Question

The attractive electrostatic force between the point charges +8.46 ✕ 10-6 and q has a magnitude of 0.967 n when the separation between the charges is 0.52 m. find the sign and magnitude of the charge q. c

Answer 1

The electrostatic force between two charges is given by
$F=k_e \frac{Q q}{r^2}$
where
$k_e = 8.99 \cdot 10^9 N m^2 C^{-2}$ is the Coulomb's constant
$Q=+8.46 \cdot 10^{-6} C$ is one of the charges
q is the other charge
r=0.52 m is the distance between the two charges

We know the intensity of the force between the two charges, F=0.967 N, so we can re-arrange the formula to find the value of the charge q:
$q= \frac{F r^2}{k_e Q}= \frac{(0.967 N)(0.52 m)^2}{(8.99 \cdot 10^9 N m^2 C^{-2})(+8.46 \cdot 10^{-6} C)} =3.44 \cdot 10^{-6} C$

And the sign of the charge is positive, because Q is positive and F is positive as well.