Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
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
[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
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
