Question

A positive point charge q1 = 1.5 μC is located at x = 0, y = 0.30 m. A negative point charge q2 = -1.5 μC is located at x = 0, y = -0.30 m. A third point charge q = 5.0 μC is located at x = 0.40 m, y = 0. What is the net electrostatic force acting on the charge q?

Answer 1

The net electrostatic force on charge q is approximately 0.38 N in the positive y-direction, as calculated using Coulomb's Law and vector addition.

Step-by-step explanation:

To determine the net electrostatic force acting on the third charge q, we must consider the forces due to both q1 and q2 on q. Using Coulomb's Law, which states that the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them, we can calculate the separate forces from q1 and q2 on q. First, we find the distance between each pair of charges using the Pythagorean theorem. The distance between q and q1 (as well as between q and q2) is the hypotenuse of a right triangle with sides 0.40 m and 0.30 m. Thus, the distance r for both pairs is √((0.40 m)² + (0.30 m)²) = 0.50 m. Next, we calculate the force due to q1 on q (Fq1) and the force due to q2 on q (Fq2). Since q1 is positive and q is positive, the force will be repulsive, and since q2 is negative and q is positive, the force will be attractive. The magnitudes of these forces will be equal (due to the charges being equal and the distances being the same), but the directions will be opposite. The magnitude of the force can be calculated as F = k * |q1 * q| / r², with k being the Coulomb constant (8.988 × 10⁹ N·m²/C²). Inserting the magnitudes of the charges and the distance calculated above, we get F = (8.988 × 10⁹ N·m²/C²) * |(1.5 × 10⁻⁶ C) * (5.0 × 10⁻⁶ C)| / (0.50 m)² = 0.27 N.