The dimension of $\left(\frac{1}{2}\right)\varepsilon_0 E^2$ ($\varepsilon_0$: permittivity of free space, $E$ electric field)

Step 1: Find the dimension of the electric field $E$.

The electric field is defined as force per unit charge.

$$E = \frac{F}{q}$$

The dimensions of force $F$ and charge $q$ are given by:

$$[F] = MLT^{-2}$$ $$[q] = AT$$

Substituting these dimensions yields the dimension of $E$:

$$[E] = \frac{MLT^{-2}}{AT}$$ $$[E] = MLT^{-3}A^{-1}$$

Step 2: Find the dimension of the permittivity of free space $\varepsilon_0$.

Using Coulomb's Law, the electrostatic force between two charges is:

$$F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}$$

Rearranging for $\varepsilon_0$ gives:

$$[\varepsilon_0] = \frac{[q_1][q_2]}{[F][r^2]}$$

Substituting the dimensions of charge, force, and distance:

$$[\varepsilon_0] = \frac{(AT)^2}{(MLT^{-2})(L^2)}$$ $$[\varepsilon_0] = M^{-1}L^{-3}T^{4}A^{2}$$

Step 3: Calculate the dimension of the given expression.

Since the constant $\frac{1}{2}$ is dimensionless, the dimension of the expression depends only on $\varepsilon_0$ and $E$.

$$\left[\frac{1}{2}\varepsilon_0 E^2\right] = [\varepsilon_0][E]^2$$

Substitute the dimensions calculated in the previous steps:

$$\left[\frac{1}{2}\varepsilon_0 E^2\right] = (M^{-1}L^{-3}T^{4}A^{2})(MLT^{-3}A^{-1})^{2}$$ $$\left[\frac{1}{2}\varepsilon_0 E^2\right] = (M^{-1}L^{-3}T^{4}A^{2})(M^{2}L^{2}T^{-6}A^{-2})$$ $$\left[\frac{1}{2}\varepsilon_0 E^2\right] = ML^{-1}T^{-2}$$

Step 4: Verify using energy density.

The expression $\frac{1}{2}\varepsilon_0 E^2$ represents electrostatic energy density, which is energy per unit volume.

$$u = \frac{\text{Energy}}{\text{Volume}}$$ $$[u] = \frac{ML^{2}T^{-2}}{L^{3}}$$ $$[u] = ML^{-1}T^{-2}$$

This confirms the dimensional analysis.

Answer: $ML^{-1}T^{-2}$