Q100 / 180 · AIIMS 2004 · Physics
Using mass $(M)$, length $(L)$, time $(T)$ and current $(A)$ as fundamental quantities, the dimension of permittivity is
- A$\left[\mathrm{ML}^{-2} \mathrm{~T}^2 \mathrm{~A}\right]$
- Correct answerB$\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]$
- C$\left[\mathrm{MLT}^{-2} \mathrm{~A}\right]$
- D$\left[\mathrm{ML}^2 \mathrm{~T}^{-1} \mathrm{~A}^2\right]$.
Solution
Force of attraction between two charges is given by
$$
\begin{aligned}
F & =\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} \\
\Rightarrow \quad \varepsilon_0 & =\frac{1}{4 \pi} \frac{1}{F} \frac{q_1 q_2}{r^2}=\frac{\mathrm{A}^2 \mathrm{~T}^2}{\mathrm{MLT}^{-2} \mathrm{~L}^2} \\
& =\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right] .
\end{aligned}
$$
(where $4 \pi$ is dimensionless)
$=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]$
Hence, the correct answer is option: 'B'.