Question

Using mass ($M$), length ($L$), time ($T$) and current ($A$) as fundamental quantities, the dimensional formula of permittivity is

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

Answer 1

Answer: (B)

$\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$

$F=\frac{1}{4\pi {ϵ}_0}\cdot \frac{q_1{q}_2}{r^2}\Rightarrow {ϵ}_0=\frac{q_1{q}_2}{4\pi F{r}^2}$

$\frac{AT\cdot AT}{ML{T}^{-2}{L}^2}=\frac{A^2{T}^2}{M{L}^3{T}^{-2}}=M^{-1}{L}^{-3}{A}^2{T}^4$