The electrostatic force between two charges is given by
[tex]F=k_e \frac{Q q}{r^2} [/tex]
where
[tex]k_e = 8.99 \cdot 10^9 N m^2 C^{-2}[/tex] is the Coulomb's constant
[tex]Q=+8.46 \cdot 10^{-6} C[/tex] 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:
[tex]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[/tex]
And the sign of the charge is positive, because Q is positive and F is positive as well.
