Question

The dimension of $1/2{\epsilon }_o{E}^2$(${\epsilon }_o$ permittivity of free space; E: electric field) is?

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

Answer 1

The correct option is C $M{L}^{-1}{T}^{-2}$

Dimension of $E_0=M^{-1}{L}^{-3}{T}^4{A}^2$

Dimension of E(Electric field)=$ML{T}^{-2}/AT=ML{T}^{-1}{A}^{-1}$
Dimension of $\frac{1}{2{\epsilon }_0{E}^2}=\frac{1}{\left(M^{-1}{L}^{-3}{T}^4{A}^2\right)(ML{T}^{-1}{A}^{-1}{)}^2}=\frac{1}{M{L}^{-1}{T}^2}=M{L}^{-1}{T}^{-2}$