Answer :
To determine which diagram correctly represents the electrical force between the charges, we need to calculate the magnitude of the force using Coulomb's Law and understand its direction. Coulomb's Law states that the electrical force ([tex]\(F\)[/tex]) between two point charges is given by the formula:
[tex]\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \][/tex]
where:
- [tex]\( k \)[/tex] is Coulomb's constant ([tex]\(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2\)[/tex]),
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the magnitudes of the charges,
- and [tex]\( r \)[/tex] is the distance between the charges.
In this problem:
- [tex]\( q_1 = 2 \mu C = 2 \times 10^{-6} \, \text{C} \)[/tex],
- [tex]\( q_2 = 3 \mu C = 3 \times 10^{-6} \, \text{C} \)[/tex],
- [tex]\( r = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m} \)[/tex].
First, we calculate the force [tex]\( F \)[/tex]:
[tex]\[ F = 8.99 \times 10^9 \times \frac{|2 \times 10^{-6} \cdot 3 \times 10^{-6}|}{(4 \times 10^{-3})^2} \][/tex]
By substituting the numbers, we get:
[tex]\[ F = 8.99 \times 10^9 \times \frac{6 \times 10^{-12}}{16 \times 10^{-6}} \][/tex]
[tex]\[ F = 8.99 \times 10^9 \times \frac{6}{16} \times 10^{-6} \][/tex]
[tex]\[ F = 8.99 \times 10^9 \times 0.375 \times 10^{-6} \][/tex]
[tex]\[ F = 3,371.250 \, \text{N} \][/tex]
This calculation confirms that the magnitude of the electrical force is [tex]\( 3371.25 \, \text{N} \)[/tex].
Next, we consider the direction of the force. Since both charges are positive, they repel each other due to the electrostatic force. This means the direction of the force on each charge is away from the other charge.
Given this information, the correct diagram should reflect the following:
- The two charges are separated by [tex]\( 4 \, \text{mm} \)[/tex].
- The force vectors on each charge point away from the other charge, indicating repulsion.
Reviewing the diagrams:
- [tex]$W$[/tex] represents forces that are attractive (force vectors towards each other).
- [tex]$X$[/tex] represents forces that are repulsive (force vectors away from each other).
- [tex]$Y$[/tex] represents forces that are attractive.
- [tex]$Z$[/tex] represents forces that are repulsive.
Between [tex]$X$[/tex] and [tex]$Z$[/tex], either could be correct depending on the specific positioning and if no weird confusions in the scaling or vectors' lengths exist.
Hence, without seeing the diagrams, the configuration where force vectors point away from each other due to repulsive interaction (likely represented by [tex]$X$[/tex] or [tex]$Z$[/tex]) would be correct.
[tex]\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \][/tex]
where:
- [tex]\( k \)[/tex] is Coulomb's constant ([tex]\(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2\)[/tex]),
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the magnitudes of the charges,
- and [tex]\( r \)[/tex] is the distance between the charges.
In this problem:
- [tex]\( q_1 = 2 \mu C = 2 \times 10^{-6} \, \text{C} \)[/tex],
- [tex]\( q_2 = 3 \mu C = 3 \times 10^{-6} \, \text{C} \)[/tex],
- [tex]\( r = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m} \)[/tex].
First, we calculate the force [tex]\( F \)[/tex]:
[tex]\[ F = 8.99 \times 10^9 \times \frac{|2 \times 10^{-6} \cdot 3 \times 10^{-6}|}{(4 \times 10^{-3})^2} \][/tex]
By substituting the numbers, we get:
[tex]\[ F = 8.99 \times 10^9 \times \frac{6 \times 10^{-12}}{16 \times 10^{-6}} \][/tex]
[tex]\[ F = 8.99 \times 10^9 \times \frac{6}{16} \times 10^{-6} \][/tex]
[tex]\[ F = 8.99 \times 10^9 \times 0.375 \times 10^{-6} \][/tex]
[tex]\[ F = 3,371.250 \, \text{N} \][/tex]
This calculation confirms that the magnitude of the electrical force is [tex]\( 3371.25 \, \text{N} \)[/tex].
Next, we consider the direction of the force. Since both charges are positive, they repel each other due to the electrostatic force. This means the direction of the force on each charge is away from the other charge.
Given this information, the correct diagram should reflect the following:
- The two charges are separated by [tex]\( 4 \, \text{mm} \)[/tex].
- The force vectors on each charge point away from the other charge, indicating repulsion.
Reviewing the diagrams:
- [tex]$W$[/tex] represents forces that are attractive (force vectors towards each other).
- [tex]$X$[/tex] represents forces that are repulsive (force vectors away from each other).
- [tex]$Y$[/tex] represents forces that are attractive.
- [tex]$Z$[/tex] represents forces that are repulsive.
Between [tex]$X$[/tex] and [tex]$Z$[/tex], either could be correct depending on the specific positioning and if no weird confusions in the scaling or vectors' lengths exist.
Hence, without seeing the diagrams, the configuration where force vectors point away from each other due to repulsive interaction (likely represented by [tex]$X$[/tex] or [tex]$Z$[/tex]) would be correct.
