Answer :
To solve this problem, we need to apply Coulomb's law, which states that the force between two charged particles is given by the equation:
[tex]\[ F_e = \frac{k q_1 q_2}{r^2} \][/tex]
where
- [tex]\( k = 9.00 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)[/tex] is Coulomb's constant,
- [tex]\( q_1 = -6.25 \times 10^{-9} \, \text{C} \)[/tex] is the charge of the first particle,
- [tex]\( q_2 = 2.91 \times 10^{-9} \, \text{C} \)[/tex] is the charge of the second particle,
- [tex]\( r = 0.38 \, \text{m} \)[/tex] is the initial distance between the particles.
First, we calculate the initial force [tex]\( F_{\text{initial}} \)[/tex] with the given distance [tex]\( r = 0.38 \, \text{m} \)[/tex]:
[tex]\[ F_{\text{initial}} = \frac{(9.00 \times 10^9) \cdot (-6.25 \times 10^{-9}) \cdot (2.91 \times 10^{-9})}{(0.38)^2} \][/tex]
Carrying out these calculations, we get:
[tex]\[ F_{\text{initial}} \approx -1.13 \times 10^{-6} \, \text{N} \][/tex]
Next, we need to find the force when the distance is cut in half. The new distance is:
[tex]\[ r_{\text{new}} = \frac{0.38}{2} = 0.19 \, \text{m} \][/tex]
We now calculate the new force [tex]\( F_{\text{new}} \)[/tex] with the reduced distance [tex]\( r_{\text{new}} = 0.19 \, \text{m} \)[/tex]:
[tex]\[ F_{\text{new}} = \frac{(9.00 \times 10^9) \cdot (-6.25 \times 10^{-9}) \cdot (2.91 \times 10^{-9})}{(0.19)^2} \][/tex]
Carrying out these calculations, we get:
[tex]\[ F_{\text{new}} \approx -4.53 \times 10^{-6} \, \text{N} \][/tex]
The two forces we calculated are:
- [tex]\( F_{\text{initial}} \approx -1.13 \times 10^{-6} \, \text{N} \)[/tex]
- [tex]\( F_{\text{new}} \approx -4.53 \times 10^{-6} \, \text{N} \)[/tex]
Based on the given options, the correct answers are:
B. [tex]\( -1.13 \times 10^{-6} \, \text{N} \)[/tex]
C. [tex]\( -4.53 \times 10^{-6} \, \text{N} \)[/tex]
These values match the calculated forces, thus confirming the predictions according to Coulomb's law.
[tex]\[ F_e = \frac{k q_1 q_2}{r^2} \][/tex]
where
- [tex]\( k = 9.00 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)[/tex] is Coulomb's constant,
- [tex]\( q_1 = -6.25 \times 10^{-9} \, \text{C} \)[/tex] is the charge of the first particle,
- [tex]\( q_2 = 2.91 \times 10^{-9} \, \text{C} \)[/tex] is the charge of the second particle,
- [tex]\( r = 0.38 \, \text{m} \)[/tex] is the initial distance between the particles.
First, we calculate the initial force [tex]\( F_{\text{initial}} \)[/tex] with the given distance [tex]\( r = 0.38 \, \text{m} \)[/tex]:
[tex]\[ F_{\text{initial}} = \frac{(9.00 \times 10^9) \cdot (-6.25 \times 10^{-9}) \cdot (2.91 \times 10^{-9})}{(0.38)^2} \][/tex]
Carrying out these calculations, we get:
[tex]\[ F_{\text{initial}} \approx -1.13 \times 10^{-6} \, \text{N} \][/tex]
Next, we need to find the force when the distance is cut in half. The new distance is:
[tex]\[ r_{\text{new}} = \frac{0.38}{2} = 0.19 \, \text{m} \][/tex]
We now calculate the new force [tex]\( F_{\text{new}} \)[/tex] with the reduced distance [tex]\( r_{\text{new}} = 0.19 \, \text{m} \)[/tex]:
[tex]\[ F_{\text{new}} = \frac{(9.00 \times 10^9) \cdot (-6.25 \times 10^{-9}) \cdot (2.91 \times 10^{-9})}{(0.19)^2} \][/tex]
Carrying out these calculations, we get:
[tex]\[ F_{\text{new}} \approx -4.53 \times 10^{-6} \, \text{N} \][/tex]
The two forces we calculated are:
- [tex]\( F_{\text{initial}} \approx -1.13 \times 10^{-6} \, \text{N} \)[/tex]
- [tex]\( F_{\text{new}} \approx -4.53 \times 10^{-6} \, \text{N} \)[/tex]
Based on the given options, the correct answers are:
B. [tex]\( -1.13 \times 10^{-6} \, \text{N} \)[/tex]
C. [tex]\( -4.53 \times 10^{-6} \, \text{N} \)[/tex]
These values match the calculated forces, thus confirming the predictions according to Coulomb's law.
