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A negative charge, [tex]\( q_1 \)[/tex], of [tex]\( 6 \mu C \)[/tex] is [tex]\( 0.002 m \)[/tex] north of a positive charge, [tex]\( q_2 \)[/tex], of [tex]\( 3 \mu C \)[/tex]. What is the magnitude and direction of the electrical force, [tex]\( F_e \)[/tex], applied by [tex]\( q_1 \)[/tex] on [tex]\( q_2 \)[/tex]?

A. Magnitude: [tex]\( 8 \times 10^1 N \)[/tex]; Direction: south
B. Magnitude: [tex]\( 8 \times 10^1 N \)[/tex]; Direction: north
C. Magnitude: [tex]\( 4 \times 10^4 N \)[/tex]; Direction: south
D. Magnitude: [tex]\( 4 \times 10^4 N \)[/tex]; Direction: north

Sagot :

GinnyAnswer
To solve this problem, you need to apply Coulomb's law, which quantifies the electrostatic force between two point charges. Coulomb's law is given 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, [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 charges,
- [tex]\( r \)[/tex] is the distance between the charges.

Let's plug in the given values.

1. Convert the charges from microcoulombs (μC) to coulombs (C):
- [tex]\( q_1 = -6 \, \mu C = -6 \times 10^{-6} \, \text{C} \)[/tex]
- [tex]\( q_2 = 3 \, \mu C = 3 \times 10^{-6} \, \text{C} \)[/tex]

2. Substitute the known values into Coulomb's law:
- [tex]\( k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \)[/tex]
- [tex]\( r = 0.002 \, \text{m} \)[/tex]

[tex]\[ F = 8.99 \times 10^9 \cdot \frac{|-6 \times 10^{-6} \cdot 3 \times 10^{-6}|}{(0.002)^2} \][/tex]

3. Calculate the product of the charges:
[tex]\[ |-6 \times 10^{-6} \cdot 3 \times 10^{-6}| = 18 \times 10^{-12} \][/tex]

4. Calculate the distance squared:
[tex]\[ (0.002)^2 = 4 \times 10^{-6} \][/tex]

5. Substitute these values into the formula:
[tex]\[ F = 8.99 \times 10^9 \cdot \frac{18 \times 10^{-12}}{4 \times 10^{-6}} \][/tex]

6. Simplify the fraction:
[tex]\[ \frac{18 \times 10^{-12}}{4 \times 10^{-6}} = 4.5 \times 10^{-6} \][/tex]

7. Calculate the final force magnitude:
[tex]\[ F = 8.99 \times 10^9 \cdot 4.5 \times 10^{-6} = 40455 \, \text{N} \][/tex]

The magnitude of the force is approximately [tex]\( 40455 \, \text{N} \)[/tex].

To determine the direction:
- [tex]\( q_1 \)[/tex] is negative.
- [tex]\( q_2 \)[/tex] is positive.
- Since opposite charges attract, the force on [tex]\( q_2 \)[/tex] due to [tex]\( q_1 \)[/tex] will be towards [tex]\( q_1 \)[/tex]. Since [tex]\( q_1 \)[/tex] is north of [tex]\( q_2 \)[/tex], the force on [tex]\( q_2 \)[/tex] will be directed south.

Therefore, the correct answer is:

- Magnitude: [tex]\( 4 \times 10^4 \, \text{N} \)[/tex]
- Direction: South
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