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To calculate the electrostatic force between two charged water molecules, we use Coulomb's Law. Coulomb's Law describes the electrostatic force between two point charges and is given by the formula:
[tex]\[ F = k_e \frac{|q_1 \cdot q_2|}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the electrostatic force,
- [tex]\( k_e \)[/tex] is Coulomb's constant ([tex]\( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \)[/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.
Let's identify and plug in the values:
1. Given charges:
- [tex]\( q_1 = 6.4 \times 10^{-10} \, \text{C} \)[/tex]
- [tex]\( q_2 = 1.6 \times 10^{-10} \, \text{C} \)[/tex]
2. Distance between the charges:
- [tex]\( r = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m} \)[/tex]
Using Coulomb's Law:
[tex]\[ F = 8.99 \times 10^9 \times \frac{|6.4 \times 10^{-10} \cdot 1.6 \times 10^{-10}|}{(5 \times 10^{-3})^2} \][/tex]
Let's first calculate the numerator:
[tex]\[ |6.4 \times 10^{-10} \cdot 1.6 \times 10^{-10}| = 6.4 \times 1.6 \times 10^{-10} \times 10^{-10} = 10.24 \times 10^{-20} \, \text{C}^2 \][/tex]
Now, calculate the denominator:
[tex]\[ (5 \times 10^{-3})^2 = 25 \times 10^{-6} \, \text{m}^2 \][/tex]
Next, plug these values back into the formula:
[tex]\[ F = 8.99 \times 10^9 \times \frac{10.24 \times 10^{-20}}{25 \times 10^{-6}} \][/tex]
[tex]\[ F = 8.99 \times 10^9 \times \frac{10.24 \times 10^{-20}}{2.5 \times 10^{-5}} \][/tex]
Simplify the fraction:
[tex]\[ \frac{10.24 \times 10^{-20}}{2.5 \times 10^{-5}} = 4.096 \times 10^{-15} \][/tex]
Now, multiply by [tex]\( 8.99 \times 10^9 \)[/tex]:
[tex]\[ F = 8.99 \times 10^9 \times 4.096 \times 10^{-15} \][/tex]
[tex]\[ F = 3.682304 \times 10^{-5} \, \text{N} \][/tex]
Thus, the electrostatic force between the two water molecules is approximately:
[tex]\[ 3.682304 \times 10^{-5} \, \text{N} \][/tex]
[tex]\[ F = k_e \frac{|q_1 \cdot q_2|}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the electrostatic force,
- [tex]\( k_e \)[/tex] is Coulomb's constant ([tex]\( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \)[/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.
Let's identify and plug in the values:
1. Given charges:
- [tex]\( q_1 = 6.4 \times 10^{-10} \, \text{C} \)[/tex]
- [tex]\( q_2 = 1.6 \times 10^{-10} \, \text{C} \)[/tex]
2. Distance between the charges:
- [tex]\( r = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m} \)[/tex]
Using Coulomb's Law:
[tex]\[ F = 8.99 \times 10^9 \times \frac{|6.4 \times 10^{-10} \cdot 1.6 \times 10^{-10}|}{(5 \times 10^{-3})^2} \][/tex]
Let's first calculate the numerator:
[tex]\[ |6.4 \times 10^{-10} \cdot 1.6 \times 10^{-10}| = 6.4 \times 1.6 \times 10^{-10} \times 10^{-10} = 10.24 \times 10^{-20} \, \text{C}^2 \][/tex]
Now, calculate the denominator:
[tex]\[ (5 \times 10^{-3})^2 = 25 \times 10^{-6} \, \text{m}^2 \][/tex]
Next, plug these values back into the formula:
[tex]\[ F = 8.99 \times 10^9 \times \frac{10.24 \times 10^{-20}}{25 \times 10^{-6}} \][/tex]
[tex]\[ F = 8.99 \times 10^9 \times \frac{10.24 \times 10^{-20}}{2.5 \times 10^{-5}} \][/tex]
Simplify the fraction:
[tex]\[ \frac{10.24 \times 10^{-20}}{2.5 \times 10^{-5}} = 4.096 \times 10^{-15} \][/tex]
Now, multiply by [tex]\( 8.99 \times 10^9 \)[/tex]:
[tex]\[ F = 8.99 \times 10^9 \times 4.096 \times 10^{-15} \][/tex]
[tex]\[ F = 3.682304 \times 10^{-5} \, \text{N} \][/tex]
Thus, the electrostatic force between the two water molecules is approximately:
[tex]\[ 3.682304 \times 10^{-5} \, \text{N} \][/tex]
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