Question

The attractive electrostatic force between the point charges +8.44 ✕ 10-6 and q has a magnitude of 0.961 n when the separation between the charges is 0.67 m. find the sign and magnitude of the charge q.

Answer 1

The electrostatic force between two charges Q1 and q is given by
$F=k_e \frac{Q_1 q}{r^2}$
where 
ke is the Coulomb's constant
Q1 is the first charge
q is the second charge
r is the distance between the two charges

Re-arranging the formula, we have
$q= \frac{F r^2}{k_e Q_1}$
and since we know the value of the force F, of the charge Q1 and the distance r between the two charges, we can calculate the value of q:
$q= \frac{(0.961 N)(0.67 m)^2}{(8.99 \cdot 10^9 N m^2 C^{-2})(+8.44\cdot 10^{-6}C)}=5.69 \cdot 10^{-6} C$

And since the force is attractive, the two charges must have opposite sign, so the charge q must have negative sign.