Answer :
Certainly! Let's solve this problem step-by-step using Coulomb's law.
Coulomb's law states that the magnitude of the electric force [tex]\( F \)[/tex] between two point charges is given by:
[tex]\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \][/tex]
Where:
- [tex]\( k \)[/tex] is Coulomb's constant [tex]\((8.988 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2)\)[/tex]
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the magnitudes of the charges (for protons, [tex]\( q_1 = q_2 = 1.60 \times 10^{-19} \, \text{C} \)[/tex])
- [tex]\( r \)[/tex] is the distance between the charges
Let's substitute the given values into the formula to calculate the electric force:
1. Identify the constants and given values:
[tex]\[ k = 8.988 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \][/tex]
[tex]\[ q = 1.60 \times 10^{-19} \, \text{C} \quad (\text{each proton}) \][/tex]
[tex]\[ r = 1.00 \times 10^{-15} \, \text{m} \][/tex]
2. Substitute the values into Coulomb's law equation:
[tex]\[ F = \frac{k \cdot q^2}{r^2} \][/tex]
[tex]\[ F = \frac{8.988 \times 10^9 \cdot (1.60 \times 10^{-19})^2}{(1.00 \times 10^{-15})^2} \][/tex]
3. Calculate the numerator:
[tex]\[ (1.60 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \text{C}^2 \][/tex]
[tex]\[ 8.988 \times 10^9 \cdot 2.56 \times 10^{-38} = 2.299968 \times 10^{-28} \, \text{N} \cdot \text{m}^2 \][/tex]
4. Calculate the denominator:
[tex]\[ (1.00 \times 10^{-15})^2 = 1.00 \times 10^{-30} \, \text{m}^2 \][/tex]
5. Divide the numerator by the denominator:
[tex]\[ \frac{2.299968 \times 10^{-28} \, \text{N} \cdot \text{m}^2}{1.00 \times 10^{-30} \, \text{m}^2} = 229.9968 \, \text{N} \][/tex]
6. Consider significant figures (three significant figures):
[tex]\[ \approx 230.09 \, \text{N} \][/tex]
So, the electric force that the two protons exert on each other is approximately [tex]\( 230.09 \)[/tex] Newtons.
Therefore, the final answer is:
[tex]\[ F \approx 230.09 \, \text{N} \][/tex]
Coulomb's law states that the magnitude of the electric force [tex]\( F \)[/tex] between two point charges is given by:
[tex]\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \][/tex]
Where:
- [tex]\( k \)[/tex] is Coulomb's constant [tex]\((8.988 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2)\)[/tex]
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the magnitudes of the charges (for protons, [tex]\( q_1 = q_2 = 1.60 \times 10^{-19} \, \text{C} \)[/tex])
- [tex]\( r \)[/tex] is the distance between the charges
Let's substitute the given values into the formula to calculate the electric force:
1. Identify the constants and given values:
[tex]\[ k = 8.988 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \][/tex]
[tex]\[ q = 1.60 \times 10^{-19} \, \text{C} \quad (\text{each proton}) \][/tex]
[tex]\[ r = 1.00 \times 10^{-15} \, \text{m} \][/tex]
2. Substitute the values into Coulomb's law equation:
[tex]\[ F = \frac{k \cdot q^2}{r^2} \][/tex]
[tex]\[ F = \frac{8.988 \times 10^9 \cdot (1.60 \times 10^{-19})^2}{(1.00 \times 10^{-15})^2} \][/tex]
3. Calculate the numerator:
[tex]\[ (1.60 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \text{C}^2 \][/tex]
[tex]\[ 8.988 \times 10^9 \cdot 2.56 \times 10^{-38} = 2.299968 \times 10^{-28} \, \text{N} \cdot \text{m}^2 \][/tex]
4. Calculate the denominator:
[tex]\[ (1.00 \times 10^{-15})^2 = 1.00 \times 10^{-30} \, \text{m}^2 \][/tex]
5. Divide the numerator by the denominator:
[tex]\[ \frac{2.299968 \times 10^{-28} \, \text{N} \cdot \text{m}^2}{1.00 \times 10^{-30} \, \text{m}^2} = 229.9968 \, \text{N} \][/tex]
6. Consider significant figures (three significant figures):
[tex]\[ \approx 230.09 \, \text{N} \][/tex]
So, the electric force that the two protons exert on each other is approximately [tex]\( 230.09 \)[/tex] Newtons.
Therefore, the final answer is:
[tex]\[ F \approx 230.09 \, \text{N} \][/tex]
