Question

The unit of permittivity of free space, ${ϵ}_0$ is

• A. $coulomb/newton-metre$
• B. $newton-$$metr{e}^2$/$coulom{b}^2$
• C. $coulom{b}^2/newton-metr{e}^2$
• D. $coulom{b}^2/{(newton-metre)}^2$

Answer 1

Answer: (C)

$coulom{b}^2/newton-metr{e}^2$

We know that Force, $F=\frac{1}{4\pi {ϵ}_0}\frac{Q_1{Q}_2}{r^2}$

Thus, $Newton={\left[{ϵ}_0\right]}^{-1}.{\left(Coulomb\right)}^2{\left(metre\right)}^{-2}$
or, $\left[{ϵ}_0\right]=\frac{{\left(Coulomb\right)}^2}{Newton-{\left(metre\right)}^2}$