Question

The dimension of $\frac{1}{2}{\epsilon }_0{E}^2$, where ${\epsilon }_0$ is permittivity of free space and $E$ is electric field, is

• A. $M{L}^2{T}^{-2}$
• B. $M{L}^2{T}^{-1}$
• C. $ML{T}^{-1}$
• D. $M{L}^{-1}{T}^{-2}$

Answer 1

The correct option is D $M{L}^{-1}{T}^{-2}$

Here, $\left[\frac{1}{2}{\epsilon }_0{E}^2\right]=$ [Energy density]

$=\frac{M{L}^2{T}^{-2}}{L^3}=M{L}^{-1}{T}^{-2}$

Hence, $\left(B\right)$ is the correct answer.