Final answer:
Using Coulomb's Law, the force on the test charge due to each of the other charges was found to be 4.32 N and 2.88 N, respectively. The net force is the difference between the two because the forces are both repulsive and act in opposite directions. The answer is a net force of 1.44 N away from the +6 μC charge. So, the correct answer is (b) (1.44 N), away from the +6 μC charge.
Step-by-step explanation:
To calculate the magnitude of the force on the test charge when it is placed halfway between two charges, we can use Coulomb's Law, which states that the force (F) between two charges is proportional to the product of the charges (q1 and q2) and inversely proportional to the square of the distance (r) between them:
F = k * |q1 * q2| / r^2
where k is Coulomb's constant (9.0 × 10^9 N·m^2/C^2).
Since the test charge (+2 μC) is placed halfway between +6 μC and +4 μC charges that are 10 cm apart, each distance is 5 cm (0.05 m). The forces due to each charge can be calculated separately and then combined using vector addition.
Force due to +6 μC charge:
F1 = k * |2 μC * 6 μC| / (0.05 m)^2
Force due to +4 μC charge:
F2 = k * |2 μC * 4 μC| / (0.05 m)^2
Next, we calculate the forces:
F1 = 9 × 10^9 * 12 × 10^-12 / 2.5 × 10^-3 = 4.32 N
F2 = 9 × 10^9 * 8 × 10^-12 / 2.5 × 10^-3 = 2.88 N
Since both forces are repulsive and the charges on the right side (+4 μC) are smaller, the net force will be toward the right, away from the +6 μC charge. Therefore, the magnitude of the net force on the test charge will be the difference between F1 and F2, which is 4.32 N - 2.88 N = 1.44 N.
The correct answer is (b) (1.44 N), away from the +6 μC charge.