Answer to Question #109042 in Electricity and Magnetism for Jafar Abbas

Question

Two point charges are located on the y-axis as follows: charge q1 = - 1.50 nC at y = - 0.600 m, and charge q2 = +3.20 nC at the origin (y = 0). What is the total force (magnitude and direction) exerted by these two charges on a third charge q3 = +5.00 nC located at y = -0.400 m?

Answer 1

Given:- Two point charges on the y-axis

 q₁ = - 1.50 nC  at ( y₁ = - 0.600 m)

q₂ = +3.20 nC at the origin (y₂ = 0m)

q₃ = +5.00 nC located at (y₃= -0.400 m)

Find the total force (F) and (magnitude and direction)

Now,

$\\From\,Coulomb's\,law\\[5pt] F=K\frac{|q_{1}q_{2}|}{r^{2}}\\[5pt] The\,force\,exerted\,by\,the\,first\,charge\,on\,the\,third\,one\\[5pt] r_{1}=y_{3}-y_{1}=-0.4m-(-0.6m)=0.2m\\[5pt] \\The\,force\,is\, attractive\, in\, the\,negative \,y\,axis\\ F_{1}=K\frac{|q_{1}q_{2}|}{r^{2}}\\[5pt] \\F_{1}=\frac{8.987\times 10^{9}N.m^{2}/C^{2}|(-1.5\times 10^{-9}C)(5\times 10^{-9}C)|}{(0.2m)^{2}}\\[5pt] F_{1}=1.685\times 10^{-6}N\\[5pt] Now, \\The\,force\,exerted\,by\,the\,second\,charge\,on\,the\,third\,one, \\r_{2}=y_{3}-y_{2}=-0.4m-0m=-0.4m\\[5pt] \\The\,force\,is\, repulsive\, in\, the\,negative \,y\,axis\\ \\F_{2}=K\frac{|q_{2}q_{3}|}{r^{2}}\\[5pt] F_{2}=\frac{8.987\times 10^{9}N.m^{2}/C^{2}|(3.2\times 10^{-9}C)(5\times 10^{-9}C)|}{(-0.4m)^{2}}\\[5pt] F_{1}=8.99\times 10^{-7}N\\[5pt] therefore,\\ Total\,force\,(F)\\[5pt] \\F=F_{1}+F_{2}\\[5pt] F=1.685\times 10^{-6}N+8.99\times 10^{-7}N\\[5pt] F=2.59\times 10^{-6}N\\[5pt] Direction:-(The \,negative\,y\,direction)$