Answer:
[tex]7.0\cdot 10^{-13}C[/tex]
Explanation:
The magnitude of the electrostatic force between two charged objects is
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
The force is attractive if the charges have opposite sign and repulsive if the charges have same sign.
In this problem, we have:
[tex]r=0.070 cm =7\cdot 10^{-4} m[/tex] is the distance between the charges
[tex]q_1=q_2=q[/tex] since the charges are identical
[tex]F=9.0\cdot 10^{-9}N[/tex] is the force between the charges
Re-arranging the equation and solving for q, we find the charge on each drop:
[tex]F=\frac{kq^2}{r^2}\\q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(9.0\cdot 10^{-9})(7\cdot 10^{-4})^2}{8.99\cdot 10^9}}=7.0\cdot 10^{-13}C[/tex]