Question

A proton (+q) is 50nm away from an electron (-q) which is 30nm away from another proton (+2q) with

a charge +2q. What is the net electric force on +2q?

Answer 1

The net force

$$F=F_{12}-F_{13}=k\frac{q*2q}{r_{12}^2}-k\frac{q*2q}{r_{13}^2}$$

$$F=9*10^9*2*(1.6*10^{-19})\\ \times\left(\frac{1}{(30*10^{-9})^2}-\frac{1}{(80*10^{-9})^2}\right)=4.4*10^{-13}\:\rm N$$