Question

The dimension of $\left(\frac{1}{2}\right){ϵ}_0{E}^2$ are(${ϵ}_0:$ permittivity of free space, $E:$ electric field)

Answer 1

The value of ${\epsilon }_0$ can be obtained as:

$F_e=\frac{1}{4\pi {\epsilon }_0}\frac{q.q}{r^2}$

So, ${\epsilon }_0=\frac{q^2}{4\pi F_e{r}^2}$

${\epsilon }_0=\frac{\left[A^2{T}^2\right]}{\left[ML{T}^{-2}\right]\left[L^2\right]}$

${\epsilon }_0=M^{-1}{L}^{-3}{T}^4{A}^2$

The dimension of the electric field is:

$\overrightarrow{E}=ML{T}^{-3}{A}^{-1}$

So,

$\frac{1}{2}{\epsilon }_0{E}^2=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]{\left[ML{T}^{-3}{A}^{-1}\right]}^2=M{L}^{-1}{T}^{-2}$