Question

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

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

Answer 1

Answer: (B)

$M{L}^{-1}{T}^{-2}$

$dimensionsof\frac{1}{2}{\epsilon }_0{E}^2\left[E\right]=ML{T}^{-2}{Q}^{-1}\left[{\epsilon }_0\right]=\left[M^{-1}{L}^{-3}{T}^2{Q}^2\right]{\epsilon }_0{E}^2=\left[M^{-1}{L}^{-3}{T}^2{Q}^2\right]\times {\left[ML{T}^{-2}{Q}^{-1}\right]}^2=M^{-1+2}{L}^{-3+2}{T}^{2-4}{Q}^{2-2}=M{L}^{-1}{T}^{-2}$