Question

What must be the distance between point charge $q_1=26.0\mu C$ and point charge $q_2=47.0\mu C$ for the electrostatic force between them to have a magnitude of $5.70N$?

Answer 1

Equation $21-1$ gives Coulomb's law $F=k\frac{|q_1||q_2|}{r^2}$, which we solve the for the distance:

$r=\sqrt{\frac{k|q_1||q_2|}{F}}=\sqrt{\frac{(8.99\times {10}^{9}N-m^2/C^2)(26.0\times {10}^{-6}C)(47.0\times {10}^{-6}C)}{5.70N}}=1.39m$