Q: Let [ε_{0} ] denote the dimensional formula of permittivity of vacuum. If M is mass, L is length, T is time and A is electric current, then

Sol: From Coulomb’s Law $\large F = \frac{q_1 q_2}{4\pi \epsilon_0 r^2}$

$\large \epsilon_0 = \frac{q_1 q_2}{4\pi F r^2}$

$\large [ \epsilon_0 ] = \frac{(A T )^2}{[MLT^{-2}] [L^2]}$

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