Answer :
To determine the electrical force between charges [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex], we utilize Coulomb's Law. Coulomb's Law states that the magnitude of the electrical force [tex]\( F \)[/tex] between two charges is given by:
[tex]\[ F = k \frac{|q_1 q_2|}{r^2} \][/tex]
Here:
- [tex]\( k \)[/tex] is the Coulomb constant, [tex]\( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)[/tex]
- [tex]\( q_1 \)[/tex] is the magnitude of the first charge, [tex]\( q_1 = 6 \, \text{C} \)[/tex]
- [tex]\( q_2 \)[/tex] is the magnitude of the second charge, [tex]\( q_2 = -4 \, \text{C} \)[/tex]
- [tex]\( r \)[/tex] is the distance between the charges, [tex]\( r = 3 \, \text{m} \)[/tex]
First, we note that Coulomb's Law uses the absolute value of the product of charges, thus:
[tex]\[ |q_1 q_2| = |6 \cdot (-4)| = | -24 | = 24 \, \text{C}^2 \][/tex]
Next, we substitute these values into the equation:
[tex]\[ F = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \cdot \frac{24 \, \text{C}^2}{(3 \, \text{m})^2} \][/tex]
We calculate the denominator first:
[tex]\[ r^2 = (3 \, \text{m})^2 = 9 \, \text{m}^2 \][/tex]
Now, substituting this back into the equation:
[tex]\[ F = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \cdot \frac{24 \, \text{C}^2}{9 \, \text{m}^2} \][/tex]
Simplify the fraction:
[tex]\[ \frac{24 \, \text{C}^2}{9 \, \text{m}^2} = \frac{24}{9} = 2.6667 \, \text{C}^2/\text{m}^2 \][/tex]
Then:
[tex]\[ F = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \cdot 2.6667 \, \text{C}^2/\text{m}^2 \][/tex]
Multiplying these together gives us:
[tex]\[ F \approx 2.3973333333 \times 10^{10} \, \text{N} \][/tex]
Therefore, the electrical force between [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] is approximately [tex]\( 2.397 \times 10^{10} \, \text{N} \)[/tex], which can be written as:
[tex]\[ F \approx 23973333333.333332 \, \text{N} \][/tex]
Among the given options, this result does not exactly match any provided choices, so there might be a mistake or a different context in the options list. However, the calculated value stands as accurate for the data given.
[tex]\[ F = k \frac{|q_1 q_2|}{r^2} \][/tex]
Here:
- [tex]\( k \)[/tex] is the Coulomb constant, [tex]\( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)[/tex]
- [tex]\( q_1 \)[/tex] is the magnitude of the first charge, [tex]\( q_1 = 6 \, \text{C} \)[/tex]
- [tex]\( q_2 \)[/tex] is the magnitude of the second charge, [tex]\( q_2 = -4 \, \text{C} \)[/tex]
- [tex]\( r \)[/tex] is the distance between the charges, [tex]\( r = 3 \, \text{m} \)[/tex]
First, we note that Coulomb's Law uses the absolute value of the product of charges, thus:
[tex]\[ |q_1 q_2| = |6 \cdot (-4)| = | -24 | = 24 \, \text{C}^2 \][/tex]
Next, we substitute these values into the equation:
[tex]\[ F = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \cdot \frac{24 \, \text{C}^2}{(3 \, \text{m})^2} \][/tex]
We calculate the denominator first:
[tex]\[ r^2 = (3 \, \text{m})^2 = 9 \, \text{m}^2 \][/tex]
Now, substituting this back into the equation:
[tex]\[ F = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \cdot \frac{24 \, \text{C}^2}{9 \, \text{m}^2} \][/tex]
Simplify the fraction:
[tex]\[ \frac{24 \, \text{C}^2}{9 \, \text{m}^2} = \frac{24}{9} = 2.6667 \, \text{C}^2/\text{m}^2 \][/tex]
Then:
[tex]\[ F = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \cdot 2.6667 \, \text{C}^2/\text{m}^2 \][/tex]
Multiplying these together gives us:
[tex]\[ F \approx 2.3973333333 \times 10^{10} \, \text{N} \][/tex]
Therefore, the electrical force between [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] is approximately [tex]\( 2.397 \times 10^{10} \, \text{N} \)[/tex], which can be written as:
[tex]\[ F \approx 23973333333.333332 \, \text{N} \][/tex]
Among the given options, this result does not exactly match any provided choices, so there might be a mistake or a different context in the options list. However, the calculated value stands as accurate for the data given.
