Question

Three point charges are located on the x-axis at the following positions: Q1 = +7.00 μC is at x = 1.00 m, Q2 = +3.00 μC is at x = 0.00, and Q3 = -5.00 μC is at x = -3.00 m. What is the magnitude of the electric force on Q2? (k = 1/4nɛO = 9.0 x 10⁹ N·m²/C²)

a) 1.08 x 10⁹ N
b) 2.70 x 10⁹ N
c) 4.32 x 10⁹ N
d) 5.40 x 10⁹ N

Answer 1

The magnitude of the electric force on Q2 is 6.30 x 10^9 N.

Step-by-step explanation:

To find the magnitude of the electric force on Q2, we can use Coulomb's Law. Coulomb's Law states that the magnitude of the electric force between two point charges is given by F = k * (|Q1| * |Q2|) / r^2, where k is the constant, Q1 and Q2 are the charges, and r is the distance between the charges.

In this case, Q1 = +7.00 μC, Q2 = +3.00 μC, and the distance between them is 1.00 m.

Plugging these values into the formula, we get:

F = (9.0 x 10^9 N·m²/C²) * ((7.00 x 10^-6 C) * (3.00 x 10^-6 C)) / (1.00 m)^2 = 6.30 x 10^9 N

Therefore, the magnitude of the electric force on Q2 is 6.30 x 10^9 N.