The dimension of $(1/2){\epsilon }_0{E}^2$${\epsilon }_0$ is the permittivity of free space, $E$ is the electric field

The correct option is C

$\left[M{L}^{-1}{T}^{-2}\right]$

Step 1: Given :

Formula of electric field

$(1/2){\epsilon }_0{E}^2$,${\epsilon }_0$ is the permittivity of free space

Step 2: Find the dimension of the ${\epsilon }_0$ :

For the dimension of ${\epsilon }_0$

We know that, $F=\frac{1}{4\mathrm{\pi }{\in }_0}\frac{q_1{q}_2}{r^2}$, where $F=$Force, $q=$charge, and $r=$distance between the charges

So, ${\in }_0=\frac{1}{4\mathrm{\pi F}}\frac{q_1{q}_2}{r^2}$

Dimensional Formula will be ${\in }_0=\frac{\left[AT\right]\left[AT\right]}{\left[ML{T}^{-2}\right]\left[L^2\right]}=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$

Step 3: Dimension of electric field:

And the dimension of electric field $E,$

We know that $E=\frac{F}{Q}\dots \dots \dots \dots \dots \left(1\right)$, where $E=$Electric Field

Also, $F=m\times a$, where $m=$mass, and $a=$acceleration

Char$\left(q\right)=i\times t$

Dimension of Electric Field, using $\left(1\right),$

$\begin{array}{rcl}E& =& \frac{F}{q}\\ & =& \frac{\left[ML{T}^{-2}\right]}{\left[A\right]\left[T\right]}\\ & =& \left[ML{T}^{-3}{A}^{-1}\right]\end{array}$

Step 4: Dimension of $(1/2){\epsilon }_0{E}^2$

$$\begin{gathered} =\frac{{\epsilon }_0{E}^2}{2} \\ =\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]{\left[ML{T}^{-3}{A}^{-1}\right]}^2 \\ =\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]\left[M^2{L}^2{T}^{-6}{A}^{-2}\right] \\ =\left[M{L}^{-1}{T}^{-2}\right] \end{gathered}$$

Hence, the correct option is C.