Q159 / 180 · COMEDK 2013 · Physics
$\mathrm{SI}$ unit of permittivity is
- A$\mathrm{C}^{2} \mathrm{~m}^{2} \mathrm{~N}^{2}$
- Correct answerB$\mathrm{C}^{2} \mathrm{~m}^{-2} \mathrm{~N}^{-1}$
- C$\mathrm{C}^{2} \mathrm{~m}^{2} \mathrm{~N}^{-1}$
- D$\mathrm{C}^{-1} \mathrm{~m}^{2} \mathrm{~N}^{-2}$
Solution
From Coulomb's law, force between two charges,
$$
F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}
$$
$$
\begin{aligned}
&\Rightarrow \quad \varepsilon_{0}=\frac{q_{1} q_{2}}{4 \pi r^{2} \bar{F}} \\
&\therefore \quad \text { Unit of } \quad \varepsilon_{0}=\frac{\mathrm{C} \cdot \mathrm{C}}{\mathrm{m}^{2} \mathrm{~N}}=\frac{\mathrm{C}^{2}}{\mathrm{~m}^{2} \mathrm{~N}}=\mathrm{C}^{2} \mathrm{~m}^{-2} \mathrm{~N}^{-1}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.