Question

The dimension of $\frac{1}{2}{ϵ}_o{E}^2$ permittivity of free space and E:
Intensity of electric field) is

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

Answer 1

Answer: (B)

Energy density of an electric field $E$ is ,

${\mu }_E=\frac{1}{2}{ϵ}_0{E}^2$

where ${ϵ}_0$ is permitivity of free space

${\mu }_E=\frac{Energy}{Volume}=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[L^3\right]}=\left[M{L}^{-1}{T}^{-2}\right]$

Hence, the dimension of $\frac{1}{2}{ϵ}_0{E}^2$ is $\left[M{L}^{-1}{T}^{-2}\right]$