec{r}_2)}{|\vec{r}_1 - \vec{r}_2|^3} = -\vec{F}_{21} \quad \dots (iv)$
Thus, forces on the two charges are forces of action and reaction.