Two point charges of $+4\ \mu$C each are separated by $0.6$ m in vacuum. The magnitude of the electrostatic force between them is approximately: (Take $k = 9 \times 10^9\ \mathrm{N\,m^2/C^2}$.)
A$0.4$ N
B$4.0$ N
C$0.1$ N
D$1.0$ N
Answer & Solution
Correct answer: A. $0.4$ N
Coulomb's law: $F = \dfrac{k q_1 q_2}{r^2} = \dfrac{9 \times 10^9 \times (4 \times 10^{-6})^2}{(0.6)^2} = \dfrac{9 \times 10^9 \times 16 \times 10^{-12}}{0.36} = \dfrac{0.144}{0.36} = 0.4$ N.
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