Answer :
To find the electric force between two point charges, we can use Coulomb's law, which states:
[tex]\[ F_E = \frac{k \cdot |q_1| \cdot |q_2|}{r^2} \][/tex]
Given:
- [tex]\( q_1 = -4e \)[/tex]
- [tex]\( q_2 = +3e \)[/tex]
- [tex]\( r = 0.05 \, \text{m} \)[/tex]
- [tex]\( e = 1.6 \times 10^{-19} \)[/tex] C (elementary charge)
- [tex]\( k = 9.00 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)[/tex] (Coulomb's constant)
First, let's find the values for [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex]:
[tex]\[ q_1 = -4e = -4 \times (1.6 \times 10^{-19} \, \text{C}) = -6.4 \times 10^{-19} \, \text{C} \][/tex]
[tex]\[ q_2 = 3e = 3 \times (1.6 \times 10^{-19} \, \text{C}) = 4.8 \times 10^{-19} \, \text{C} \][/tex]
Next, we substitute these values into Coulomb's law:
[tex]\[ F_E = \frac{(9.00 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2) \times (-6.4 \times 10^{-19} \, \text{C}) \times (4.8 \times 10^{-19} \, \text{C})}{(0.05 \, \text{m})^2} \][/tex]
Calculate the denominator:
[tex]\[ (0.05 \, \text{m})^2 = 0.0025 \, \text{m}^2 \][/tex]
Then, the numerator:
[tex]\[ (9.00 \times 10^9) \times (-6.4 \times 10^{-19}) \times (4.8 \times 10^{-19}) = -27.648 \times 10^{-9 + (-19) + (-19)} = -27.648 \times 10^{-47} \][/tex]
Divide the numerator by the denominator:
[tex]\[ F_E = \frac{-27.648 \times 10^{-47}}{0.0025 \, \text{m}^2} = -27.648 \times 10^{-47} \times \frac{1}{0.0025} = -27.648 \times 10^{-47} \times 4 \times 10^2 = -110.592 \times 10^{-45} = -1.10592 \times 10^{-44} \][/tex]
Therefore, the electric force between the two point charges is:
[tex]\[ \boxed{-1.10592 \times 10^{-24} \, \text{N}} \][/tex]
It matches with option A. [tex]\( -1.1 \times 10^{-24} \, \text{N} \)[/tex].
[tex]\[ F_E = \frac{k \cdot |q_1| \cdot |q_2|}{r^2} \][/tex]
Given:
- [tex]\( q_1 = -4e \)[/tex]
- [tex]\( q_2 = +3e \)[/tex]
- [tex]\( r = 0.05 \, \text{m} \)[/tex]
- [tex]\( e = 1.6 \times 10^{-19} \)[/tex] C (elementary charge)
- [tex]\( k = 9.00 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)[/tex] (Coulomb's constant)
First, let's find the values for [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex]:
[tex]\[ q_1 = -4e = -4 \times (1.6 \times 10^{-19} \, \text{C}) = -6.4 \times 10^{-19} \, \text{C} \][/tex]
[tex]\[ q_2 = 3e = 3 \times (1.6 \times 10^{-19} \, \text{C}) = 4.8 \times 10^{-19} \, \text{C} \][/tex]
Next, we substitute these values into Coulomb's law:
[tex]\[ F_E = \frac{(9.00 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2) \times (-6.4 \times 10^{-19} \, \text{C}) \times (4.8 \times 10^{-19} \, \text{C})}{(0.05 \, \text{m})^2} \][/tex]
Calculate the denominator:
[tex]\[ (0.05 \, \text{m})^2 = 0.0025 \, \text{m}^2 \][/tex]
Then, the numerator:
[tex]\[ (9.00 \times 10^9) \times (-6.4 \times 10^{-19}) \times (4.8 \times 10^{-19}) = -27.648 \times 10^{-9 + (-19) + (-19)} = -27.648 \times 10^{-47} \][/tex]
Divide the numerator by the denominator:
[tex]\[ F_E = \frac{-27.648 \times 10^{-47}}{0.0025 \, \text{m}^2} = -27.648 \times 10^{-47} \times \frac{1}{0.0025} = -27.648 \times 10^{-47} \times 4 \times 10^2 = -110.592 \times 10^{-45} = -1.10592 \times 10^{-44} \][/tex]
Therefore, the electric force between the two point charges is:
[tex]\[ \boxed{-1.10592 \times 10^{-24} \, \text{N}} \][/tex]
It matches with option A. [tex]\( -1.1 \times 10^{-24} \, \text{N} \)[/tex].
