# What is the Vector form of Coulomb's Law?

We know that two like or similar charges repel each other and two opposite charges attract each other. Coulomb's law reveals these forces that are working in the charges ( To know the details of Coulomb's law click here... ).

But force is a vector quantity, so it has both magnitude and direction. That is why Coulomb's law can be re-written in the form of vectors.

# What is the Vector form of Coulomb's Law?

Consider in a Cartesian coordinate system there are two similar point charges q1 and q2 are in positions A and B respectively. Here r1 is the position vector of points A and r2 is the position vector of point B relative to the origin O of the system.

We will determine - 1. The force that is acting on the charge q1 due to the chare q2 and 2. The force that is acting on the charge q2 due to the charge q1.

1. The force that is acting on the charge q1 due to the chare q2 :

$\vec{F}_{12}=\frac{1}{4\pi \varepsilon }\frac{q_{1}q_{2}}{r_{12}^{2}}\frac{\vec{r}_{12}}{r_{12}}=\frac{1}{4\pi \varepsilon }\frac{q_{1}q_{2}}{r_{12}^{3}}\vec{r}_{12}$

$Where, \;\vec{r}_{12}=\vec{r}_{1}-\vec{r}_{2}\; (The \;position\;vector\;from\;q_{2}\;to\;q_{1})$

This is the vector form of Coulomb's Law.

We can also write this vector form of coulomb's law in terms of unit vector notation

$\vec{F}_{12}=\frac{1}{4\pi \varepsilon }\frac{q_{1}q_{2}}{r_{12}^{2}}\hat{r}_{12}$

$Where,\;\hat{r}_{12} \;is \;the \;unit \;vector \;that \;shows \;the \;direction \;of \;the \;force \;(q_{2} \;to \;q_{1} \;direction).$

2. The force that is acting on the charge q2 due to the charge q1 :

$\vec{F}_{21}=\frac{1}{4\pi \varepsilon }\frac{q_{1}q_{2}}{r_{21}^{2}}\frac{\vec{r}_{21}}{r_{21}}=\frac{1}{4\pi \varepsilon }\frac{q_{1}q_{2}}{r_{21}^{3}}\vec{r}_{21}$

$Where, \;\vec{r}_{21}=\vec{r}_{2}-\vec{r}_{1}\; (The \;position\;vector\;from\;q_{1}\;to\;q_{2})$

We can also write this vector form of coulomb's law in terms of unit vector notation

$\vec{F}_{21}=\frac{1}{4\pi \varepsilon }\frac{q_{1}q_{2}}{r_{21}^{2}}\hat{r}_{21}$

$Where,\;\hat{r}_{21} \;is \;the \;unit \;vector \;that \;shows \;the \;direction \;of \;the \;force \;(q_{1} \;to \;q_{2} \;direction).$

$Here,\;\vec{r}_{12}=\vec{r}_{1}-\vec{r}_{2}\;and\;\vec{r}_{21}=\vec{r}_{2}-\vec{r}_{1}$

$So, \;\vec{r}_{12}=-(\vec{r}_{2}-\vec{r}_{1})=-\vec{r}_{21}$

$\therefore \vec{F}_{12}=-\vec{F}_{21}$

This says that the forces that are working in the electrical charge particle are equal but opposite, which is Newton's third law of motion. Thus, Coulomb’s law agrees with Newton’s third law.

Read Next: Forces Between Multiple Charges