A charge 4.98 nC is placed at the origin of an xy-coordinate system, and a charge -1.96 nC is placed on the positive x-axis at x = 3.97 cm . A third particle, of charge 6.00 nC is now placed at the point x = 3.97 cm , y = 3.00 cm .
Part A
Find the x-component of the total force exerted on the third charge by the other two.
Part B
Find the y-component of the total force exerted on the third charge by the other two.
Part C
Find the magnitude of the total force acting on the third charge.
Part D
Find the direction of the total force acting on the third charge.
force between q1 and q3
F13 = k q1 q3 / r13^2 = 9*10^9* 4.98* 6*10^-18 / (0.0397^2 + 0.03^2)
F13 = 1.086*10^-4 N
direction of F1 with horizontal
x = arctan ( 3/ 3.97) = 37.08
force between q2 and q3
F23 = 9* 10^9* 1.96*10^-18* 6 / 0.03^2
F23 = 1.176*10^-4 N
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a)
net horizontal component of force
Fx = F13 cos x = 8.66*10^-5 N
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b)
net vertical component of force
Fy = F13 sin x - F23
Fy = - 5.21 *10^-5 N
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c)
net force acting
F^2 = Fx^2 + Fy^2
F = 10.1 *10^-5 N
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d)
direction measured from +x in cw way
phi = arctan ( Fy/ Fx)
phi = arctan ( 5.21 / 8.66)
phi = 31
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Comment in case any doubt, will reply for sure.. Goodluck
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