Question

Let $\varepsilon_{0}$ 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 $\left[M^{p} L^{q} T^{r} A^{s}\right]$. Find the value of $\frac{p- q+ r}{s}$.

Answer 1

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