The electrostatic force on a small sphere of charge \( 0.4\,\mu\mathrm{C} \) due to another small sphere of charge \( -0.8\,\mu\mathrm{C} \) in air is 0.2 N. What is the distance between the two spheres?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 12 cm
By Coulomb’s law, two point charges \( q_1 \) and \( q_2 \) at a distance \( r \) apart in air exert a force of magnitude \( F=k|q_1q_2|/r^{2} \) on each other. Here \( k=9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \). The opposite signs only make the force attractive. The magnitudes of the charges fix the distance.
In SI units the data are:
\( |q_1|=0.4\,\mu\mathrm{C}=0.4\times10^{-6}\,\mathrm{C} \)
\( |q_2|=0.8\,\mu\mathrm{C}=0.8\times10^{-6}\,\mathrm{C} \)
\( F=0.2\,\mathrm{N} \)
From Coulomb’s law, \( r^{2} \) is:
\( \begin{aligned} r^{2} &= \frac{k\,|q_1q_2|}{F} \\ &= \frac{9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}\times0.4\times10^{-6}\,\mathrm{C}\times0.8\times10^{-6}\,\mathrm{C}}{0.2\,\mathrm{N}} \\ &= 0.0144\,\mathrm{m^{2}} \end{aligned} \)
\( r=0.12\,\mathrm{m}=12\,\mathrm{cm} \)