Answer to Question #316392 in Physics for Cael Goo

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

The net force

$$\vec F_{net}=k\frac{q_1q_3}{r_{13}^3}\vec r_{13}+k\frac{q_2q_3}{r_{23}^3}\vec r_{23}$$ $$\vec r_{13}=(2,-10),\quad r_{13}=\sqrt{2^2+(-10)^2}=\sqrt{104},\\ \vec r_{23}=(2,-4),\quad r_{23}=\sqrt{2^2+(-4)^2}=\sqrt{20}.$$ $$\vec F_{net}=9*10^9*\frac{(-1.5*10^{-9})*(-5.0*10^{-9})}{(\sqrt{104})^3}*(2,-10)\\ +9*10^9*\frac{(3.20*10^{-9})*(-5.0*10^{-9})}{(\sqrt{20})^3}*(2,-4)$$ $$\vec F_{net}=6.36*10^{-11}*(2,-10)\\ -1.61*10^{-9}*(2,-4)=(-3.1*10^{-9},5.8*10^{-9})$$

The magnitude of force

$$F_{net}=\sqrt{(-3.1)^2+5.8^2}*10^{-9}=6.6*10^{-9}\:\rm N$$

Direction

$$\theta=\tan^{-1}\frac{F_y}{F_x}=\tan^{-1}\frac{5.8}{-3.1}=118^\circ\: \rm from\; +x$$