Physics - Unit & Dimension Question with Solution | TestHub

The dimensional formulae of $$\begin{array}{rl}& \mathrm{e}=\left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^1{\mathrm{A}}^1\right]\\ & {\epsilon }_0=\left[{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^4{\mathrm{A}}^2\right]\\ & \mathrm{G}=\left[{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\right]\\ & \text{ and }{\mathrm{m}}_{\mathrm{e}}=\left[{\mathrm{M}}^1{\mathrm{L}}^0{\mathrm{T}}^0\right]\\ & \text{ Now, }\frac{{\mathrm{e}}^2}{2\pi {\epsilon }_0{\mathrm{Gm}}_{\mathrm{e}}^2}\\ & =\frac{{\left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^1{\mathrm{A}}^1\right]}^2}{2\pi \left[{\mathrm{M}}^{-1}{\mathrm{L}}^{-3}{\mathrm{T}}^4{\mathrm{A}}^2\right]\left[{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\right]{\left[{\mathrm{M}}^1{\mathrm{L}}^0{\mathrm{T}}^0\right]}^2}\end{array}$$ $$\begin{array}{rl}& =\frac{\left[{\mathrm{T}}^2{\mathrm{A}}^2\right]}{2\pi \left[{\mathrm{M}}^{-1-1+2}{\mathrm{L}}^{-3+3}{\mathrm{T}}^{4-2}{\mathrm{A}}^2\right]}\\ & =\frac{\left[{\mathrm{T}}^2{\mathrm{A}}^2\right]}{2\pi \left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^2{\mathrm{A}}^2\right]}=\frac{1}{2\pi }\end{array}$$ $\because \frac{1}{2\pi }$ is dimensionless thus the combination $\frac{{\mathrm{e}}^2}{2\pi {\epsilon }_0{\mathrm{Gm}}_{\mathrm{e}}^2}$ would have the same value in different systems of units.