Answer :
To solve this problem, we will use Coulomb's Law, which describes the electric force between two charged particles. The formula for Coulomb's Law is:
[tex]\[ F = k \frac{q_1 q_2}{r^2} \][/tex]
Where:
- [tex]\( F \)[/tex] is the magnitude of the electric force between the charges.
- [tex]\( k \)[/tex] is Coulomb's constant, approximately [tex]\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)[/tex].
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the magnitudes of the charges.
- [tex]\( r \)[/tex] is the distance between the charges.
### Step 1: Calculate the Initial Force
First, we need to find the initial force when the distance between the particles is 44.0 mm (which is 0.044 meters).
Given:
- [tex]\( q_1 = 1.6 \times 10^{-19} \, \text{C} \)[/tex] (charge of particle 1)
- [tex]\( q_2 = 1.6 \times 10^{-19} \, \text{C} \)[/tex] (charge of particle 2)
- [tex]\( r = 0.044 \, \text{m} \)[/tex] (initial distance)
Now, plug these values into Coulomb's Law:
[tex]\[ F_{\text{initial}} = 8.99 \times 10^9 \, \frac{(1.6 \times 10^{-19})(1.6 \times 10^{-19})}{(0.044)^2} \][/tex]
### Step 2: Adjust the Distance by a Factor of 3
Next, we increase the distance by a factor of 3. This means the new distance [tex]\( r_{\text{new}} \)[/tex] is:
[tex]\[ r_{\text{new}} = 3 \times 0.044 = 0.132 \, \text{m} \][/tex]
### Step 3: Calculate the New Force
Now, we will calculate the new force with the new distance:
[tex]\[ F_{\text{new}} = 8.99 \times 10^9 \, \frac{(1.6 \times 10^{-19})(1.6 \times 10^{-19})}{(0.132)^2} \][/tex]
### Step 4: Compare the Forces
To determine the effect of increasing the distance on the electric force, we compare [tex]\( F_{\text{initial}} \)[/tex] and [tex]\( F_{\text{new}} \)[/tex].
Given that the initial force [tex]\( F_{\text{initial}} \)[/tex] is approximately [tex]\( 1.18876 \times 10^{-25} \, \text{N} \)[/tex] and the new force [tex]\( F_{\text{new}} \)[/tex] is approximately [tex]\( 1.32084 \times 10^{-26} \, \text{N} \)[/tex], we can observe the change.
### Conclusion
When the distance between the two charged particles is increased by a factor of 3, the electric force between them decreases. Specifically, the initial force was [tex]\( 1.18876 \times 10^{-25} \, \text{N} \)[/tex], and the new force at three times the initial distance is [tex]\( 1.32084 \times 10^{-26} \, \text{N} \)[/tex]. This shows that the force decreases by a significant factor when the distance is increased by three times.
[tex]\[ F = k \frac{q_1 q_2}{r^2} \][/tex]
Where:
- [tex]\( F \)[/tex] is the magnitude of the electric force between the charges.
- [tex]\( k \)[/tex] is Coulomb's constant, approximately [tex]\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)[/tex].
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the magnitudes of the charges.
- [tex]\( r \)[/tex] is the distance between the charges.
### Step 1: Calculate the Initial Force
First, we need to find the initial force when the distance between the particles is 44.0 mm (which is 0.044 meters).
Given:
- [tex]\( q_1 = 1.6 \times 10^{-19} \, \text{C} \)[/tex] (charge of particle 1)
- [tex]\( q_2 = 1.6 \times 10^{-19} \, \text{C} \)[/tex] (charge of particle 2)
- [tex]\( r = 0.044 \, \text{m} \)[/tex] (initial distance)
Now, plug these values into Coulomb's Law:
[tex]\[ F_{\text{initial}} = 8.99 \times 10^9 \, \frac{(1.6 \times 10^{-19})(1.6 \times 10^{-19})}{(0.044)^2} \][/tex]
### Step 2: Adjust the Distance by a Factor of 3
Next, we increase the distance by a factor of 3. This means the new distance [tex]\( r_{\text{new}} \)[/tex] is:
[tex]\[ r_{\text{new}} = 3 \times 0.044 = 0.132 \, \text{m} \][/tex]
### Step 3: Calculate the New Force
Now, we will calculate the new force with the new distance:
[tex]\[ F_{\text{new}} = 8.99 \times 10^9 \, \frac{(1.6 \times 10^{-19})(1.6 \times 10^{-19})}{(0.132)^2} \][/tex]
### Step 4: Compare the Forces
To determine the effect of increasing the distance on the electric force, we compare [tex]\( F_{\text{initial}} \)[/tex] and [tex]\( F_{\text{new}} \)[/tex].
Given that the initial force [tex]\( F_{\text{initial}} \)[/tex] is approximately [tex]\( 1.18876 \times 10^{-25} \, \text{N} \)[/tex] and the new force [tex]\( F_{\text{new}} \)[/tex] is approximately [tex]\( 1.32084 \times 10^{-26} \, \text{N} \)[/tex], we can observe the change.
### Conclusion
When the distance between the two charged particles is increased by a factor of 3, the electric force between them decreases. Specifically, the initial force was [tex]\( 1.18876 \times 10^{-25} \, \text{N} \)[/tex], and the new force at three times the initial distance is [tex]\( 1.32084 \times 10^{-26} \, \text{N} \)[/tex]. This shows that the force decreases by a significant factor when the distance is increased by three times.
