Answer :
To find the electric force acting between two charges, we use Coulomb's Law, which states that the electric force ([tex]\(F_e\)[/tex]) between two point charges is given by:
[tex]\[ F_e = \frac{k q_1 q_2}{r^2} \][/tex]
where:
- [tex]\( k \)[/tex] is Coulomb's constant ([tex]\(9.00 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)[/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.
Given:
- [tex]\( q_1 = -0.0085 \, \text{C} \)[/tex]
- [tex]\( q_2 = -0.0025 \, \text{C} \)[/tex]
- [tex]\( r = 0.0020 \, \text{m} \)[/tex]
Let's plug these values into the formula:
[tex]\[ F_e = \frac{(9.00 \times 10^9) (-0.0085) (-0.0025)}{(0.0020)^2} \][/tex]
First, we calculate the numerator:
[tex]\[ 9.00 \times 10^9 \times -0.0085 \times -0.0025 \][/tex]
The product of the charges ([tex]\(-0.0085 \times -0.0025\)[/tex]) is:
[tex]\[ -0.0085 \times -0.0025 = 2.125 \times 10^{-5} \][/tex]
Therefore, the numerator becomes:
[tex]\[ 9.00 \times 10^9 \times 2.125 \times 10^{-5} = 191.25 \times 10^4 = 1.9125 \times 10^6 \][/tex]
Next, we calculate the denominator:
[tex]\[ (0.0020)^2 = 4.0 \times 10^{-6} \][/tex]
Now, we divide the numerator by the denominator:
[tex]\[ F_e = \frac{1.9125 \times 10^6}{4.0 \times 10^{-6}} \][/tex]
Dividing the values:
[tex]\[ F_e = 47812500000.0 \][/tex]
So, the electric force is:
[tex]\[ F_e = 4.78125 \times 10^{10} \, \text{N} \][/tex]
Since charges are negative, and we are looking for the magnitude of force:
[tex]\[ F_e = 4.8 \times 10^{10} \, \text{N} \][/tex]
Thus, the correct answer is:
B. [tex]\( 4.8 \times 10^{10} \, \text{N} \)[/tex]
[tex]\[ F_e = \frac{k q_1 q_2}{r^2} \][/tex]
where:
- [tex]\( k \)[/tex] is Coulomb's constant ([tex]\(9.00 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)[/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.
Given:
- [tex]\( q_1 = -0.0085 \, \text{C} \)[/tex]
- [tex]\( q_2 = -0.0025 \, \text{C} \)[/tex]
- [tex]\( r = 0.0020 \, \text{m} \)[/tex]
Let's plug these values into the formula:
[tex]\[ F_e = \frac{(9.00 \times 10^9) (-0.0085) (-0.0025)}{(0.0020)^2} \][/tex]
First, we calculate the numerator:
[tex]\[ 9.00 \times 10^9 \times -0.0085 \times -0.0025 \][/tex]
The product of the charges ([tex]\(-0.0085 \times -0.0025\)[/tex]) is:
[tex]\[ -0.0085 \times -0.0025 = 2.125 \times 10^{-5} \][/tex]
Therefore, the numerator becomes:
[tex]\[ 9.00 \times 10^9 \times 2.125 \times 10^{-5} = 191.25 \times 10^4 = 1.9125 \times 10^6 \][/tex]
Next, we calculate the denominator:
[tex]\[ (0.0020)^2 = 4.0 \times 10^{-6} \][/tex]
Now, we divide the numerator by the denominator:
[tex]\[ F_e = \frac{1.9125 \times 10^6}{4.0 \times 10^{-6}} \][/tex]
Dividing the values:
[tex]\[ F_e = 47812500000.0 \][/tex]
So, the electric force is:
[tex]\[ F_e = 4.78125 \times 10^{10} \, \text{N} \][/tex]
Since charges are negative, and we are looking for the magnitude of force:
[tex]\[ F_e = 4.8 \times 10^{10} \, \text{N} \][/tex]
Thus, the correct answer is:
B. [tex]\( 4.8 \times 10^{10} \, \text{N} \)[/tex]
