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A positive charge, [tex]\( q_1 \)[/tex], of [tex]\( 5 \mu C \)[/tex] is [tex]\( 3 \times 10^{-2} \, m \)[/tex] west of a positive charge, [tex]\( q_2 \)[/tex], of [tex]\( 2 \mu C \)[/tex].

What is the magnitude and direction of the electrical force, [tex]\( F_e \)[/tex], applied by [tex]\( q_1 \)[/tex] on [tex]\( q_2 \)[/tex]?

A. Magnitude: [tex]\( 3 \, N \)[/tex], Direction: east
B. Magnitude: [tex]\( 3 \, N \)[/tex], Direction: west
C. Magnitude: [tex]\( 100 \, N \)[/tex], Direction: east
D. Magnitude: [tex]\( 100 \, N \)[/tex], Direction: west



Answer :

GinnyAnswer
To determine the magnitude and direction of the electrical force, [tex]\( F_e \)[/tex], between two charges [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] using Coulomb's law, follow these steps:

1. Understand the Problem:
- [tex]\( q_1 = 5 \mu C \)[/tex] (microcoulombs) is a positive charge.
- [tex]\( q_2 = 2 \mu C \)[/tex] (microcoulombs) is also a positive charge.
- The distance between them is [tex]\( 3 \times 10^{-2} \)[/tex] meters (or 0.03 meters).
- We need to determine the magnitude and direction of the force between these charges.

2. Convert the Charges to Coulombs:
- The charge [tex]\( q_1 = 5 \mu C = 5 \times 10^{-6} \)[/tex] Coulombs.
- The charge [tex]\( q_2 = 2 \mu C = 2 \times 10^{-6} \)[/tex] Coulombs.

3. Use Coulomb's Law:
Coulomb's law states that the magnitude of the electrical force between two charges is given by:
[tex]\[ F_e = k \frac{|q_1 \cdot q_2|}{r^2} \][/tex]
where:
- [tex]\( k \)[/tex] is Coulomb's constant ([tex]\( k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 \cdot \text{C}^{-2} \)[/tex]).
- [tex]\( |q_1 \cdot q_2| \)[/tex] is the product of the magnitudes of the charges.
- [tex]\( r \)[/tex] is the distance between the charges.

4. Calculate the Force:
[tex]\[ F_e = 8.99 \times 10^9 \times \frac{(5 \times 10^{-6})(2 \times 10^{-6})}{(0.03)^2} \][/tex]
[tex]\[ F_e = 8.99 \times 10^9 \times \frac{10 \times 10^{-12}}{(0.03)^2} \][/tex]
[tex]\[ F_e = 8.99 \times 10^9 \times \frac{10 \times 10^{-12}}{9 \times 10^{-4}} \][/tex]
[tex]\[ F_e = 8.99 \times 10^9 \times \frac{10 \times 10^{-12}}{9 \times 10^{-4}} \][/tex]
[tex]\[ F_e = 8.99 \times 10^9 \times 1.111 \times 10^{-8} \][/tex]
[tex]\[ F_e \approx 99.89 \, \text{N} \][/tex]

5. Determine the Direction:
Since both [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are positive charges, they repel each other. Therefore, the force on [tex]\( q_2 \)[/tex] due to [tex]\( q_1 \)[/tex] will be directed away from [tex]\( q_1 \)[/tex]. Given that [tex]\( q_1 \)[/tex] is west of [tex]\( q_2 \)[/tex], the force on [tex]\( q_2 \)[/tex] will be directed east.

Therefore, the magnitude of the electrical force is approximately [tex]\( 100 \, \text{N} \)[/tex] and the direction is east.

So, the correct choice is:
- Magnitude: [tex]\( 100 \, \text{N} \)[/tex]
- Direction: East

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