The dimension of $\dfrac{1}{2}\varepsilon_{0}E^{2}$, where $ \varepsilon_{0}$ is permittivity of free space and $E$ is electric field, is: [PMT/NEET-2010]
a. $ML^{2}T^{-2}$
b. $ML^{-1}T^{-2}$
c. $ML^{2}T^{-1}$
d. $MLT^{-1}$

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

Explanation:
To find the dimensions of $ \frac{1}{2}\varepsilon_{0}E^{2} $, we need to determine the dimensions of permittivity of free space $ \varepsilon_{0} $ and electric field $ E $ separately.

Dimensions of Electric Field (E)

Electric field is defined as force per unit charge:

• Force $ F $: $ [MLT^{-2}] $
• Electric charge $ q $: $ [IT] $ (current × time)
• Electric field $ E = \frac{F}{q} $: $ \frac{[MLT^{-2}]}{[IT]} = [MLT^{-3}I^{-1}] $

Dimensions of Permittivity ($ \varepsilon_{0} $)

From Coulomb’s law: $ F = \frac{1}{4\pi\varepsilon_{0}} \frac{q_1 q_2}{r^2} $

Rearranging: $ \varepsilon_{0} = \frac{1}{4\pi} \frac{q_1 q_2}{Fr^2} $

• Product of charges $ q_1 q_2 $: $ [I^2T^2] $
• Force $ F $: $ [MLT^{-2}] $
• Distance squared $ r^2 $: $ [L^2] $

Therefore: $ [\varepsilon_{0}] = \frac{[I^2T^2]}{[MLT^{-2}][L^2]} = [M^{-1}L^{-3}T^4I^2] $

Combining the Dimensions

For $ \varepsilon_{0}E^2 $:

• $ [\varepsilon_{0}] = [M^{-1}L^{-3}T^4I^2] $
• $ [E^2] = ([MLT^{-3}I^{-1}])^2 = [M^2L^2T^{-6}I^{-2}] $

$
[\varepsilon_{0}E^2] = [M^{-1}L^{-3}T^4I^2][M^2L^2T^{-6}I^{-2}] = [ML^{-1}T^{-2}] $

The factor $ \frac{1}{2} $ is dimensionless and doesn’t affect the result.

This expression $ \frac{1}{2}\varepsilon_{0}E^{2} $ represents energy density (energy per unit volume) of an electric field, which has the same dimensions as pressure or stress.