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Object [tex]$A$[/tex] has a positive charge of [tex]$4.0 \times 10^{-6} C$[/tex]. Object [tex]$B$[/tex] has a positive charge of [tex]$2.0 \times 10^{-6} C$[/tex]. If the distance between [tex]$A$[/tex] and [tex]$B$[/tex] is [tex]$0.04 \, m$[/tex], what is the force on [tex]$A$[/tex]?

Note: the constant of proportionality is [tex]$9.0 \times 10^9 \, N \cdot m^2 / C^2$[/tex].

A. [tex]$180 \, N$[/tex]
B. [tex]$45 \, N$[/tex]
C. [tex]$90 \, N$[/tex]
D. [tex]$360 \, N$[/tex]


Sagot :

To solve for the force between two charged objects using Coulomb's law, we can follow these steps:

1. Identify the given values:
- Charge of object \( A \), \( q_1 = 4.0 \times 10^{-6} \, C \)
- Charge of object \( B \), \( q_2 = 2.0 \times 10^{-6} \, C \)
- Distance between the objects, \( r = 0.04 \, m \)
- Coulomb's constant, \( k = 9.0 \times 10^9 \, N \cdot m^2 / C^2 \)

2. Write down Coulomb's law:
[tex]\[ F = k \cdot \frac{q_1 \cdot q_2}{r^2} \][/tex]
Where:
- \( F \) is the force between the two charges
- \( k \) is Coulomb's constant
- \( q_1 \) and \( q_2 \) are the magnitudes of the two charges
- \( r \) is the distance between the charges

3. Substitute the given values into Coulomb's law:
[tex]\[ F = 9.0 \times 10^9 \, \frac{N \cdot m^2}{C^2} \cdot \frac{(4.0 \times 10^{-6} \, C) \cdot (2.0 \times 10^{-6} \, C)}{(0.04 \, m)^2} \][/tex]

4. Calculate the product of the charges:
[tex]\[ q_1 \cdot q_2 = (4.0 \times 10^{-6}) \cdot (2.0 \times 10^{-6}) = 8.0 \times 10^{-12} \, C^2 \][/tex]

5. Calculate the square of the distance:
[tex]\[ r^2 = (0.04 \, m)^2 = 0.0016 \, m^2 \][/tex]

6. Combine the values and solve for \( F \):
[tex]\[ F = 9.0 \times 10^9 \, \frac{N \cdot m^2}{C^2} \cdot \frac{8.0 \times 10^{-12} \, C^2}{0.0016 \, m^2} \][/tex]

7. Divide the product of the charges by the distance squared:
[tex]\[ \frac{8.0 \times 10^{-12} \, C^2}{0.0016 \, m^2} = 5.0 \times 10^{-9} \, C^2 / m^2 \][/tex]

8. Multiply by Coulomb’s constant:
[tex]\[ F = 9.0 \times 10^9 \cdot 5.0 \times 10^{-9} = 45 \, N \][/tex]

Therefore, the force on \( A \) is \( 45 \, N \). The correct answer is:
[tex]\[ \boxed{45 \, N} \][/tex]
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