Answer :
To solve this problem, we need to understand that the electric force between two point charges is determined by Coulomb's Law, which states that the electric force [tex]\( F \)[/tex] between two charges is inversely proportional to the square of the distance [tex]\( r \)[/tex] between them:
[tex]\[ F \propto \frac{1}{r^2} \][/tex]
Given:
1. The initial electric force [tex]\( F_1 \)[/tex] is [tex]\( 9.50 \, \mu \text{N} \)[/tex] (or [tex]\( 9.50 \times 10^{-6} \, \text{N} \)[/tex]).
2. The initial distance [tex]\( r_1 \)[/tex] is [tex]\( 44.1 \, \text{cm} \)[/tex] (or [tex]\( 0.441 \, \text{m} \)[/tex]).
3. The new distance [tex]\( r_2 \)[/tex] is [tex]\( 55.3 \, \text{cm} \)[/tex] (or [tex]\( 0.553 \, \text{m} \)[/tex]).
We need to find the new electric force [tex]\( F_2 \)[/tex] when the distance between the charges changes.
### Step-by-Step Solution:
1. Establish Coulomb's Law Relationship:
Coulomb's Law indicates that the force is inversely proportional to the square of the separation distance:
[tex]\[ F \propto \frac{1}{r^2} \][/tex]
2. Write the Ratio of Forces:
The ratio of the new force [tex]\( F_2 \)[/tex] to the initial force [tex]\( F_1 \)[/tex] can be expressed in terms of the distances:
[tex]\[ \frac{F_2}{F_1} = \left( \frac{r_1}{r_2} \right)^2 \][/tex]
3. Substitute the Given Values:
Let’s substitute the initial and new distances into the ratio:
[tex]\[ \frac{F_2}{F_1} = \left( \frac{0.441 \, \text{m}}{0.553 \, \text{m}} \right)^2 \][/tex]
4. Calculate the Ratio:
[tex]\[ \frac{0.441}{0.553} \approx 0.797 \][/tex]
[tex]\[ \left(0.797\right)^2 \approx 0.636 \][/tex]
So:
[tex]\[ \frac{F_2}{F_1} \approx 0.636 \][/tex]
5. Determine the New Force:
Now, multiply the initial force [tex]\( F_1 \)[/tex] by the ratio to find the new force [tex]\( F_2 \)[/tex]:
[tex]\[ F_2 = F_1 \times 0.636 \][/tex]
[tex]\[ F_2 = 9.50 \times 10^{-6} \, \text{N} \times 0.636 \][/tex]
[tex]\[ F_2 \approx 6.04 \times 10^{-6} \, \text{N} \][/tex]
[tex]\[ F_2 \approx 6.04 \, \mu \text{N} \][/tex]
### Conclusion:
After the charges are moved apart to a distance of [tex]\( 55.3 \, \text{cm} \)[/tex], the magnitude of the electric force they exert on each other becomes approximately [tex]\( 6.04 \, \mu \text{N} \)[/tex].
[tex]\[ F \propto \frac{1}{r^2} \][/tex]
Given:
1. The initial electric force [tex]\( F_1 \)[/tex] is [tex]\( 9.50 \, \mu \text{N} \)[/tex] (or [tex]\( 9.50 \times 10^{-6} \, \text{N} \)[/tex]).
2. The initial distance [tex]\( r_1 \)[/tex] is [tex]\( 44.1 \, \text{cm} \)[/tex] (or [tex]\( 0.441 \, \text{m} \)[/tex]).
3. The new distance [tex]\( r_2 \)[/tex] is [tex]\( 55.3 \, \text{cm} \)[/tex] (or [tex]\( 0.553 \, \text{m} \)[/tex]).
We need to find the new electric force [tex]\( F_2 \)[/tex] when the distance between the charges changes.
### Step-by-Step Solution:
1. Establish Coulomb's Law Relationship:
Coulomb's Law indicates that the force is inversely proportional to the square of the separation distance:
[tex]\[ F \propto \frac{1}{r^2} \][/tex]
2. Write the Ratio of Forces:
The ratio of the new force [tex]\( F_2 \)[/tex] to the initial force [tex]\( F_1 \)[/tex] can be expressed in terms of the distances:
[tex]\[ \frac{F_2}{F_1} = \left( \frac{r_1}{r_2} \right)^2 \][/tex]
3. Substitute the Given Values:
Let’s substitute the initial and new distances into the ratio:
[tex]\[ \frac{F_2}{F_1} = \left( \frac{0.441 \, \text{m}}{0.553 \, \text{m}} \right)^2 \][/tex]
4. Calculate the Ratio:
[tex]\[ \frac{0.441}{0.553} \approx 0.797 \][/tex]
[tex]\[ \left(0.797\right)^2 \approx 0.636 \][/tex]
So:
[tex]\[ \frac{F_2}{F_1} \approx 0.636 \][/tex]
5. Determine the New Force:
Now, multiply the initial force [tex]\( F_1 \)[/tex] by the ratio to find the new force [tex]\( F_2 \)[/tex]:
[tex]\[ F_2 = F_1 \times 0.636 \][/tex]
[tex]\[ F_2 = 9.50 \times 10^{-6} \, \text{N} \times 0.636 \][/tex]
[tex]\[ F_2 \approx 6.04 \times 10^{-6} \, \text{N} \][/tex]
[tex]\[ F_2 \approx 6.04 \, \mu \text{N} \][/tex]
### Conclusion:
After the charges are moved apart to a distance of [tex]\( 55.3 \, \text{cm} \)[/tex], the magnitude of the electric force they exert on each other becomes approximately [tex]\( 6.04 \, \mu \text{N} \)[/tex].
