Question

Two points are placed as follows: Charge q¹= -1.50 nC is at y=+6.00m 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 (2.00m, -4.00 m)?

Answer 1

Make a sketch:

Find the force from the first on the third and the angle:

$$|F_{13}|=\frac{kq_1q_3}{r_{13}^2}=\frac{kq_1q_3}{(x_1-x_3)^2+(y_1-y_3)^2}=6.48\text{ N}.\\\space\\ \theta_{13}=\arctan\frac{x_1-x_3}{y_1-y_3}=11.3°.$$

Force from the second on the third charge and the angle:

$$|F_{23}|=\frac{kq_2q_3}{r_{23}^2}=\frac{kq_2q_3}{(x_2-x_3)^2+(y_2-y_3)^2}=7.2\text{ N}.\\\space\\ \theta_{23}=\arctan\frac{x_2-x_3}{y_2-y_3}=26.6°.$$

Find the components of the forces:

$$F_{13x}=1.27\text{N},\\ F_{13y}=6.35\text{N},\\ F_{23x}=3.22\text{N},\\ F_{23y}=6.44\text{N}.\\\space\\ |F_R|=\sqrt{(F_{13x}-F_{23x})^2+(F_{13y}-F_{23y})^2},\\ |F_R|=1.95\text{ N}.\\\space\\ \theta_R=\arctan\frac{F_{13x}-F_{23x}}{F_{13y}-F_{23y}}=87°$$