To determine the magnitude of the force one charge exerts on another, we use Coulomb's law. Coulomb's law is expressed by the formula:
[tex]\[ F = k \cdot \frac{|q_1 \cdot q_2|}{r^2} \][/tex]
where:
- [tex]\(F\)[/tex] is the magnitude of the force between the charges,
- [tex]\(k\)[/tex] is Coulomb's constant, which is [tex]\(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2\)[/tex],
- [tex]\(q_1\)[/tex] and [tex]\(q_2\)[/tex] are the magnitudes of the two charges,
- [tex]\(r\)[/tex] is the distance between the centers of the two charges.
Given:
- [tex]\(q_1 = 29 \, \mu\text{C}\)[/tex] (microcoulombs), which is [tex]\(29 \times 10^{-6} \, \text{C}\)[/tex],
- [tex]\(q_2 = 3.2 \, \text{mC}\)[/tex] (millicoulombs), which is [tex]\(3.2 \times 10^{-3} \, \text{C}\)[/tex],
- [tex]\(r = 45 \, \text{cm}\)[/tex], which is [tex]\(0.45 \, \text{m}\)[/tex].
Insert these values into the Coulomb's law formula:
1. Convert the charges to Coulombs:
[tex]\[ q_1 = 29 \times 10^{-6} \, \text{C} \][/tex]
[tex]\[ q_2 = 3.2 \times 10^{-3} \, \text{C} \][/tex]
2. The distance between the charges is 0.45 meters.
3. Calculate the force:
[tex]\[ F = 8.99 \times 10^9 \cdot \frac{|29 \times 10^{-6} \cdot 3.2 \times 10^{-3}|}{(0.45)^2} \][/tex]
4. Simplify the expression inside the absolute value:
[tex]\[ |29 \times 10^{-6} \cdot 3.2 \times 10^{-3}| = 9.28 \times 10^{-8} \][/tex]
5. Calculate the square of the distance:
[tex]\[ (0.45)^2 = 0.2025 \][/tex]
6. Plugging all values into the formula:
[tex]\[ F = 8.99 \times 10^9 \cdot \frac{9.28 \times 10^{-8}}{0.2025} \][/tex]
7. Performing the division inside the parenthesis:
[tex]\[ \frac{9.28 \times 10^{-8}}{0.2025} \approx 4.58 \times 10^{-7} \][/tex]
8. Finally, multiply by Coulomb's constant:
[tex]\[ F = 8.99 \times 10^9 \cdot 4.58 \times 10^{-7} \][/tex]
[tex]\[ F \approx 4119.86 \, \text{N} \][/tex]
To express the final answer to two significant figures:
[tex]\[ F \approx 4100 \, \text{N} \][/tex]
Thus, the magnitude of the force that a +29 μC charge exerts on a +3.2 mC charge 45 cm away is approximately [tex]\( \mathbf{4.1 \times 10^3 \, \text{N}} \)[/tex].
