The electrostatic force on a small sphere of charge $0.4 \, \mu ext{C}$ due to another small sphere of charge

Question

The electrostatic force on a small sphere of charge $0.4 \, \mu	ext{C}$ due to another small sphere of charge $-0.8 \, \mu	ext{C}$ in the air is $0.2 \, 	ext{N}$. The distance between the two spheres is:

Accepted Answer

The magnitude of the electrostatic force between two point charges is given by Coulomb's Law: $F = k \frac{|q_1 q_2|}{r^2}$.

Given:
$F = 0.2 \, 	ext{N}$
$q_1 = 0.4 \, \mu	ext{C} = 0.4 	imes 10^{-6} \, 	ext{C}$
$|q_2| = |-0.8 \, \mu	ext{C}| = 0.8 	imes 10^{-6} \, 	ext{C}$
$k = 9 	imes 10^9 \, 	ext{N m}^2	ext{C}^{-2}$

Rearranging the formula to solve for distance $r$:
$r^2 = \frac{k |q_1 q_2|}{F} = \frac{(9 	imes 10^9)(0.4 	imes 10^{-6})(0.8 	imes 10^{-6})}{0.2}$
$r^2 = \frac{2.88 	imes 10^{-3}}{0.2} = 14.4 	imes 10^{-3} = 144 	imes 10^{-4} \, 	ext{m}^2$
Taking the square root:
$r = \sqrt{144 	imes 10^{-4}} = 12 	imes 10^{-2} \, 	ext{m} = 12 \, 	ext{cm}$.

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