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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, we will use Coulomb's law, which states that the magnitude of the electrical force between two point charges is given by the formula:

[tex]\[ F_e = k \frac{|q_1 q_2|}{r^2} \][/tex]

where:
- [tex]\( k \)[/tex] is Coulomb's constant, approximately [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 charges.

Given:
- [tex]\( q_1 = -6 \times 10^{-6} \, \text{C} \)[/tex] (negative charge),
- [tex]\( q_2 = 3 \times 10^{-6} \, \text{C} \)[/tex] (positive charge),
- [tex]\( r = 0.002 \, \text{m} \)[/tex].

First, let's substitute these values into Coulomb's law to find the magnitude of the electric force:

[tex]\[ F_e = k \frac{|q_1 q_2|}{r^2} \][/tex]

Substitute the constants and values of [tex]\( q_1 \)[/tex], [tex]\( q_2 \)[/tex], and [tex]\( r \)[/tex] into the formula:

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

Calculate the product of the charges [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex]:

[tex]\[ (-6 \times 10^{-6}) (3 \times 10^{-6}) = -18 \times 10^{-12} \][/tex]

Taking the absolute value, we get:

[tex]\[ |(-18 \times 10^{-12})| = 18 \times 10^{-12} \][/tex]

Now, calculate the square of the distance [tex]\( r^2 \)[/tex]:

[tex]\[ (0.002)^2 = 4 \times 10^{-6} \][/tex]

Now, substitute these values back into the formula:

[tex]\[ F_e = (8.99 \times 10^9) \frac{18 \times 10^{-12}}{4 \times 10^{-6}} \][/tex]

Next, simplify the fraction:

[tex]\[ \frac{18 \times 10^{-12}}{4 \times 10^{-6}} = 4.5 \times 10^{-6} \][/tex]

So, now calculate the force:

[tex]\[ F_e = (8.99 \times 10^9) (4.5 \times 10^{-6}) = 40455 \, \text{N} \][/tex]

Thus, the magnitude of the electrical force is:

[tex]\[ F_e = 40455 \, \text{N} \][/tex]

To determine the direction, we consider the signs of the charges:
- [tex]\( q_1 \)[/tex] is negative.
- [tex]\( q_2 \)[/tex] is positive.

Since opposite charges attract, the force direction will be towards [tex]\( q_1 \)[/tex]. Hence, the force on [tex]\( q_2 \)[/tex] is directed south (towards the location of [tex]\( q_1 \)[/tex]).

So, the correct answer is:

- Magnitude: [tex]\( 40455 \, \text{N} \)[/tex]
- Direction: South

So the answer is:
magnitude: [tex]\( 4 \times 10^4 \, N \)[/tex] direction: south
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