Three charged particles form a triangle: particle 1 with charge Q 1 = 93.0 nC is

Question

Three charged particles form a triangle: particle 1 with charge Q1 = 93.0 nC is at xy coordinates (0, 4.20 mm), particle 2 with charge Q2 is at (0, -4.20 mm), and particle 3 with charge q = 46.0 nC is at (5.60 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 93.0 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 -93.0 nC, ?

Explanation / Answer

(a)(b) If Q2 = 93.0 nC, then the force on particle 3 is in the x direction only,
since the y-components from Q1 and Q2 will cancel each other out.
The x-component of the force on 3 will be twice the contribution of either Q1 or Q2, so it's
2 Q1 q k cos(theta) / r^2
= (2) (93 x 10^(-9) C)(46 x 10^(-9) C) (9.0 x 10^9 Nm^2/C^2) [4.20/sqrt(31.36+17.64)] divided by
(0.00003136 m^2 + 0.00001764 m^2)
= 0.943 N

(c)(d) If Q2 = -93 nC, then the force on particle 3 is in the (negative) y direction only,
since the x-components from Q1 and Q2 will cancel each other out.
The magnitude of the downward force on particle 3 would be
2 Q1 q k sin(theta) /r^2
= same answer as (a)(b) except that sin(theta) replaces cos(theta),
= 0.943 N tan(theta)
= 0.943 N (4.20/5.60) = 0.707 N

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