Question

Write Coulomb’s law in vector form and show that it obeys Newton’s third law of motion.

Answer 1

Coulomb’s law defines the electrostatic force that acts between the two charges. As a force is a vector, thus, Coulomb's law also becomes a vector. As, Newton’s third law of motion applies to the property of forces, thus, the Coulomb’s law and Newton’s third law of motion are interrelated to each other.

Formula used:
$\overset{\to }{\mathop{F}}\,=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{|\overset{\to }{\mathop{d}}\,{{|}^{2}}}\overset{\hat{\ }}{\mathop{r}}\,$

Complete answer:
Let us understand the concepts of Coulomb’s law and Newton’s third law of motion.

Coulomb’s law states that “The electrostatic force of attraction/repulsion between the charged particles is directly proportional to the product of the magnitude of the charges and is inversely proportional to the distance between the same two charged particles.

Newton’s third law of motion is defined as, “For every action, there is an equal and opposite reaction.” The quantities, equal and opposite refers to the magnitude and the direction as, equal in magnitude and opposite in direction.

The Coulomb’s law of force deals with the electrostatic force that is generated between the charged particles whereas Newton's law of motion deals with the action and the related reaction between the bodies.
Now consider Coulomb's law.
Using the statement, we have,
$F\propto {{q}_{1}}{{q}_{2}}$and $F\propto \dfrac{1}{{{d}^{2}}}$
$\Rightarrow F\propto \dfrac{{{q}_{1}}{{q}_{2}}}{{{d}^{2}}}$
$\Rightarrow \overset{\to }{\mathop{F}}\,=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{|\overset{\to }{\mathop{d}}\,{{|}^{2}}}\overset{\hat{\ }}{\mathop{r}}\,$
Where r represents the unit vector, ${{q}_{1}},{{q}_{2}}$represents the charges and d represents the distance between the charges.

Hence the vector form of Coulomb's law.

Let ${{F}_{1}}$be the electrostatic force applied by the charge ${{q}_{1}}$on to the charge ${{q}_{2}}$. So, we have,
$\overset{\to }{\mathop{{{F}_{12}}}}\,=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{|\overset{\to }{\mathop{d}}\,{{|}^{2}}}\overset{\hat{\ }}{\mathop{{{r}_{12}}}}\,$ …… (1)
Let ${{F}_{2}}$be the electrostatic force applied by the charge ${{q}_{2}}$on to the charge ${{q}_{1}}$. So, we have,
$\overset{\to }{\mathop{{{F}_{21}}}}\,=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{|\overset{\to }{\mathop{d}}\,{{|}^{2}}}\overset{\hat{\ }}{\mathop{{{r}_{21}}}}\,$…… (2)

Comparing the equations (1) and (2), we get,
$\overset{\to }{\mathop{{{F}_{12}}}}\,=-\overset{\to }{\mathop{{{F}_{21}}}}\,$……. (3)

As both the equations, that is, (1) and (2), represent the same force in terms of the magnitude and the equation (3) represents the opposite directions of forces, thus, implies that the Coulomb’s law obeys Newton’s third law of motion.

Note:
The things to be on your finger-tips for further information on solving these types of problems are: A tip to understand this type of concept is, to remember the scalar and vector quantities. When both the quantities, that is, the magnitude and the direction are involved, then those quantities will be vector quantities, but not all.