Two particles are located on the x axis of a Cartesian coordinate system. Partic

Question

Two particles are located on the x axis of a Cartesian coordinate system. Particle 1 carries a charge of +4.0 nC and is at x = -30 mm and particle 2 carries a charge of -4.0 nC and is at x = 30 mm . Particle 3, which carries a charge of +8.0 C , is located on the positive y axis 90 mm from the origin.

What is the magnitude of the vector sum of the electric forces exerted on particle 3?

What is the direction angle of the vector sum of the electric forces exerted on particle 3 measured counterclockwise from the positive x axis?

Explanation / Answer

1. This will create an image of triangle. by the use of pythogorus theorem the distance between particle 3 & particle 1 or 2 equal to be =

r = sqrt(902 + 302) = 94.87 mm

hence force on particle 3 due to particle 1

F = kq1.q2/r2 = 9 x 109* 4 x 10-6 * 8 x 10-6/ (94.87 x 10-3)2

F = 32 N

force on particle 3 due to particle 2

F = kq1.q2/r2 = 9 x 109* -4 x 10-6 * 8 x 10-6/ (94.87 x 10-3)2

F = -32 N

Angle between these two force vectors = 2 * tan-1 (90/30) = 143.13 deg

hence sum of the se two force vectors is

sqrt(322 + 322 - 2*-32*32*cos(180-143.13)) =60.715 N ..........Ans.

Let denote the angle between uu and u+vu+v . Then using the Law of Cosines

cos= 60.715 / ( 64*64)

angle = 89.15 deg

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