Question

If ${\epsilon }_0$ is the permittivity of free space and $E$ is the electric field, then the dimensions of $\frac{1}{2}{\epsilon }_0{E}^2$ is

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

Answer 1

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

Here, $\frac{1}{2}{\epsilon }_0{E}^2$ represetns energy per unit volume
So the dimensions of
$\left[{\epsilon }_0\right]\left[E^2\right]=\frac{\left[DimensionsofEnergy\right]}{\left[DimensionsofVolume\right]}$
$=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[L^3\right]}=\left[M{L}^{-1}{T}^{-2}\right]$