High School

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Two tiny particles have charges of [tex]+5.00 \times 10^{-6}[/tex] C and [tex]+7.00 \times 10^{-6}[/tex] C. They are placed along the x-axis. The [tex]+5.00 \, \mu \text{C}[/tex] particle is at [tex]x = 0.00 \, \text{cm}[/tex].

A: True
B: False

Answer :

Final answer:

The magnitude of the force on the test charge is determined by calculating the force from each nearby charge using Coulomb's Law and understanding that the force is repulsive due to the positive nature of all charges involved. The direction of the net force on the test charge will be away from the larger +6µC charge.

Explanation:

The problem involves calculating the magnitude of the force on a test charge placed halfway between two charged particles and determining the direction of this force. We use Coulomb's Law, which states the force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.

To solve part (a), the magnitude of the force exerted on the test charge by each of the two charges (+6µC and +4µC) needs to be calculated separately and then compared to understand the net effect. Since the test charge is equidistant from both charges and they are both positive, implying repulsion, the forces will add up in magnitude but have different directions.

For part (b), given the charges are both positive, the test charge will be repelled by both. However, since the force magnitude depends on both the amount of charge and the distance, and both distances are the same, the test charge will experience a greater force away from the +6µC charge due to its larger magnitude. Therefore, the net force direction will be away from the +6µC charge.

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