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Sagot :
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
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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