physics question?

Question

The charges +2.0uC, +3.0uc, and +4.0 uC are placed at points A, B and C of an equlateral triangle with each side of .20 m. Determine the force on the charge +4.0 uC placed at point C?

Answer 1

The force on 4 uC charge can be found by vector addition
Fr=Fbc + Fac where
Fr- resultant force
Fbc – force between +3uC and +4uC charges (repulsive)
Fac - force between +2uC and +4uC charges (repulsive)

We know that since we have a equilateral triangle the angle between the forces acting on +4 uC charge at point C must be 60 degrees. To fix the point of reference relative to the angle we must assume that the force Fbc lies on the x-axis.

Then we have
Fr(x – component) =Fbc + Fac cos(60)
Fr(y – component) = Fac sin(60)

|Fr|= sqrt(Fr(x – component)^2 + Fr(y – component)^2)
Angle relative to the x-axis= arcTan(Fr(y – component)/ Fr(x – component))
And I almost forgot

Fbc=k(3 uC x 4 uC)/(0.20)^2
Fbc=k(2 uC x 4 uC)/(0.20)^2

Where k = 8.988×109 N m^2 /C^2

Now just let the calculator do the work.