# Let $\varepsilon_0$ denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length and T = time and A = electric current then,

$\begin {array} {1 1} (A)\;[ \varepsilon_0 ]=[ M^{-1}L^{-3}T^4A^2] & \quad (B)\;[ \varepsilon_0 ]=[ M^{-1}L^{2}T^{-1}A^{-2}] \\ (C)\;[ \varepsilon_0 ]=[ M^{-1}L^{2}T^{-1}A] & \quad (D)\;[ \varepsilon_0 ]=[ M^{-1}L^{-3}T^2A ] \end {array}$

$\large\frac{1}{4 \pi \varepsilon_0}$ $\large\frac{q^2}{r^2}$ $= F$
$\varepsilon_0 = \large\frac{[A^2T^2]}{[MLT^{-2}L^2]}$$= M^{-1}L^{-3}T^4A^2$
Ans : (A)
edited Mar 14, 2014