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A negative charge, [tex]$q_1$[/tex], of [tex]$6 \mu C$[/tex] is 0.002 m north of a positive charge, [tex][tex]$q_2$[/tex][/tex]. What is the magnitude and direction of the electrical force, [tex]$F_e$[/tex], applied by [tex]$q_1$[/tex] on [tex][tex]$q_2$[/tex][/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][tex]$4 \times 10^4 N$[/tex][/tex], direction: south
D. magnitude: [tex]$4 \times 10^4 N$[/tex], direction: north



Answer :

GinnyAnswer
Let's solve this problem step-by-step using Coulomb's law, which describes the force between two point charges.

1. Identify the given data:
- Charge [tex]\( q_1 \)[/tex] = [tex]\( -6 \mu C \)[/tex] (negative charge)
- Charge [tex]\( q_2 \)[/tex] = [tex]\( +3 \mu C \)[/tex] (positive charge)
- Distance [tex]\( r \)[/tex] = 0.002 m (2 mm)

2. Convert the charges to Coulombs:
[tex]\[ q_1 = -6 \times 10^{-6} \text{ C} \][/tex]
[tex]\[ q_2 = 3 \times 10^{-6} \text{ C} \][/tex]

3. Coulomb's constant [tex]\( k \)[/tex]:
[tex]\[ k = 8.99 \times 10^9 \text{ N·m}^2/\text{C}^2 \][/tex]

4. Apply Coulomb's law to find the magnitude of the electrical force:
[tex]\[ F_e = k \frac{|q_1 q_2|}{r^2} \][/tex]

5. Substitute the given values into the equation:
[tex]\[ F_e = (8.99 \times 10^9) \frac{|(-6 \times 10^{-6})(3 \times 10^{-6})|}{(0.002)^2} \][/tex]

6. Calculate the absolute product of charges:
[tex]\[ |q_1 q_2| = |(-6 \times 10^{-6}) \times (3 \times 10^{-6})| \][/tex]
[tex]\[ |q_1 q_2| = 18 \times 10^{-12} \text{ C}^2 \][/tex]

7. Calculate the square of the distance:
[tex]\[ r^2 = (0.002)^2 \][/tex]
[tex]\[ r^2 = 4 \times 10^{-6} \text{ m}^2 \][/tex]

8. Calculate the force:
[tex]\[ F_e = (8.99 \times 10^9) \frac{18 \times 10^{-12}}{4 \times 10^{-6}} \][/tex]
[tex]\[ F_e = (8.99 \times 10^9) \times (4.5 \times 10^{-6}) \][/tex]
[tex]\[ F_e = 40455 \text{ N} \][/tex]

So the magnitude of the electrical force [tex]\( F_e \)[/tex] is:
[tex]\[ F_e \approx 40455 \text{ N} \][/tex]

9. Determine the direction of the force:
Since [tex]\( q_1 \)[/tex] is a negative charge and [tex]\( q_2 \)[/tex] is a positive charge, the force between them will be attractive. This means [tex]\( q_1 \)[/tex] will exert a force on [tex]\( q_2 \)[/tex] pulling it towards [tex]\( q_1 \)[/tex].

Given that [tex]\( q_1 \)[/tex] is north of [tex]\( q_2 \)[/tex], the force on [tex]\( q_2 \)[/tex] will be directed towards [tex]\( q_1 \)[/tex], which is south.

10. State the solution:
- Magnitude of the force: [tex]\( 4 \times 10^4 \)[/tex] N
- Direction of the force: south

Therefore, the correct answer is:

[tex]\[ \text{magnitude: } 4 \times 10^4 \text{ N} \][/tex]
[tex]\[ \text{direction: } \text{south} \][/tex]

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