If \(\epsilon_{ o }\) is the permittivity of free space and \(E\) is the electric field, then \(\epsilon_{ o } E ^2\) ha…

$$\begin{aligned}& \text { } E=\frac{Q}{4 \pi \varepsilon_0 R^2} \Rightarrow \varepsilon_0=\frac{Q}{4 \pi R^2 E} \\& \text { Now, } \varepsilon_0 E^2=\frac{Q}{4 \pi R^2 E} \cdot E^2=\frac{Q}{4 \pi R^2} \cdot E \\& {\left[\varepsilon_0 E^2\right]=\left[\frac{Q E}{R^2}\right]=\frac{[Q][E]}{\left[R^2\right]}=\frac{[Q]}{\left[R^2\right][Q][R]}[W]} \\& =\frac{[W]}{\left[R^3\right]}=\frac{M L^2 T^{-2}}{L^3}=M L^{-1} T^{-2}\end{aligned}$$