Question

Let $\epsilon _{0}$ denote the dimensional formula of the permittivity of vacuum. If $M=$ mass, $L=$ length, $T=$ time and $A=$ electric current, then dimensions of permittivity is given as $\left[\text{M}^{\text{p}} \text{L}^{\text{q}} \text{T}^{\text{r}} \text{A}^{\text{s}}\right]$ . Find the value of $\frac{p - q + r}{s}$

Answer 1

From Coulomb's law,
$F=\frac{q_{1} q_{2}}{4 \pi \epsilon _{0} r^{2}}$
$\therefore \varepsilon_{0}=\frac{ q _{1} q _{2}}{4 \pi Fr ^{2}}=\frac{\left( A ^{1} T ^{1}\right)\left( A ^{1} T ^{1}\right)}{\left[ M ^{1} L ^{1} T ^{-2}\right]\left[ L ^{2}\right]}=\left[M^{-1} L^{-3} T^{4} A^{2}\right]$
$\therefore p=-1,q=-3,r=4,s=2$
$\therefore \frac{p - q + r}{s}=\frac{- 1 - \left(\right. - 3 \left.\right) + 4}{2}=\frac{6}{2}=3$