Question

Dimensions of ${\epsilon }_0$ are

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

Answer 1

Answer: (C)

From Coulombs law, $F=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q_1{q}_2}{r^2}$ $\therefore$ ${\epsilon }_0=\frac{1}{4\pi F}\cdot \frac{q_1{q}_2}{r^2}$ $\therefore$ Dimension of ${\epsilon }_0=\frac{1}{[ML{T}^{-2}]}\times \frac{{[AT]}^2}{{[L]}^2}$ $=[M^{-1}{L}^{-3}{T}^4{A}^2]$