To solve this problem, let's follow each step methodically:
1. Identify the given values:
- Charge of the first particle ([tex]\(q_1\)[/tex]) = [tex]\(1.25 \times 10^{-9}\)[/tex] C
- Charge of the second particle ([tex]\(q_2\)[/tex]) = [tex]\(1.92 \times 10^{-9}\)[/tex] C
- Original distance between the particles ([tex]\(r_{\text{original}}\)[/tex]) = 0.38 m
- Coulomb's constant ([tex]\(k\)[/tex]) = [tex]\(9.00 \times 10^9\)[/tex] N·m²/C²
2. Apply Coulomb's Law for the original distance:
Coulomb's Law states:
[tex]\[
F_e = \frac{k q_1 q_2}{r^2}
\][/tex]
Substitute the given values for the original distance:
[tex]\[
F_{\text{original}} = \frac{(9.00 \times 10^9) \times (1.25 \times 10^{-9}) \times (1.92 \times 10^{-9})}{(0.38)^2}
\][/tex]
3. Double the distance and calculate the new force:
The new distance ([tex]\(r_{\text{new}}\)[/tex]) = 2 \times 0.38 m = 0.76 m. Now, calculate the force with the new distance:
[tex]\[
F_{\text{new}} = \frac{k q_1 q_2}{(r_{\text{new}})^2}
\][/tex]
Substitute the new distance:
[tex]\[
F_{\text{new}} = \frac{(9.00 \times 10^9) \times (1.25 \times 10^{-9}) \times (1.92 \times 10^{-9})}{(0.76)^2}
\][/tex]
4. Evaluate the forces:
- The original force calculated:
[tex]\[
F_{\text{original}} = 1.50 \times 10^{-7} \, \text{N}
\][/tex]
- The new force calculated:
[tex]\[
F_{\text{new}} = 3.74 \times 10^{-8} \, \text{N}
\][/tex]
The force between the particles when the distance is doubled is [tex]\(3.74 \times 10^{-8} \, \text{N}\)[/tex]. Therefore, the correct answer is:
A. [tex]\(3.74 \times 10^{-8} \, \text{N}\)[/tex]
