Question

The unit of electrostatic constant $‘k^{\prime }$ in Coulomb’s law is

• A. $\frac{{\text{kgm}}^3{\text{s}}^4}{{\text{A}}^2}$
• B. $\frac{{\text{kgm}}^3}{{\text{A}}^2{\text{s}}^4}$
• C. $\frac{{\text{kgm}}^2{\text{S}}^2}{{\text{A}}^2}$
• D. $\frac{{\text{kgm}}^2}{{\text{A}}^2{\text{S}}^4}$

Answer 1

The correct option is B $\frac{{\text{kgm}}^3}{{\text{A}}^2{\text{s}}^4}$

According to Coulomb’s law:
$F=\frac{k{q}_1{q}_2}{r^2}$
$\Rightarrow k=\frac{F{r}^2}{q_1{q}_2}$ [Unit of $F={\text{kgms}}^{-2}$, unit of $q=As$]
$\Rightarrow$ Unit of $k={\text{kgms}}^{-2}{\text{m}}^2{\text{(As)}}^{-2}$
$={\text{kgm}}^3{\text{A}}^{-2}{\text{s}}^{-4}$
$=\frac{{\text{kg m}}^3}{{\text{A}}^2{\text{s}}^4}$
Hence, the correct answer is option (b).