Question

charges of +3 μc and –5 μc are 2 mm from each other. the –5 μc charge is replaced with a +5 μc charge. how will the electrical force between the charges compare with the original force? ○ same force, but in the opposite direction ○ same force, but in the same direction ○ greater force, but in the opposite direction ○ greater force, but in the same direction

Explanation:

1. Recall Coulomb's Law: \( F = k\frac{q_1q_2}{r^2} \), where \( k \) is Coulomb's constant, \( q_1, q_2 \) are charges, and \( r \) is the distance between them.
2. Original charges: \( q_1 = +3\ \mu\text{C} \), \( q_2 = -5\ \mu\text{C} \), distance \( r = 2\ \text{mm} \) (constant). The force is attractive (opposite charges).
3. After replacement: \( q_2' = +5\ \mu\text{C} \). Now, \( q_1 \) and \( q_2' \) are like charges (both positive), so the force is repulsive (opposite direction of original attractive force).
4. Calculate the magnitude of the force: Original magnitude \( |F| = k\frac{|3\mu\text{C} \times (-5\mu\text{C})|}{r^2} = k\frac{15\mu\text{C}^2}{r^2} \). After replacement, \( |F'| = k\frac{|3\mu\text{C} \times 5\mu\text{C}|}{r^2} = k\frac{15\mu\text{C}^2}{r^2} \). So the magnitude of the force is the same.

Answer:

A. same force, but in the opposite direction