The electrostatic force between two charges is given by
[tex]F=k_e \frac{q_1 q_2}{r^2} [/tex]
where
ke is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the charges
In our problem,
[tex]q_1 = 25 \mu C=25 \cdot 10^{-6} C[/tex]
[tex]q_2 = 3 mC = 3 \cdot 10^{-3} C[/tex]
[tex]r=35 cm=0.35 m[/tex]
Therefore the electrostatic force is
[tex]F=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(25 \cdot 10^{-6}C)(3 \cdot 10^{-3}C)}{(0.35 m)^2}= 5510 N[/tex]
