Question

The dimension of $\left(\frac{1}{2}\right){\epsilon }_0{E}^2$ is(${\epsilon }_0$: permittivity of free space, Electric field)

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

Answer 1

Answer: (C)

${ML}^{-1}{T}^{-2}$

As we know that

Energy Density$=\frac{energy}{volume}=\left(\frac{1}{2}\right){\epsilon }_0{E}^2$

Hence ,

Dimesion of $\left(\frac{1}{2}\right){\epsilon }_0{E}^2$ is, =$\frac{\left[M{L}^2{T}^{-2}\right]}{L^3}=M{L}^{-1}{T}^{-2}$