Physics - Unit & Dimension Question with Solution | TestHub

$$\begin{array}{rl}& \text{ Dimensions of }{\epsilon }_0=\left[{\mathrm{M}}^{-1}{\mathrm{L}}^{-3}{\mathrm{T}}^4{\mathrm{A}}^2\right]\\ & \text{ Dimensions of }\mathrm{E}=\left[{\mathrm{MLT}}^{-3}{\mathrm{A}}^{-1}\right]\\ & \therefore \text{ Dimensions of }\frac{1}{2}{\epsilon }_0{\mathrm{E}}^2=\left[{\mathrm{M}}^{-1}{\mathrm{L}}^{-3}{\mathrm{T}}^4{\mathrm{A}}^2\right]\\ & =\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-3}\right]\times \left[{\mathrm{M}}^2{\mathrm{L}}^2{\mathrm{T}}^{-6}{\mathrm{A}}^{-2}\right]\end{array}$$