In the figure particle 1 (of charge +1.43 mC), particle 2 (of charge +1.43 mC), and particle 3 (of charge Q) form an equilateral triangle of edge length a.

Question

In the figure particle 1 (of charge +1.43 mC), particle 2 (of charge +1.43 mC), and particle 3 (of charge Q) form an equilateral triangle of edge length a. For what value of Q (both sign and magnitude) does the net electric field produced by the particles at the center of the triangle vanish? Tolerance is +-2%

Answer 1

According to superposition, the net electric field seen at the center of the triangle is the summation of fields seen by every individual point. The equation for electric field contributed by a point source with charge Q is

E = kQ/r²

For an equilateral triangle of side length a, the distance from every corner to the center is r = a/√3 .

Thus there are 3 vectors of electric field that run parallel to the corner-center connecting lines. Each of these vectors has a x and y component. In order to cancel out the fields at the centers, the net force in x and y must be 0.

The electric field magnitude by each point source is then accordingly. (It is assumed that q = 1.43 mC as stated in the question)

|E₁| = kq/(a/√3)² = 3kq/a²

|E₂| = kq/(a/√3)² = 3kq/a²

|E₃| = kQ/(a/√3)² = 3kQ/a²

However, we need the individual x and y components of every vector, not just the magnitude. We know that the vectors arise from the corners of the triangle. Given that this is an equilateral triangle, every corner has an angle of 60°. Thus the resulting vector comes from the center of the corner with an angle of 30° with respect to the x-axis.

Thus the resulting vector for each component can be calculated as. Take note that the vector notation is <x component, y component>

E₁ = < 3kq/a² cos(30) , 3kq/a² sin(30) >

E₂ = < -3kq/a² cos(30) , 3kq/a² sin(30) >

E₃ = < 0 , -3kQ/a² >

Thus the net electric field in the x-direction is E_x = 3kq/a² cos(30) - 3kq/a² cos(30) = 0

the net electric field in the y-direction is E_y = 3kq/a² sin(30) + 3kq/a² sin(30) -3kQ/a² = 3kq/a² - 3kQ/a²

To have no electric field the y-field must be zero. Thus

3kq/a² - 3kQ/a² = 0 ⇒q = Q

Thus Q = +1.43 mC