A physical quantity of the dimensions of time that can be formed out of $c, G$ and $\frac{e^{2}}{4 \

Question

A physical quantity of the dimensions of time that can be formed out of $c, G$ and $\frac{e^{2}}{4 \pi \varepsilon_{0}}$ is [ $c$ is velocity of light, $G$ is the universal constant of gravitation and $e$ is charge]

Professor PC Thomas Classes · Answer

(a) : Dimensions of $\frac{e^{2}}{4 \pi \varepsilon_{0}}=\left[F 	imes d^{2}ight]=\left[M L^{3} \mathrm{~T}^{-2}ight]$ \\ Dimensions of $G=\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}ight]$, Dimensions of $c=\left[\mathrm{LT}^{-1}ight]$\\ $t \propto\left(\frac{e^{2}}{4 \pi \varepsilon_{0}}ight)^{p} G^{q} c^{r} 	herefore\left[\mathrm{T}^{1}ight]=\left[\mathrm{ML}^{3} \mathrm{~T}^{-2}ight]^{p}\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}ight]^{q}\left[\mathrm{LT}^{-1}ight]^{r}$\\ On comparing both sides, $p-q=0 \Rightarrow p=q$\\ $3 p+3 q+r=0 \Rightarrow 6 p+r=0$\\ $-2 p-2 q-r=1 \Rightarrow -4 p-r=1$\\ Adding the length and time equations, $2 p=1 \Rightarrow p=\frac{1}{2}, q=\frac{1}{2}$ and $r=-3$\\ $	herefore t \propto \frac{1}{c^{3}}\left[\frac{G e^{2}}{4 \pi \varepsilon_{0}}ight]^{1 / 2}$