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Sagot :
To determine the force between two charged particles at an initial distance and then when the distance is doubled, we will apply Coulomb's law. The formula for the electrostatic force between two charges is given by:
[tex]\[ F_e = \frac{k \cdot q_1 \cdot q_2}{r^2} \][/tex]
where:
- [tex]\( F_e \)[/tex] is the electrostatic force,
- [tex]\( k \)[/tex] is Coulomb's constant ([tex]\( 9.00 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)[/tex]),
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the charges of the particles,
- [tex]\( r \)[/tex] is the separation distance between the charges.
Given data:
- Charge [tex]\( q_1 = 1.25 \times 10^{-9} \, \text{C} \)[/tex]
- Charge [tex]\( q_2 = 1.92 \times 10^{-9} \, \text{C} \)[/tex]
- Initial separation [tex]\( r = 0.38 \, \text{m} \)[/tex]
1. Calculating the force at the initial distance [tex]\( r = 0.38 \, \text{m} \)[/tex]:
[tex]\[ F_{\text{initial}} = \frac{9.00 \times 10^9 \cdot 1.25 \times 10^{-9} \cdot 1.92 \times 10^{-9}}{(0.38)^2} \][/tex]
The calculated result is:
[tex]\[ F_{\text{initial}} \approx 1.50 \times 10^{-7} \, \text{N} \][/tex]
2. Doubling the distance [tex]\( r_{\text{new}} = 2 \times 0.38 \, \text{m} = 0.76 \, \text{m} \)[/tex]:
[tex]\[ F_{\text{new}} = \frac{9.00 \times 10^9 \cdot 1.25 \times 10^{-9} \cdot 1.92 \times 10^{-9}}{(0.76)^2} \][/tex]
The calculated result is:
[tex]\[ F_{\text{new}} \approx 3.74 \times 10^{-8} \, \text{N} \][/tex]
So, the initial electrostatic force is [tex]\( 1.50 \times 10^{-7} \, \text{N} \)[/tex], and the new force when the distance is doubled is [tex]\( 3.74 \times 10^{-8} \, \text{N} \)[/tex].
Among the given choices:
- A. [tex]\( -3.74 \times 10^{-8} \, \text{N} \)[/tex]
- B. [tex]\( 1.50 \times 10^{-7} \, \text{N} \)[/tex]
- C. [tex]\( -1.50 \times 10^{-7} \, \text{N} \)[/tex]
- D. [tex]\( 3.74 \times 10^{-8} \, \text{N} \)[/tex]
The correct numerical value for the new force is [tex]\( 3.74 \times 10^{-8} \, \text{N} \)[/tex].
Thus, the correct answer is:
D. [tex]\( 3.74 \times 10^{-8} \, \text{N} \)[/tex]
[tex]\[ F_e = \frac{k \cdot q_1 \cdot q_2}{r^2} \][/tex]
where:
- [tex]\( F_e \)[/tex] is the electrostatic force,
- [tex]\( k \)[/tex] is Coulomb's constant ([tex]\( 9.00 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)[/tex]),
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the charges of the particles,
- [tex]\( r \)[/tex] is the separation distance between the charges.
Given data:
- Charge [tex]\( q_1 = 1.25 \times 10^{-9} \, \text{C} \)[/tex]
- Charge [tex]\( q_2 = 1.92 \times 10^{-9} \, \text{C} \)[/tex]
- Initial separation [tex]\( r = 0.38 \, \text{m} \)[/tex]
1. Calculating the force at the initial distance [tex]\( r = 0.38 \, \text{m} \)[/tex]:
[tex]\[ F_{\text{initial}} = \frac{9.00 \times 10^9 \cdot 1.25 \times 10^{-9} \cdot 1.92 \times 10^{-9}}{(0.38)^2} \][/tex]
The calculated result is:
[tex]\[ F_{\text{initial}} \approx 1.50 \times 10^{-7} \, \text{N} \][/tex]
2. Doubling the distance [tex]\( r_{\text{new}} = 2 \times 0.38 \, \text{m} = 0.76 \, \text{m} \)[/tex]:
[tex]\[ F_{\text{new}} = \frac{9.00 \times 10^9 \cdot 1.25 \times 10^{-9} \cdot 1.92 \times 10^{-9}}{(0.76)^2} \][/tex]
The calculated result is:
[tex]\[ F_{\text{new}} \approx 3.74 \times 10^{-8} \, \text{N} \][/tex]
So, the initial electrostatic force is [tex]\( 1.50 \times 10^{-7} \, \text{N} \)[/tex], and the new force when the distance is doubled is [tex]\( 3.74 \times 10^{-8} \, \text{N} \)[/tex].
Among the given choices:
- A. [tex]\( -3.74 \times 10^{-8} \, \text{N} \)[/tex]
- B. [tex]\( 1.50 \times 10^{-7} \, \text{N} \)[/tex]
- C. [tex]\( -1.50 \times 10^{-7} \, \text{N} \)[/tex]
- D. [tex]\( 3.74 \times 10^{-8} \, \text{N} \)[/tex]
The correct numerical value for the new force is [tex]\( 3.74 \times 10^{-8} \, \text{N} \)[/tex].
Thus, the correct answer is:
D. [tex]\( 3.74 \times 10^{-8} \, \text{N} \)[/tex]
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