The value of electric permittivity of free sp | vedclass

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Question

From $F=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{ I ^2}$

If Force, $F=1 N$

charge, $q=1 C$

distance, $r=1 m$

Then, $q_0=\frac{q^2}{F 	ime

Vedclass · Accepted Answer

$8.85 	imes {10^{ - 12}}\,{C^2}/N{m^2}$

Vedclass · Answer

From $F=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{ I ^2}$

If Force, $F=1 N$

charge, $q=1 C$

distance, $r=1 m$

Then, $q_0=\frac{q^2}{F 	imes 4 \pi 	imes r^2}$

$=\frac{(1)^2}{1 N 	imes 4 	imes 3.14 	imes(1 m )^2}$

$\varepsilon_0=8.854 	imes 10^{-12} \frac{ C ^2}{ Nm ^2}$

The value of electric permittivity of free space is

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