Question

The unit of permittivity is
(a) C² N⁻¹ m⁻²
(b) M m² C⁻²
(c) H m⁻¹
(d) N C⁻² m⁻²

Answer 1

Coulomb's Law states that:


$\boxed{F = \frac{1}{4\pi\varepsilon_{\circ}}\frac{q_1q_2}{r^2}}$


Here,


$F$ = Force [Unit: newton (N) ]

$q_1$ and $q_2$ = Charge [Unit: coulomb (C) ]

$r$ = Distance between Charges [Unit: metre (m) ]

$\varepsilon_{\circ}$ = Absolute Permittivity


We can write:

$F = \frac{1}{4\pi\varepsilon_{\circ}}\frac{q_1q_2}{r^2} \\ \\ \\ \implies \varepsilon_{\circ} = \frac{1}{4\pi F} \frac{q_1q_2}{r^2} \\ \\ \\ \implies Unit \, \, of \, \, \varepsilon_{\circ} = \frac{Unit \, \, of \, \, q_1q_2}{Unit \, \, of \, \, F r^2} \\ \\ \\ \implies Unit \, \, of \, \, \varepsilon_{\circ} = frac{C \times C}{N \times m^2} \\ \\ \\ \implies \boxed{Unit \, \, of \, \, \varepsilon_{\circ} = C^2 \, N^{-1} \, m^{-2}}$


Thus, Unit of Permittivity is Option (a) $\bold{C^2 \, \, N^{-1} \, m^{-2}}$