Question

The constant of proportionality $(1/4\pi {\epsilon }_0)$ in Coulomb's law has the following unit.

• A. $C^{-2}N{m}^2$
• B. $C^2{N}^{-1}{m}^{-2}$
• C. $C^{2}N{m}^2$
• D. $C^{-2}{N}^{-1}{m}^{-2}$

Answer 1

The correct option is A

$C^{-2}N{m}^2$

Coulomb's law

1. According to the coulomb's law, the force of attraction or repulsion between two charges is directly proportional to the product of their charges and inversely proportional to the square of the separation distance between them.
2. The constant of proportionality in coulomb's law is $(1/4\pi {\epsilon }_0)$

$\begin{array}{l}\Rightarrow F=\left(\frac{1}{4\pi {\in }_0}\right)\frac{q_1{q}_2}{r^2}\\ \Rightarrow \frac{\left[F\times r^2\right]}{{\left[q\right]}^2}=\left(\frac{1}{4\pi {\in }_0}\right)\end{array}$

Where,

$F$ is the force between charges $q_{1}and{q}_2$,

${\epsilon }_o$ is absolute permittivity,

$r$ is separation distance,

$(1/4\pi {\epsilon }_0)$ is constant of proportionality.

Unit of constant in coulomb's law

Putting the units of each physical quantities

$\begin{array}{l}\Rightarrow \left(\frac{1}{4\pi {\in }_0}\right)=\frac{\left[F\times r^2\right]}{{\left[q\right]}^2}\\ \Rightarrow \left(\frac{1}{4\pi {\in }_0}\right)=\frac{\left[newton\right]{\left[metre\right]}^2}{{\left[Coulomb\right]}^2}\\ \Rightarrow \left(\frac{1}{4\pi {\in }_0}\right)=\frac{N{m}^2}{C^2}\\ \Rightarrow \left(\frac{1}{4\pi {\in }_0}\right)=N{m}^2{C}^{-2}\end{array}$

Hence, option A is the correct answer.