If $$\epsilon_0$$ is the permittivity of free space and E is the electric field, then $$\epsilon_0 E^2$$ has the dimensions :

We need to find the dimensions of $$\epsilon_0 E^2$$, where $$\epsilon_0$$ is the permittivity of free space and $$E$$ is the electric field.

The electrostatic energy density stored in an electric field is $$u = \frac{1}{2}\epsilon_0 E^2$$. Since $$\frac{1}{2}$$ is dimensionless, $$\epsilon_0 E^2$$ has the same dimensions as energy density.

Energy density is defined as energy per unit volume, so

$$[u] = \frac{[E]}{[V]} = \frac{ML^2T^{-2}}{L^3} = ML^{-1}T^{-2}$$

Alternatively, we can verify this result by considering the dimensions of $$\epsilon_0$$ and $$E$$ separately. From Coulomb’s law, $$F = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}$$, so

$$\epsilon_0 = \frac{q^2}{Fr^2} \implies [\epsilon_0] = \frac{A^2T^2}{MLT^{-2}\cdot L^2} = M^{-1}L^{-3}T^4A^2$$

The electric field has dimensions

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

Therefore,

$$[\epsilon_0 E^2] = M^{-1}L^{-3}T^4A^2 \times M^2L^2T^{-6}A^{-2} = ML^{-1}T^{-2}$$

This confirms that $$[\epsilon_0 E^2] = ML^{-1}T^{-2}$$. The correct answer is Option (4): $$[ML^{-1}T^{-2}]$$.