The unit of permittivity of free space {varep | vedclass

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Question

(D) According to Coulomb's Law,the force between two point charges is given by:$F = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{{Q_1}{Q_2}}}{{{r^2}

Vedclass · Accepted Answer

$Coulomb^2/Newton-metre^2$

Vedclass · Answer

(D) According to Coulomb's Law,the force between two point charges is given by:$F = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{{Q_1}{Q_2}}}{{{r^2}}}$Rearranging the formula to solve for permittivity ${\varepsilon _0}$:${\varepsilon _0} = \frac{{{Q_1}{Q_2}}}{{4\pi F{r^2}}}$Substituting the $SI$ units for each quantity:$Q$ (Charge) is in $Coulomb$ $(C)$$F$ (Force) is in $Newton$ $(N)$$r$ (Distance) is in $metre$ $(m)$Therefore,the unit of ${\varepsilon _0}$ is:$\frac{C \cdot C}{N \cdot m^2} = C^2 / (N \cdot m^2)$Thus,the correct option is $D$.

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

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