Answer :
To find the electric force acting between two charges, we use Coulomb's law, which states:
[tex]\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \][/tex]
Where:
- \( F \) is the magnitude of the electric force.
- \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \).
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
- \( r \) is the distance between the charges.
Given values:
- \( q_1 = 0.0042 \, \text{C} \)
- \( q_2 = -0.0050 \, \text{C} \)
- \( r = 0.0030 \, \text{m} \)
Now, calculate the magnitude of the force:
1. Calculate the product of the charges:
[tex]\[ q_1 \times q_2 = 0.0042 \times -0.0050 = -0.000021 \, \text{C}^2 \][/tex]
2. Square the distance \( r \):
[tex]\[ r^2 = (0.0030)^2 = 0.000009 \, \text{m}^2 \][/tex]
3. Apply Coulomb's law:
[tex]\[ F = 8.99 \times 10^9 \times \frac{-0.000021}{0.000009} \][/tex]
4. Simplify the fraction:
[tex]\[ \frac{-0.000021}{0.000009} = -2.3333 \][/tex]
5. Continue with the calculation:
[tex]\[ F = 8.99 \times 10^9 \times (-2.3333) \][/tex]
[tex]\[ F \approx -2.1 \times 10^{10} \, \text{N} \][/tex]
The calculated electric force is approximately \( -2.1 \times 10^{10} \, \text{N} \). The negative sign indicates that the force is attractive since one of the charges is positive and the other is negative.
Therefore, the correct answer is:
C. [tex]\(-2.1 \times 10^{10} \, \text{N} \)[/tex]
[tex]\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \][/tex]
Where:
- \( F \) is the magnitude of the electric force.
- \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \).
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
- \( r \) is the distance between the charges.
Given values:
- \( q_1 = 0.0042 \, \text{C} \)
- \( q_2 = -0.0050 \, \text{C} \)
- \( r = 0.0030 \, \text{m} \)
Now, calculate the magnitude of the force:
1. Calculate the product of the charges:
[tex]\[ q_1 \times q_2 = 0.0042 \times -0.0050 = -0.000021 \, \text{C}^2 \][/tex]
2. Square the distance \( r \):
[tex]\[ r^2 = (0.0030)^2 = 0.000009 \, \text{m}^2 \][/tex]
3. Apply Coulomb's law:
[tex]\[ F = 8.99 \times 10^9 \times \frac{-0.000021}{0.000009} \][/tex]
4. Simplify the fraction:
[tex]\[ \frac{-0.000021}{0.000009} = -2.3333 \][/tex]
5. Continue with the calculation:
[tex]\[ F = 8.99 \times 10^9 \times (-2.3333) \][/tex]
[tex]\[ F \approx -2.1 \times 10^{10} \, \text{N} \][/tex]
The calculated electric force is approximately \( -2.1 \times 10^{10} \, \text{N} \). The negative sign indicates that the force is attractive since one of the charges is positive and the other is negative.
Therefore, the correct answer is:
C. [tex]\(-2.1 \times 10^{10} \, \text{N} \)[/tex]
