Given the problem, we are provided with the initial conditions and asked to use Coulomb's law to predict how the force between the two particles changes when the distance between them is doubled.
### Step-by-Step Solution
1. Identify the given constants:
- Coulomb's constant, [tex]\( k = 9.00 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \)[/tex]
- Charge of the first particle, [tex]\( q_1 = 1.25 \times 10^{-9} \, \text{C} \)[/tex]
- Charge of the second particle, [tex]\( q_2 = 1.92 \times 10^{-9} \, \text{C} \)[/tex]
- Initial distance between the charges, [tex]\( r = 0.38 \, \text{m} \)[/tex]
2. Calculate the initial force using Coulomb's Law:
[tex]\[
F_e = \frac{k q_1 q_2}{r^2}
\][/tex]
Substituting the given values:
[tex]\[
F_{\text{initial}} = \frac{(9.00 \times 10^9) \times (1.25 \times 10^{-9}) \times (1.92 \times 10^{-9})}{(0.38)^2}
\][/tex]
This results in an initial force:
[tex]\[
F_{\text{initial}} = 1.50 \times 10^{-7} \, \text{N}
\][/tex]
3. Determine the new distance when it is doubled:
[tex]\[
r_{\text{doubled}} = 2 \times 0.38 \, \text{m} = 0.76 \, \text{m}
\][/tex]
4. Calculate the new force using Coulomb's Law with the doubled distance:
[tex]\[
F_{\text{new}} = \frac{k q_1 q_2}{r_{\text{doubled}}^2}
\][/tex]
Substituting 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]
This results in a new force:
[tex]\[
F_{\text{new}} = 3.74 \times 10^{-8} \, \text{N}
\][/tex]
### Conclusion
Given the above calculations:
- The initial force between the particles is [tex]\( 1.50 \times 10^{-7} \, \text{N} \)[/tex].
- When the distance between the particles is doubled, the new force is [tex]\( 3.74 \times 10^{-8} \, \text{N} \)[/tex].
Thus, the correct answer from the given options is D. [tex]\( 3.74 \times 10^{-8} \, \text{N} \)[/tex].
