To determine the electromagnetic force between two charged particles, we will use Coulomb's law, which is given by the equation:
[tex]\[ F_e = \frac{k \cdot q_1 \cdot q_2}{r^2} \][/tex]
Where:
- [tex]\( F_e \)[/tex] is the electromagnetic force.
- [tex]\( k \)[/tex] is Coulomb's constant ([tex]\( 9.00 \times 10^9 \, N \cdot m^2 / C^2 \)[/tex]).
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the charges of the particles.
- [tex]\( r \)[/tex] is the distance between the two charges.
Given data:
- [tex]\( q_1 = 2.15 \times 10^{-9} \, C \)[/tex]
- [tex]\( q_2 = 3.22 \times 10^{-9} \, C \)[/tex]
- [tex]\( r = 0.015 \, m \)[/tex]
Now, let's substitute these values into the Coulomb's law equation and solve for [tex]\( F_e \)[/tex].
[tex]\[ F_e = \frac{(9.00 \times 10^9 \, N \cdot m^2 / C^2) \cdot (2.15 \times 10^{-9} \, C) \cdot (3.22 \times 10^{-9} \, C)}{(0.015 \, m)^2} \][/tex]
First, calculate [tex]\( r^2 \)[/tex]:
[tex]\[ r^2 = (0.015 \, m)^2 = 0.000225 \, m^2 \][/tex]
Next, calculate the numerator:
[tex]\[ (9.00 \times 10^9) \cdot (2.15 \times 10^{-9}) \cdot (3.22 \times 10^{-9}) = 6.2145 \times 10^{-8} \, N \cdot m^2 / C^2 \][/tex]
Now, divide the numerator by [tex]\( r^2 \)[/tex]:
[tex]\[ F_e = \frac{6.2145 \times 10^{-8}}{0.000225} \][/tex]
Finally, compute the force:
[tex]\[ F_e = 0.0002769200000000001 \, N \][/tex]
Rounding to two significant figures, as given in the options:
[tex]\[ F_e \approx 2.77 \times 10^{-4} \, N \][/tex]
Therefore, the electromagnetic force between the two particles is closest to the choice:
[tex]\[ \boxed{2.77 \times 10^{-4} \, N} \][/tex]
The correct answer is:
B. [tex]\( 2.77 \times 10^{-4} \, N \)[/tex]
