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 $Coulom{b^2}/Newton{m{ - }}{m^2}$

LIST$-I$	LIST$-II$
$(A)$ Surface tension	$(I)$ $Kg m ^{-1} s ^{-1}$
$(B)$ Pressure	$(II)$ $Kg ms^{-1 }$
$(C)$ Viscosity	$(III)$ $Kg m ^{-1} s ^{-2}$
$(D)$ Impulse	$(IV)$ $Kg s ^{-2}$

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

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$(A)$ Surface tension $(I)$ $Kg m ^{-1} s ^{-1}$

$(B)$ Pressure $(II)$ $Kg ms^{-1 }$

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