Chapter 21, Problem 015 The charges and coordinates of t o charged particles hel

Question

Chapter 21, Problem 015 The charges and coordinates of t o charged particles held fixed in an xy plane area -2.99 _ 4.44 y 0 941 cm and Q2 -5,56 uc.x2 -1 61 on, y2-1 26 a Find the a) magnitu de and (b) direction (with respect to +x axis in the ange ( 180° 100° of the electrostatic Torce on partide 2 due to partide 1. At what () x and (d) y coordinates should a third particle of charge o3 - 4.61 C eplaced such that the net electrestatic force on particle 2 due to particles 1 and 3 is ero? (a) Number (b) Number (c) Number (d) Number

Explanation / Answer

Solution) q1=2.99UC=2.99×10^(-6)C

X1=4.44cm=4.44×10^(-2)=0.0444m

Y1=0.941cm=0.941×10^(-2)=0.00941m

q2=-5.56UC=-5.56×10^(-6)C

X2=-1.61cm=-1.61×10^(-2)=-0.0161m

Y2=1.26cm=1.26×10^(-2)=0.0126m

(a) Magnitude of force F

F=Kq1q2/r^2

r^2=(X1-X2)^2+(Y1-Y2)^2

r^2=(0.0444+0.0161)^2+(0.00941-0.0126)^2

r=0.06m

F=(9×10^(9))×(2.99×10^(-6))×(5.56×10^(-6))/(0.06)^2

F=41.561N

(b) Direction theeta

Theeta = tan inverse(y/x)

y=0.00941-0.0126

y=-0.00319m

x=0.0444+0.0161

x=0.0605m

Theeta =tan inverse(-0.00319/0.0605)

Theeta=-3.01°

(c)we have

F23=F21

Kq2q3/L^2 = Kq2q1/r^2

L^2 = r^2q3/q1

L=r(q3/q1)^(1÷2)

L=0.06(4.61÷2.99)

L=0.074m

X=Lcos(Theeta)

X=0.074cos(-3.01)=0.073m

Y=Lsin(Theeta)

Y=0.074sin(-3.01)

Y=-0.0038m

(X,Y) =(-0.0038,0.073)

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