Answer:
[tex]F=-12896N[/tex], attractive.
Explanation:
For calculating this force we use the Coulomb Law:
[tex]F=\frac{kq_1q_1}{r^2}[/tex]
Where [tex]k=9\times10^9Nm^2/C^{-2}[/tex] is the Coulomb's constant, [tex]q_1[/tex] and [tex]q_2[/tex] the values of each charge and r the distance between them.
Since the Coulomb's constant as I wrote it is in S.I. we have to write all the magnitudes in that system of units, and substitute:
[tex]F=\frac{(9\times10^9Nm^2/C^{-2})(-18.8\times10^{-6}C)(46.5\times10^{-6}C)}{(0.0247m)^2}=-12896N[/tex]
This force is attractive since both charges are of opposite sign.
