Question

Three point charges are arranged as shown: q1=4.0x10^-9C; q2=5.0x10^-9C; and q3=-6.0x10^-9 C.

Find the net force on q1 due to q2 and q3. Draw the direction of the force.

Answer 1

$$F_{12}=9\cdot10^9\frac{(4\cdot10^{-9})(5\cdot10^{-9})}{1^2}=180\cdot10^{-9}N\\ F_{13}=-9\cdot10^9\frac{(4\cdot10^{-9})(6\cdot10^{-9})}{1^2}=-216\cdot10^{-9}N\\ F=180\cdot10^{-9}-216\cdot10^{-9}=-3.6\cdot10^{-8}N$$