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]
