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# 9:57 AM My Notes Ask Your Teac -113 points 10 22.3.0P007.MI.SA This question has several parts...

9:57 AM My Notes Ask Your Teac -113 points 10 22.3.0P007.MI.SA This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise Three point charges are arranged as shown in the figure below. Find the magnitude and direction of the electric force on the particle q -4.71 nC at the origin.(Let 120.206 m.) 6.00 nC 100 m nC Need Help?

#### Homework Answers

Answer #1

answer) we know the formula

F=Kq1q2/r2

so now we have F1 in horizontal direction and F2 in vertical direction

F1=8.99*109*4.71*10-9*6*10-9/0.3062=2.713*10-6N

F2=8.99*109*4.71*10-9*3*10-9/0.12=12.703*10-6N

magnitude F=$\sqrt{}$F12+F22=$\sqrt{}$(2.713*10-6)2+(12.703*10-6)2=12.99*10-6N

so the answer is 1.3*10-5N or 12.99*10-6N or 1.299*10-5N

and angle

$\Theta$=tan-1(12.703/2.713)=77.9o

since the point charge in the third quadrant the angle=180+77.9=257.9

so the answer is 257.9o or 258o( if the angle is wrong then try -257.9o or -258o)

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