Three charged particles are placed at each of three corners of an equilateral tr

Question

Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 3.3 cm . Two of the particles have a negative charge: q1 = -7.4 nC and q2 = -14.8 nC . The remaining particle has a positive charge, q3 = 8.0 nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?

Part B

Find the net force F 3 acting on particle 3 due to the presence of the other two particles. Report you answer as a magnitude F3 and a direction measured from the positive x axis.

Express the magnitude in newtons and the direction in degrees to three significant figures.

Explanation / Answer

from the Columb's law , the force between q3 and q1 is

F13 (on q3 due to q1) = [k *q3*q1]/r2

= [(9*109)(7.4*10-9)(8*10-9)]/(0.033 m)2

= 4.89*10-4 N

the x compoennt force is

F13x = F13 cos 60 = 4.89*10-4 N cos 60 = 2.445*10-4 N

the y compoennt force is

F13y = F13 sin 60 = 4.89*10-4 N sin 60 = 4.23*10-4 N

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the force between q3 and q2 is

F23 (on q3 due to q2) = [k *q3*q2]/r2

= [(9*109)(14.8*10-9)(8*10-9)]/(0.033 m)2

= 9.78*10-4 N

the x compoennt force is

F23x = F23 cos 60 = 9.78*10-4 N cos 60 = 4.89*10-4 N

Fx = F13x + F23x = 2.445*10-4 N+4.89*10-4 N=7.335*10-4 N

Fy = 4.23*10-4 N

the magnitude of force is

F = root Fx^2 + Fy^2 = root (7.335*10-4 N)^2 + (4.23*10-4 N)^2 = 8.46 * 10 ^-4 N

angle is

theta = tan^-1 ( Fy/ Fx ) = tan^-1(4.23*10-4 N/7.335*10-4 N)=29.97

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