Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m(see (Figure 1) ). The charges are Q1 = 6.9 μC , Q2 = -9.0 μC , and Q3 = -5.0 μC ....

Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m(see (Figure 1) ). The charges are Q1 = 6.9 μC , Q2 = -9.0 μC , and Q3 = -5.0 μC .

Calculate the magnitude of the net force on particle 1 due to the other two.
Express your answers using two significant figures.


Calculate the direction of the net force on particle 1 due to the other two.
Express your answer as positive angle using two significant figures


Calculate the magnitude of the net force on particle 2 due to the other two.
Express your answers using two significant figures.


Calculate the direction of the net force on particle 2 due to the other two.
Express your answer as positive angle using two significant figures.


Calculate the magnitude of the net force on particle 3 due to the other two.
Express your answers using two significant figures.

Calculate the direction of the net force on particle 3 due to the other two.
Express your answer as positive angle using two significant figures.

Asked on by yikulah

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gsenviro | College Teacher | (Level 1) Educator Emeritus

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Q1 = 6.9 `mu` C, Q2= -9 `mu` C  and Q3 = -5 `mu` C, side length, a = 1.2 m. Since the figure is not given, I am assuming Q2 and Q3 are along x-axis and Q1 is in first quadrant.

Net force on Q1 due to Q2 and Q3:

Let F21: force applied by Q2 on Q1 and F31 is force applied by Q3 on Q1. 

Using F21 = kQ1Q2/a^2 = (8.998 x10^9)(6.9 x 10^-6)(-9 x 10^-6)/(1.2^2) N

= -0.388 N (negative sign indicates attractive forces)

Similarly, F31 = kQ1Q3/a^2 = (8.998 x 10^9)(6.9x10^-6)(-5x10^-6)/(1.2^2) = -0.215 N (negative force means attractive forces).

The x and y components of the forces can be calculated as:

F21x = F21 cos 60 = -0.388 cos 60 = -0.194 N (left)

F21y =  F21 cos 30 = -0.388 cos 30 = -0.336 N (down)

F31x= F31 cos 60 = -0.215 cos 60 = -0.108 N (right)

F31y = F31 cos 30 = -0.215 cos 30 = - 0.186 N (down)

Net Force, Fx = -0.194 - (-0.108) N = -0.086 N (left) and Fy = -0.336+ (-0.186) = -0.522 N (down)

Magnitude of net force, F= `sqrt(Fx^2 + Fy^2)`  = -0.53 N (negative sign means attractive force)

and the direction is:  `theta = tan^(-1)[Fy/Fx] = tan^(-1) [-0.522/-0.086]`

= 78.3 degrees with the horizontal.

Hope this helps.  

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