In the figure below, qA = 6.0 µC, qB = 2.7 µC, qC = 1.7 µC, dAB = 4.00 cm, and dAC = 3.00 cm. How would you determine the magnitude and direction of the net force on qA?

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Answer 1 · 2017-03-20T22:24:31

#"Magnitude : "13.67*10^(-8)" "N" , angle="48.26^o#

enter image source here

#q_A=6.0 mu C=6*10^(-6)C#

#q_B=2.7 mu C=2.7*10^(-6)C#

#q_C=1.7 mu C=1.7*10^(-6)C#

#color(red)(F_("CA")=k*(q_C*q_A)/d_("AC")^2)=9*10^9(1.7*10^(-6)*6*10^(-6))/((3*10^(-2))^2)#

#color(red)(F_("CA"))=(cancel(9)*1.7*6*10^(-12))/(cancel(9)*10^(-4))=10.2*10^(-8)" "N#

#color(blue)(F_("BA")=k*(q_B*q_A)/(d_("AB")^2))=9.10^9(2.7*10^(-6)*6*10^(-6))/((4*10^(-2))^2)#

#color(blue)(F_("BA"))=(9*2.7*6*10^(-12))/(16*10^(-4))=9.1*10^(-8)" "N#

#color(green)(F)=sqrt(color(red)(F_("CA")^2)+color(blue)(F_("BA")^2))#

#color(green)(F)=sqrt((10.2*10^(-8))^2+(9.1*10^(-8))^2)#

#color(green)(F)=sqrt(104.04*10^-(16)+82.81*10^(-16)#

#color(green)(F)=sqrt(186.85*10^(-16)#

#color(green)(F)=13.67*10^(-8)" "N("Net Force")#

#tan alpha=(color(red)(F_("CA")))/(color(blue)(F_("BA")))#

#tan alpha=(10.2*cancel(10^(-8)))/(9.1*cancel(10^(-8)))#

#tan alpha=1.1208791#

#alpha=48.26 ^o#