Question: The unit of permittivity of free space, $\varepsilon_{0}$, is: [PMT/NEET-2004]
a. coulomb/newton–metre
b. newton–metre$^2$/coulomb$^2$
c. coulomb$^2$/newton–metre$^2$
d. coulomb$^2$/(newton–metre)$^2$

Answer: The correct answer is: c. coulomb² / (newton·metre²)

Explanation:
From Coulomb’s law:

$
F = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2}
$

Rearranging for $\varepsilon_0$:

$
\varepsilon_0 = \frac{q_1 q_2}{4 \pi F r^2}
$

Units:

• Charge ($q$) = coulomb (C)
• Force ($F$) = newton (N)
• Distance squared ($r^2$) = m²

So,
$
[\varepsilon_0] = \frac{C \cdot C}{N \cdot m^2} = \frac{C^2}{N \, m^2}
$

Thus, the unit of permittivity of free space is coulomb²/(newton·metre²).