Question

The dimension of $\frac{1}{2}{\epsilon }_0{E}^2$ is (with usual notations):

• A. $\left[M{L}^2{T}^{-2}\right]$
• B. $\left[ML{T}^{-1}\right]$
• C. $\left[M{L}^2{T}^{-1}\right]$
• D. $\left[M{L}^{-1}{T}^{-2}\right]$

Answer 1

The correct option is D $\left[M{L}^{-1}{T}^{-2}\right]$

Dimension of ${\epsilon }_0=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$
Dimension of $E=\left[M^1{L}^1{T}^{-3}{A}^{-1}\right]$
Therefore, the dimension of $\frac{1}{2}{\epsilon }_0{E}^2$ will be $\left[M^{-1}{L}^{-3}{T}^4{A}^2\right][M^1{L}^1{T}^{-3}{A}^{-1}{]}^2=[M^1{L}^{-1}{T}^{-2}]$