Question

The dimension of $\frac{1}{2}{\epsilon }_0{E}^2$ (${\epsilon }_0$ is the permittivity of free space and $E$ is 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)

$\left[M{L}^{-1}{T}^{-2}\right]$

We know that the given equation gives the energy density for a capacitor. Dimensions of energy density can be found as Electrical Energy divided by Volume, or $\frac{M{L}^2{T}^{-2}}{L^3}=M{L}^{-1}{T}^{-2}$