Question

The unit of permittivity of free space, $ {{\varepsilon }_{0}}, $ is

• A. coulomb/newton-metre
• B. Newton metre$^2$ / Coulomb$^2$
• C. Coulomb$^2$ /Newton metre$^2$
• D. Coulomb$^2$ / (Newton metre)$^2$

Answer 1

The Correct Option is C

Key Idea : Substitute the units for all the quantities involved in an expression written for permittivity of free space. By Coulomb's law, the electrostatic force $F =\frac{1}{4 \pi \varepsilon_{0}} \times \frac{q_{1} q_{2}}{r^{2}}$ $\Rightarrow \varepsilon_{0} =\frac{1}{4 \pi} \times \frac{q_{1} q_{2}}{r^{2} F}$ Substituting the units for $q, r$ and $F$, we obtain unit of $\varepsilon_{0}=\frac{\text { coulomb } \times \text { coulomb }}{\text { newton }-(\text { metre })^{2}} $ $=\frac{(\text { coulomb })^{2}}{\text { newton }-(\text { metre })^{2}}$ $= C ^{2} / N - m ^{2}$