The unit of permittivity of free space {varep | vedclass

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Question

(d) $F = \frac{1}{{4\pi {\varepsilon _0}}}\,.\,\frac{{{Q_1}{Q_2}}}{{{r^2}}}$

==> ${\varepsilon _0} \propto \frac{{{Q^2}}}{{F 	imes {r^2}}}$

Vedclass · Accepted Answer

$Coulom{b^2}/Newton{\rm{ - }}metr{e^2}$

Vedclass · Answer

(d) $F = \frac{1}{{4\pi {\varepsilon _0}}}\,.\,\frac{{{Q_1}{Q_2}}}{{{r^2}}}$

==> ${\varepsilon _0} \propto \frac{{{Q^2}}}{{F 	imes {r^2}}}$

So${\varepsilon _0}$ has units of $Coulom{b^2}/Newton{m{ - }}{m^2}$

The unit of permittivity of free space ${\varepsilon _0}$ is

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