Question

Two equal, but oppositely charged particles are attracted to each other electrically. The size of the force of attraction is 98.9 N when they are separated by 73.9 cm. What is the magnitude of the charges in microCoulombs ?

Answer 1

Using Coulomb's law:

$F=K\cdot\frac{|q1||q2|}{r^2}$

Where:

$\begin{gathered} K=8.988\times10^9 \\ r=0.739m \\ F=98.9N \\ |q1|=|q2| \end{gathered}$

Therefore:

$98.9=8.988\times10^9\cdot\frac{q1^2}{(0.739^2)}$

Solve for q1:

$\begin{gathered} q1=\sqrt{\frac{(98.9)(0.739^2)}{8.988\times10^9}} \\ q1\approx7.75\times10^{-5}C \end{gathered}$