Question

Using mass(M), length (L), time(T), and electric current(A) as fundamental quantities, the dimensions of permittivity will be

• A. $\left[ML{T}^{-1}{A}^{-1}\right]$
• B. $\left[ML{T}^{-2}{A}^{-2}\right]$
• C. $\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$
• D. $\left[M^2{L}^{-2}{T}^{-2}A\right]$

Answer 1

The correct option is C $\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$

We know $\overrightarrow{F}=\frac{1}{4\Pi {\epsilon }_0}.\frac{Q^2}{R^2}$ (Elctrostatic force)

$\Rightarrow$ ${\epsilon }_0\alpha \frac{Q^2}{R^2}\times \frac{1}{F}$

$\Rightarrow dim\left({\epsilon }_0\right)=\frac{\left[A^2{T}^2\right]}{\left[L^2\right]\left[{MLT}^{-2}\right]}=M^{-1}{L}^3{T}^4{A}^2$