Question

Let ${ϵ}_0$ 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

Answer: 3

From Coulomb's law,
$F=\frac{q_1{q}_2}{4\pi {ϵ}_0{r}^2}$
$\therefore {\epsilon }_0=\frac{q_1{q}_2}{4\pi F{r}^2}=\frac{\left(A^1{T}^1\right)\left(A^1{T}^1\right)}{\left[M^1{L}^1{T}^{-2}\right]\left[L^2\right]}=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$
$\therefore p=-1,q=-3,r=4,s=2$
$\therefore \frac{p-q+r}{s}=\frac{-1-(-3)+4}{2}=\frac{6}{2}=3$