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Two particles are separated by 0.38 m and have charges of [tex]$1.25 \times 10^{-9} \, C$[/tex] and [tex]$1.92 \times 10^{-9} \, C$[/tex]. Use Coulomb's law to predict the force between the particles if the distance is doubled. The equation for Coulomb's law is [tex][tex]$F_e = \frac{k q_1 q_2}{r^2}$[/tex][/tex], and the constant, [tex]k[/tex], equals [tex]$9.00 \times 10^9 \, N \cdot m^2 / C^2$[/tex].

A. [tex]$3.74 \times 10^{-8} \, N$[/tex]

B. [tex][tex]$-1.50 \times 10^{-7} \, N$[/tex][/tex]

C. [tex]$1.50 \times 10^{-7} \, N$[/tex]

D. [tex]$-3.74 \times 10^{-8} \, N$[/tex]

Sagot :

GinnyAnswer
To solve this problem, let's follow each step methodically:

1. Identify the given values:
- Charge of the first particle ([tex]\(q_1\)[/tex]) = [tex]\(1.25 \times 10^{-9}\)[/tex] C
- Charge of the second particle ([tex]\(q_2\)[/tex]) = [tex]\(1.92 \times 10^{-9}\)[/tex] C
- Original distance between the particles ([tex]\(r_{\text{original}}\)[/tex]) = 0.38 m
- Coulomb's constant ([tex]\(k\)[/tex]) = [tex]\(9.00 \times 10^9\)[/tex] N·m²/C²

2. Apply Coulomb's Law for the original distance:
Coulomb's Law states:
[tex]\[ F_e = \frac{k q_1 q_2}{r^2} \][/tex]
Substitute the given values for the original distance:
[tex]\[ F_{\text{original}} = \frac{(9.00 \times 10^9) \times (1.25 \times 10^{-9}) \times (1.92 \times 10^{-9})}{(0.38)^2} \][/tex]

3. Double the distance and calculate the new force:
The new distance ([tex]\(r_{\text{new}}\)[/tex]) = 2 \times 0.38 m = 0.76 m. Now, calculate the force with the new distance:
[tex]\[ F_{\text{new}} = \frac{k q_1 q_2}{(r_{\text{new}})^2} \][/tex]
Substitute 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]

4. Evaluate the forces:
- The original force calculated:
[tex]\[ F_{\text{original}} = 1.50 \times 10^{-7} \, \text{N} \][/tex]
- The new force calculated:
[tex]\[ F_{\text{new}} = 3.74 \times 10^{-8} \, \text{N} \][/tex]

The force between the particles when the distance is doubled is [tex]\(3.74 \times 10^{-8} \, \text{N}\)[/tex]. Therefore, the correct answer is:

A. [tex]\(3.74 \times 10^{-8} \, \text{N}\)[/tex]
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