Certainly! Let's solve this problem using Coulomb's law step-by-step. Recall Coulomb's law:
[tex]\[ F = k \frac{q_1 q_2}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the magnitude of the force between the two charges.
- [tex]\( k \)[/tex] is Coulomb's constant ([tex]\( k = 8.99 \times 10^9 \, \text{N m}^2 / \text{C}^2 \)[/tex]).
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the magnitudes of the charges.
- [tex]\( r \)[/tex] is the distance between the charges.
We are given the following values:
- The charges [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are both [tex]\( e = 1.6 \times 10^{-19} \, \text{C} \)[/tex].
- The distance [tex]\( r \)[/tex] between the charges is [tex]\( 10^{-16} \, \text{m} \)[/tex].
Step-by-Step Solution:
1. Write down the known values:
- [tex]\( q_1 = 1.6 \times 10^{-19} \, \text{C} \)[/tex]
- [tex]\( q_2 = 1.6 \times 10^{-19} \, \text{C} \)[/tex]
- [tex]\( r = 10^{-16} \, \text{m} \)[/tex]
- [tex]\( k = 8.99 \times 10^9 \, \text{N m}^2 / \text{C}^2 \)[/tex]
2. Substitute these values into Coulomb's law formula.
[tex]\[ F = 8.99 \times 10^9 \, \frac{(1.6 \times 10^{-19}) (1.6 \times 10^{-19})}{(10^{-16})^2} \][/tex]
3. Calculate the product of the charges.
[tex]\[ (1.6 \times 10^{-19}) (1.6 \times 10^{-19}) = 2.56 \times 10^{-38} \, \text{C}^2 \][/tex]
4. Calculate the square of the distance.
[tex]\[ (10^{-16})^2 = 10^{-32} \, \text{m}^2 \][/tex]
5. Substitute these values into the formula:
[tex]\[ F = 8.99 \times 10^9 \, \frac{2.56 \times 10^{-38}}{10^{-32}} \][/tex]
6. Simplify the exponentiation in the denominator:
[tex]\[ \frac{10^{-38}}{10^{-32}} = 10^{-6} \][/tex]
7. Continue the calculation:
[tex]\[ F = 8.99 \times 10^9 \, \times 2.56 \times 10^{-6} \][/tex]
8. Multiply the constants:
[tex]\[ F = 23.0144 \, \text{N} \][/tex]
After following this step-wise calculation, we find the electrical force between these two particles to be approximately [tex]\( 23,014.4 \, \text{N} \)[/tex].
Therefore, the answer is:
[tex]\[ \boxed{23,000 \, \text{N}} \][/tex]
