To determine the electric force [tex]\(F_e\)[/tex] acting between two charges [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] that are a distance [tex]\( r \)[/tex] apart, we use Coulomb's Law, given by the formula:
[tex]\[ F_e = \frac{k q_1 q_2}{r^2} \][/tex]
where:
- [tex]\( k \)[/tex] is Coulomb's constant, equal to [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 magnitudes of the charges, which are given as [tex]\( -0.0050 \, \text{C} \)[/tex] and [tex]\( 0.0050 \, \text{C} \)[/tex] respectively
- [tex]\( r \)[/tex] is the distance between the charges, which is [tex]\( 0.025 \, \text{m} \)[/tex]
Plugging in the values:
[tex]\[ k = 9.00 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \][/tex]
[tex]\[ q_1 = -0.0050 \, \text{C} \][/tex]
[tex]\[ q_2 = 0.0050 \, \text{C} \][/tex]
[tex]\[ r = 0.025 \, \text{m} \][/tex]
First, calculate [tex]\( r^2 \)[/tex]:
[tex]\[ r^2 = (0.025 \, \text{m})^2 = 0.000625 \, \text{m}^2 \][/tex]
Next, compute the product [tex]\( k q_1 q_2 \)[/tex]:
[tex]\[ k q_1 q_2 = (9.00 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2) \times (-0.0050 \, \text{C}) \times (0.0050 \, \text{C}) \][/tex]
[tex]\[ k q_1 q_2 = 9.00 \times 10^9 \times -0.0050 \times 0.0050 \][/tex]
[tex]\[ k q_1 q_2 = 9.00 \times 10^9 \times (-0.000025) \][/tex]
[tex]\[ k q_1 q_2 = -225000000 \][/tex]
Finally, divide by [tex]\( r^2 \)[/tex]:
[tex]\[ F_e = \frac{-225000000}{0.000625} \][/tex]
[tex]\[ F_e = -360000000 \, \text{N} \][/tex]
Thus, the electric force acting between the two charges is [tex]\( -360000000 \, \text{N} \)[/tex], which corresponds to option:
[tex]\[ \boxed{\text{C. } -3.6 \times 10^8 \, \text{N}} \][/tex]
