To determine the force [tex]\( F \)[/tex] between two charges, [tex]\( Q_1 \)[/tex] and [tex]\( Q_2 \)[/tex], separated by a distance [tex]\( r \)[/tex], we use Coulomb's law. The formula for Coulomb's law is:
[tex]\[ F = k \frac{Q_1 Q_2}{r^2} \][/tex]
Where:
- [tex]\( k \)[/tex] is Coulomb's constant, [tex]\( 9.00 \times 10^9 \, \text{N} \cdot \frac{\text{m}^2}{\text{C}^2} \)[/tex]
- [tex]\( Q_1 \)[/tex] is the first charge, [tex]\( 3.0 \times 10^{-5} \, \text{C} \)[/tex]
- [tex]\( Q_2 \)[/tex] is the second charge, [tex]\( 4.0 \times 10^{-5} \, \text{C} \)[/tex]
- [tex]\( r \)[/tex] is the distance between the charges, [tex]\( 3.0 \, \text{m} \)[/tex]
Let's substitute these values into the formula:
[tex]\[ F = \left(9.00 \times 10^9 \, \text{N} \cdot \frac{\text{m}^2}{\text{C}^2} \right) \frac{(3.0 \times 10^{-5} \, \text{C})(4.0 \times 10^{-5} \, \text{C})}{(3.0 \, \text{m})^2} \][/tex]
First, let's calculate the product of the charges:
[tex]\[ Q_1 \times Q_2 = (3.0 \times 10^{-5} \, \text{C})(4.0 \times 10^{-5} \, \text{C}) = 12.0 \times 10^{-10} \, \text{C}^2 \][/tex]
Next, calculate the square of the distance:
[tex]\[ r^2 = (3.0 \, \text{m})^2 = 9.0 \, \text{m}^2 \][/tex]
Now, substitute these results back into the formula:
[tex]\[ F = \left(9.00 \times 10^9 \, \text{N} \cdot \frac{\text{m}^2}{\text{C}^2} \right) \frac{12.0 \times 10^{-10} \, \text{C}^2}{9.0 \, \text{m}^2} \][/tex]
Perform the division inside the fraction:
[tex]\[ \frac{12.0 \times 10^{-10} \, \text{C}^2}{9.0 \, \text{m}^2} = 1.333... \times 10^{-10} \, \text{C}^2/\text{m}^2 \approx 1.34 \times 10^{-10} \, \text{C}^2/\text{m}^2 \][/tex]
Then multiply by Coulomb's constant [tex]\( k \)[/tex]:
[tex]\[ F = (9.00 \times 10^9) \times 1.34 \times 10^{-10} = 1.206 \times 10^0 \, \text{N} \approx 1.2 \, \text{N} \][/tex]
So, the force [tex]\( F \)[/tex] between the two charges, with the provided values, is approximately [tex]\( 1.2 \, \text{N} \)[/tex]. From the given options, this matches closest to:
[tex]\[
\boxed{1.0 \, \text{N}}
\][/tex]
