Question

Two point charges are placed as follows: charge q1 = -1.50 nC is at y = +6.00 m and charge q2 = +3.20 nC is at the origin. What is the total force (magnitude and direction) exerted by these two charges on a negative point charge q3 = -5.00 nC located at (2.00 m, -4.00 m)? 

Answer 1

$$F_{13}=k\dfrac {q_1q_3}{\sqrt{(x_3-x_1)^2+(y_3-y_1)^2}^2}=6.48·10^{-10}\text{ N}.\\\space\\ F_{23}=k\dfrac {q_2q_3}{\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}^2}=7.19·10^{-9}\text{ N}.\\\space\\ \theta_{13}=\arctan\dfrac{x_3-x_1}{y_3-y_1}=11.3°,\\\space\\ \theta_{23}=\arctan\dfrac{x_3-x_2}{y_3-y_2}=26.6°,\\\space\\ \\\space\\ F_{13x}=F_{13}\sin\theta_{13}=127\text{ pN},\\\space\\ F_{13y}=F_{13}\cos\theta_{13}=635\text{ pN},\\\space\\ F_{23x}=F_{23}\sin\theta_{23}=3.22\text{ nN},\\\space\\ F_{23y}=F_{23}\cos\theta_{23}=6.43\text{ nN},\\\space\\ F=\sqrt{(F_{13x}-F_{23x})^2+(F_{13y}-F_{23y})^2}=3.25\text{ nN},\\\space\\ \theta=\arctan\dfrac{F_{13x}-F_{23x}}{F_{13y}-F_{23y}}=9°.$$