Coulomb's Law And Superposition

Question

Two identical point charges of 2 microcoulomb each are placed 30 cm apart in vacuum, so what is the magnitude of the electrostatic force between them?

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Coulomb's law tells us that the force between two point charges is $F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}$, where the constant $\frac{1}{4\pi\varepsilon_0} = 9	imes10^9\ \mathrm{N\,m^2/C^2}$. Here both charges are $q = 2	imes10^{-6}$ C and the separation is $r = 0.30$ m. Substituting the values gives $F = 9	imes10^9 	imes \frac{(2	imes10^{-6})(2	imes10^{-6})}{(0.30)^2}$. The numerator of the charge product is $4	imes10^{-12}$, and dividing by $r^2 = 0.09$ yields $\frac{4	imes10^{-12}}{0.09} = 4.44	imes10^{-11}$. Multiplying by $9	imes10^9$ gives $F \approx 0.4$ N. The answer 0.2 N is wrong because it comes from mistakenly using $r$ instead of $r^2$ in only half the calculation. The value 4.0 N is wrong because it drops a factor of 10 by treating $r$ as 0.3 without squaring correctly. The value 0.04 N is wrong because it uses $r$ in centimetres incorrectly, shifting the magnitude by two orders. As stated in NCERT Class 12, Chapter 1 (Electric Charges and Fields), like charges repel along the line joining them. A magnitude check confirms micro-coulomb charges at tens of centimetres give sub-newton forces, so 0.4 N is physically reasonable.

Coulomb's Law And Superposition

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