Question

In coulomb’s law, the constant of proportionality, $k=\dfrac{1}{4\pi {{\varepsilon }_{0}}}$ has units?
$\begin{aligned} & A.\text{ N} \\\ & B.\text{ N-}{{\text{m}}^{2}}\text{ } \\\ & C.\text{ N}{{\text{m}}^{2}}/{{C}^{2}} \\\ & D.\text{ N}{{\text{C}}^{2}}/{{m}^{2}} \\\ \end{aligned}$

Answer 1

Question itself has a hint that we need to use coulomb’s law, to calculate unit of proportionality constant. Use expression of Coulomb's law which gives relation between charges, force, distance between charges and constant of proportionality (k). Put units of force, charge and distance in the given expression. This way you will get a unit of k.

Formula used: Expression of coulombs’ law is given as,
$F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$
Where, ${{q}_{1}}\text{ and }{{\text{q}}_{2}}$ are the charges
“r” is the distance between two charges.
F is the force acting between two charges.

Complete step by step answer:
If we consider two charged bodies and if charges are like i.e. both positive and both negative, the bodies replicate each other. If charges on the body are unlike then bodies attract each other. In 1785, coulomb’s first measured electrical attractions and repulsion quantitatively and deduced the law which governs them. We know that coulomb’s law is applicable only for rest charge. Hence Coulomb's law stated as the electrostatic force of attraction or repulsion between two charges of the body and inversely proportional to the square of the distance between them.
Consider the force in coulomb’s law always along the line joining the charges. Let ${{q}_{1}}\text{ and }{{\text{q}}_{2}}$be the two point electric charges separated by distance r and F be the magnitude of electrostatic force of attraction or repulsion between them. Then according to coulomb’s law, we have,
$\begin{aligned} & F\alpha \dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}} \\\ & \therefore F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}} \\\ \end{aligned}$
Where, k is the constant of proportionality. ‘r’ is the distance between two bodies. If the charges are kept in vacuum, then k can be expressed as$k=\dfrac{1}{4\pi {{\varepsilon }_{0}}}$
Where, ${{\varepsilon }_{o}}$ is called the permittivity of free space. Hence in vacuum, coulomb’s law can be expressed as,
$F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}.......\left( 1 \right)$
We know that the unit of force (f) is Newton denoted (N). Unit of charge is Coulomb, denoted by C. The unit of distance between two charge(r) is meter, denoted by m.
Rearrange equation (1), we get
$\begin{aligned} & \dfrac{1}{4\pi {{\varepsilon }_{0}}}=\dfrac{F{{r}^{2}}}{{{q}_{1}}{{q}_{2}}} \\\ & 1\text{ }unit\text{ }of\text{ }\dfrac{1}{4\pi {{\varepsilon }_{0}}}=\dfrac{N{{m}^{2}}}{{{C}^{2}}} \\\ & i.e\text{ }\dfrac{1}{4\pi {{\varepsilon }_{0}}}=\dfrac{N{{m}^{2}}}{{{C}^{2}}} \\\ \end{aligned}$
Hence, in coulomb’s law, the constant of proportionality, $k=\dfrac{1}{4\pi {{\varepsilon }_{0}}}$has unit $N{{m}^{2}}/{{C}^{2}}$

So, the correct answer is “Option C”.

Note: Coulomb’s law is applicable for static or steady charges only. Electric charges in coulomb’s law always exist in internal form. If the two charges ${{q}_{1}}\text{ and }{{\text{q}}_{2}}$ are situated in medium of permittivity $\varepsilon$, then coulomb’s law can be expressed as follows,
$F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$
Where, $\varepsilon ={{\varepsilon }_{0}}k$
Students usually mug the units, but in spite of mugging, you can remember formulas and can deduce units.