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Q. What are the units of $K = 1/4 \pi \varepsilon_{0}? $

AFMCAFMC 2004

Solution:

From Coulomb's law two stationary point charges $q_{1}$ and $q_{2}$ attract or repel each other with a force $F$ which is directly proportional to the product of the charges and inversely proportional to the square of distance between them,
i.e., $ F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} $
$\therefore K=\frac{1}{4 \pi \varepsilon_{0}}=\frac{F r^{2}}{q^{2}}$ (if $q_{1}=q_{2}=q )$
Hence, units of $K=\frac{ Nm ^{2}}{ C ^{2}}$
$= Nm ^{2} C ^{-2}$