Answer :
To determine the magnitude and direction of the electrical force between two point charges, we use Coulomb's law.
Coulomb's law states that the magnitude of the electrical force [tex]\( F \)[/tex] between two point charges is given by:
[tex]\[ F = k \frac{|q_1 \cdot q_2|}{d^2} \][/tex]
where:
- [tex]\( k \)[/tex] is Coulomb's constant, approximately [tex]\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)[/tex],
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the magnitudes of the two charges,
- [tex]\( d \)[/tex] is the distance between the charges.
Given data:
- Charge [tex]\( q_1 = 5 \, \mu C = 5 \times 10^{-6} \, \text{C} \)[/tex],
- Charge [tex]\( q_2 = 2 \, \mu C = 2 \times 10^{-6} \, \text{C} \)[/tex],
- Distance [tex]\( d = 3 \times 10^{-2} \, \text{m} = 0.03 \, \text{m} \)[/tex].
Substitute these values into Coulomb's law:
[tex]\[ F = (8.99 \times 10^9) \frac{|(5 \times 10^{-6}) \cdot (2 \times 10^{-6})|}{(0.03)^2} \][/tex]
Calculate the numerator:
[tex]\[ (5 \times 10^{-6}) \times (2 \times 10^{-6}) = 10 \times 10^{-12} = 1 \times 10^{-11} \, \text{C}^2 \][/tex]
Calculate the denominator:
[tex]\[ (0.03)^2 = 9 \times 10^{-4} \, \text{m}^2 \][/tex]
Now, divide the numerator by the denominator:
[tex]\[ \frac{1 \times 10^{-11}}{9 \times 10^{-4}} = 1.1111 \times 10^{-8} \, \text{C}^2/\text{m}^2 \][/tex]
Finally, multiply by Coulomb's constant:
[tex]\[ F = 8.99 \times 10^9 \times 1.1111 \times 10^{-8} = 99.8889 \, \text{N} \][/tex]
So, the magnitude of the force is approximately [tex]\( 99.8889 \, \text{N} \)[/tex].
Since both charges are positive, they repel each other. The problem states that charge [tex]\( q_1 \)[/tex] is located west of charge [tex]\( q_2 \)[/tex]. Therefore, the direction of the force on [tex]\( q_2 \)[/tex] is towards the east.
Thus, the electrical force [tex]\( F_e \)[/tex] applied by [tex]\( q_1 \)[/tex] on [tex]\( q_2 \)[/tex] has a magnitude of approximately [tex]\( 100 \, \text{N} \)[/tex] and the direction is east.
The correct answer from the provided options is:
Magnitude: 100 N, Direction: east
Coulomb's law states that the magnitude of the electrical force [tex]\( F \)[/tex] between two point charges is given by:
[tex]\[ F = k \frac{|q_1 \cdot q_2|}{d^2} \][/tex]
where:
- [tex]\( k \)[/tex] is Coulomb's constant, approximately [tex]\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)[/tex],
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the magnitudes of the two charges,
- [tex]\( d \)[/tex] is the distance between the charges.
Given data:
- Charge [tex]\( q_1 = 5 \, \mu C = 5 \times 10^{-6} \, \text{C} \)[/tex],
- Charge [tex]\( q_2 = 2 \, \mu C = 2 \times 10^{-6} \, \text{C} \)[/tex],
- Distance [tex]\( d = 3 \times 10^{-2} \, \text{m} = 0.03 \, \text{m} \)[/tex].
Substitute these values into Coulomb's law:
[tex]\[ F = (8.99 \times 10^9) \frac{|(5 \times 10^{-6}) \cdot (2 \times 10^{-6})|}{(0.03)^2} \][/tex]
Calculate the numerator:
[tex]\[ (5 \times 10^{-6}) \times (2 \times 10^{-6}) = 10 \times 10^{-12} = 1 \times 10^{-11} \, \text{C}^2 \][/tex]
Calculate the denominator:
[tex]\[ (0.03)^2 = 9 \times 10^{-4} \, \text{m}^2 \][/tex]
Now, divide the numerator by the denominator:
[tex]\[ \frac{1 \times 10^{-11}}{9 \times 10^{-4}} = 1.1111 \times 10^{-8} \, \text{C}^2/\text{m}^2 \][/tex]
Finally, multiply by Coulomb's constant:
[tex]\[ F = 8.99 \times 10^9 \times 1.1111 \times 10^{-8} = 99.8889 \, \text{N} \][/tex]
So, the magnitude of the force is approximately [tex]\( 99.8889 \, \text{N} \)[/tex].
Since both charges are positive, they repel each other. The problem states that charge [tex]\( q_1 \)[/tex] is located west of charge [tex]\( q_2 \)[/tex]. Therefore, the direction of the force on [tex]\( q_2 \)[/tex] is towards the east.
Thus, the electrical force [tex]\( F_e \)[/tex] applied by [tex]\( q_1 \)[/tex] on [tex]\( q_2 \)[/tex] has a magnitude of approximately [tex]\( 100 \, \text{N} \)[/tex] and the direction is east.
The correct answer from the provided options is:
Magnitude: 100 N, Direction: east
