The net electric force on q, resulting from the interaction with two point charges, is 1.518 N to the left, as determined by Coulomb's law and the superposition principle.
To solve this problem, you need to use the formula for electric force, which is given by Coulomb's law. According to one of the web search results, the formula is:
F = k * (q1 * q2) / r^2
where F is the electric force, k is a constant equal to about 8.99 * 10^9 N * m^2 / C^2 in a vacuum, q1 and q2 are the charges of the objects, and r is the distance between them.
To find the electric force on q due to the other two point charges, you need to apply the superposition principle, which states that the net force on a charge is the vector sum of the forces exerted by each of the other charges. Therefore, you need to find the force on q due to q and the force on q due to 2q, and then add them as vectors.
The force on q due to q is repulsive, since they have the same sign, and its magnitude is:
Fqq = k * (q^2) / d^2
The force on q due to 2q is attractive, since they have opposite signs, and its magnitude is:
Fq2q = k * (2 * q^2) / (2 * d)^2
The direction of these forces is along the line joining the charges, so they are parallel to each other. To add them as vectors, you need to consider their signs. Since the repulsive force is to the left and the attractive force is to the right, you need to subtract them to get the net force. The net force on q is:
Fnet = Fq2q - Fqq
Substituting the values of k, q, and d, you get:
Fnet = (8.99 * 10^9) * (2 * (1.2 * 10^-5)^2) / (0.32)^2 - (8.99 * 10^9) * ((1.2 * 10^-5)^2) / (0.16)^2
Simplifying, you get:
Fnet = 0.506 - 2.024 = -1.518 N
The negative sign indicates that the net force is to the left. Therefore, the electric force on q due to the other two point charges is 1.518 N to the left.