Question

Dimensional formula for ${\epsilon }_0$ is

• A. $M^{-1}{L}^{-2}{A}^2{T}^2]$
• B. $\left[M{L}^{-2}{A}^{-2}{T}^4\right]$
• C. $\left[M^{-1}{L}^{-3}{A}^2{T}^4\right]$
• D. $\left[M{L}^3{A}^{-2}{T}^{-4}\right]$

Answer 1

Answer: (C)

From Coulomb's law, $F=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q_1{q}_2}{r^2}$
$\therefore {\epsilon }_0=\frac{1}{4\pi F}\frac{q_1{q}_2}{r^2}$
$\therefore$ Dimensions of ${\epsilon }_0=\frac{1}{\left[ML{T}^{-2}\right]}\times \frac{[AT{]}^2}{\left[L^2\right]}$
$=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$