Three charged particles form a triangle: particle 1 with charge Q1 = 88.0 nC is

Question

Three charged particles form a triangle: particle 1 with charge Q1 = 88.0 nC is at xy coordinates (0, 6.30 mm), particle 2 with charge Q2 is at (0, -6.30 mm), and particle 3 with charge q = 26.0 nC is at (4.70 mm, 0). What are (a) the x component and (b) the y component of electrostatic force on particle 3 due to the other two particles if Q2 is equal to 88.0 nC? What are (c) the x component and (d) the y component of electrostatic force on particle 3 due to the other two particles if Q2 is equal to -88.0 nC, ?

Explanation / Answer

F3 = F31 + F32

F31 = kq3q1/r31^2

F32 = kq3q2/r32^2

r31 = r32 = 7.86 x 10^-3

angle = tan^-1(6.3/4.7) = 53.28 degree

F31 = 0.3333 , -53.28 angle

F31x = F31* cos(theta)

cos(-theta) = costheta

F31x = 0.1993 N

F31y = -0.2672 N

for F32 = 0.333 N , 53.28 angle

F32x = 0.1993 N

F32y = 0.2672 N

Fx = F31x + F32x = 0.667 N

Fy = F31y + F32y = 0 N

part c )

now charge q2 = -88 nC

direction of F32 now is -126.7 degree angle

F32x = -0.1991 N

F32y = -0.2671 N

Fx = F31x + F32x = 0

Fy = F31y + F32y = -0.5342 N

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