Answer :
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.
