Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
To solve this problem, let's start by understanding the concept and using Coulomb's Law.
### Step-by-Step Solution:
1. Identify the charges and their values:
- Charge \( q_1 = 7.0 \mu C \) (microcoulombs).
- Charge \( q_2 = -4.5 \mu C \) (microcoulombs).
2. Convert the charges to Coulombs:
- \( q_1 = 7.0 \times 10^{-6} \) C.
- \( q_2 = -4.5 \times 10^{-6} \) C.
3. Determine the distance between the charges:
- The distance is given as \( 50 \) cm.
- Convert the distance to meters (since SI units should be used):
[tex]\[ r = \frac{50}{100} = 0.50 \text{ meters} \][/tex]
4. Coulomb's constant (\( k \)):
- Coulomb's constant \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \).
5. Apply Coulomb's Law to find the magnitude of the force:
- Coulomb's Law formula:
[tex]\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \][/tex]
- Substitute the known values:
[tex]\[ F = 8.99 \times 10^9 \, \frac{7.0 \times 10^{-6} \times 4.5 \times 10^{-6}}{(0.50)^2} \][/tex]
- Doing the necessary calculations, we will get:
[tex]\[ F \approx 1.13274 \text{ N} \][/tex]
So, the magnitude of the force is approximately \( 1.13274 \) Newtons.
6. Determine whether the force is attractive or repulsive:
- Since one charge is positive (\( q_1 = +7.0 \mu C \)) and one charge is negative (\( q_2 = -4.5 \mu C \)), the force between them is attractive. Opposite charges attract each other.
### Final Answer:
- Magnitude of the force: Approximately \( 1.13274 \) Newtons.
- Direction of the force: The force is attractive, meaning the charges are pulling towards each other.
Therefore, the [tex]\( +7.0 \mu C \)[/tex] charge experiences an attractive force of approximately [tex]\( 1.13274 \)[/tex] N towards the [tex]\( -4.5 \mu C \)[/tex] charge.
### Step-by-Step Solution:
1. Identify the charges and their values:
- Charge \( q_1 = 7.0 \mu C \) (microcoulombs).
- Charge \( q_2 = -4.5 \mu C \) (microcoulombs).
2. Convert the charges to Coulombs:
- \( q_1 = 7.0 \times 10^{-6} \) C.
- \( q_2 = -4.5 \times 10^{-6} \) C.
3. Determine the distance between the charges:
- The distance is given as \( 50 \) cm.
- Convert the distance to meters (since SI units should be used):
[tex]\[ r = \frac{50}{100} = 0.50 \text{ meters} \][/tex]
4. Coulomb's constant (\( k \)):
- Coulomb's constant \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \).
5. Apply Coulomb's Law to find the magnitude of the force:
- Coulomb's Law formula:
[tex]\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \][/tex]
- Substitute the known values:
[tex]\[ F = 8.99 \times 10^9 \, \frac{7.0 \times 10^{-6} \times 4.5 \times 10^{-6}}{(0.50)^2} \][/tex]
- Doing the necessary calculations, we will get:
[tex]\[ F \approx 1.13274 \text{ N} \][/tex]
So, the magnitude of the force is approximately \( 1.13274 \) Newtons.
6. Determine whether the force is attractive or repulsive:
- Since one charge is positive (\( q_1 = +7.0 \mu C \)) and one charge is negative (\( q_2 = -4.5 \mu C \)), the force between them is attractive. Opposite charges attract each other.
### Final Answer:
- Magnitude of the force: Approximately \( 1.13274 \) Newtons.
- Direction of the force: The force is attractive, meaning the charges are pulling towards each other.
Therefore, the [tex]\( +7.0 \mu C \)[/tex] charge experiences an attractive force of approximately [tex]\( 1.13274 \)[/tex] N towards the [tex]\( -4.5 \mu C \)[/tex] charge.
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.