Question

Dimensions of $\epsilon_0$ are

• A. M$^{-1}$L$^{-3}$T$^{4}$A$^{2}$
• B. M$^{0}$L$^{-3}$T$^{3}$A$^{3}$
• C. M$^{-1}$L$^{-3}$T$^{3}$A
• D. M$^{-1}$L$^{-3}$T A$^{2}$

Answer 1

Answer: (A)

M$^{-1}$L$^{-3}$T$^{4}$A$^{2}$

Answer 2

From Coulomb's law, the force between two charges $q_1$ and $q_2$ separated by distance $r$ is $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$.

Rearranging for $\epsilon_0$, we get $\epsilon_0 = \frac{q_1 q_2}{4\pi F r^2}$.

The dimensions of the quantities are:

• Charge $[q] = [AT]$ (since current $I = q/t$)

• Force $[F] = [MLT^{-2}]$

• Distance $[r] = [L]$

$4\pi$ is dimensionless.

Substituting these dimensions into the expression for $\epsilon_0$:

$[\epsilon_0] = \frac{[AT][AT]}{[MLT^{-2}][L]^2} = \frac{[A^2T^2]}{[ML^3T^{-2}]}$

$[\epsilon_0] = [M^{-1}L^{-3}T^{2}A^2T^{2}] = [M^{-1}L^{-3}T^{4}A^{2}]$.