In the figure particle 1 (of charge +4.27 mC), particle 2 (ofcharge +4.27 mC), and particle 3 (of charge Q) form an equilateraltriangle of edge length a. For what value of Q (both sign andmagnitude) does the net electric field produced by the particles atthe center of the triangle vanish?

To solve this problem, we need to use Coulomb's Law, which states that the electric field E produced by a charge q at a distance r is given by E = kq/r^2, where k is Coulomb's constant.

1. First, we calculate the electric field produced by particle 1 at the center of the triangle. Since the triangle is equilateral, the distance from each corner to the center is a/sqrt(3), where a is the length of the edge. So, the electric field E1 produced by particle 1 is E1 = k*4.27mC/(a/sqrt(3))^2.

2. Similarly, the electric field E2 produced by particle 2 at the center of the triangle is E2 = k*4.27mC/(a/sqrt(3))^2.

3. The electric field E3 produced by particle 3 at the center of the triangle is E3 = k*Q/(a/sqrt(3))^2.

4. For the net electric field at the center of the triangle to vanish, the sum of the electric fields produced by the three particles must be zero. That is, E1 + E2 + E3 = 0.

5. Substituting the expressions for E1, E2, and E3 into this equation, we get k4.27mC/(a/sqrt(3))^2 + k4.27mC/(a/sqrt(3))^2 + k*Q/(a/sqrt(3))^2 = 0.

6. Simplifying this equation, we find that Q = -2*4.27mC = -8.54mC.

So, for the net electric field produced by the particles at the center of the triangle to vanish, the charge Q of particle 3 must be -8.54mC.