Answer to Question #177697 in Physics for Jo-an Balicao

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

$F_{13}=k\frac{q_1q_3}{r_{13}^2}=9\cdot10^9\cdot\frac{1.5\cdot10^{-9}\cdot5\cdot10^{-9}}{(\sqrt{10^2+2^2})^2}=6.49\cdot10^{-10}\ (N)$

$F_{23}=k\frac{q_2q_3}{r_{23}^2}=9\cdot10^9\cdot\frac{3.2\cdot10^{-9}\cdot5\cdot10^{-9}}{(\sqrt{4^2+2^2})^2}=72\cdot10^{-10}\ (N)$

$F_{13x}=6.49\cdot10^{-10}\cdot\cos\alpha=6.49\cdot10^{-10}\cdot\frac{2}{\sqrt{10^2+2^2}}=1.27\cdot10^{-10}\ (N)$

$F_{13y}=6.49\cdot10^{-10}\cdot\sin\alpha=6.49\cdot10^{-10}\cdot\frac{10}{\sqrt{10^2+2^2}}=6.36\cdot10^{-10}\ (N)$

$F_{23x}=72\cdot10^{-10}\cdot\cos\beta=72\cdot10^{-10}\cdot\frac{2}{\sqrt{4^2+2^2}}=32.2\cdot10^{-10}\ (N)$

$F_{23y}=72\cdot10^{-10}\cdot\sin\beta=72\cdot10^{-10}\cdot\frac{4}{\sqrt{4^2+2^2}}=64.4\cdot10^{-10}\ (N)$

$F_x=1.27\cdot10^{-10}-32.2\cdot10^{-10}=-30.93\cdot10^{-10}\ (N)$

$F_y=64.4\cdot10^{-10}-6.36\cdot10^{-10}=58.04\cdot10^{-10}\ (N)$

$F=\sqrt{(30.93\cdot10^{-10})^2+(58.04\cdot10^{-10})^2}=65.8\cdot10^{-10} \ (N)$ . Answer

$\gamma =\tan^{-1}\frac{58.04\cdot10^{-10}}{30.93\cdot10^{-10}}=61.94°$ or $118.06°$ relative to the positive direction of the x-axis . Answer