Physics - Unit & Dimension Question with Solution | TestHub

$\begin{array}{rl}& \mathrm{E}=\frac{\mathrm{KQ}}{{\mathrm{R}}^2}\\ & \mathrm{E}=\frac{\mathrm{Q}}{4\pi {\epsilon }_0{\mathrm{R}}^2}\\ & {\epsilon }_0=\frac{\mathrm{Q}}{4\pi {\mathrm{R}}^2\mathrm{E}}\\ & \text{ Now, }{\epsilon }_0{\mathrm{E}}^2=\frac{\mathrm{Q}}{4\pi {\mathrm{R}}^2\mathrm{E}}\cdot {\mathrm{E}}^2=\frac{\mathrm{Q}}{4\pi {\mathrm{R}}^2}\cdot \mathrm{E}\\ & \left[{\epsilon }_0{\mathrm{E}}^2\right]=\left[\frac{\mathrm{QE}}{{\mathrm{R}}^2}\right]=\frac{[\mathrm{Q}][\mathrm{E}]}{\left[{\mathrm{R}}^2\right]}=\frac{[\mathrm{Q}]}{\left[{\mathrm{R}}^2\right]}\frac{[\mathrm{W}]}{[\mathrm{Q}][\mathrm{R}]}\\ & =\frac{[\mathrm{W}]}{\left[{\mathrm{R}}^3\right]}=\frac{{\mathrm{ML}}^2{\mathrm{T}}^{-2}}{{\mathrm{L}}^3}={\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}\end{array}$