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(A) Given: Force $F = 0.144 \, N$, distance $r = 5 \, cm = 5 	imes 10^{-2} \, m$, and charges are identical $(q_1 = q_2 = q)$.According to Coulomb's

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$0.2$

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(A) Given: Force $F = 0.144 \, N$, distance $r = 5 \, cm = 5 	imes 10^{-2} \, m$, and charges are identical $(q_1 = q_2 = q)$.According to Coulomb's Law: $F = \frac{k q^2}{r^2}$.Substituting the values: $0.144 = \frac{(9 	imes 10^9) 	imes q^2}{(5 	imes 10^{-2})^2}$.$0.144 = \frac{9 	imes 10^9 	imes q^2}{25 	imes 10^{-4}}$.$q^2 = \frac{0.144 	imes 25 	imes 10^{-4}}{9 	imes 10^9}$.$q^2 = 0.016 	imes 25 	imes 10^{-13} = 0.4 	imes 10^{-13} = 4 	imes 10^{-14}$.$q = \sqrt{4 	imes 10^{-14}} = 2 	imes 10^{-7} \, C = 0.2 	imes 10^{-6} \, C = 0.2 \, \mu C$.

When two identical point charges are placed at a distance of $5 \, cm$, they experience a repulsive force of $0.144 \, N$. The value of each charge in microcoulombs $(\mu C)$ is:

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