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Sagot :
Certainly! Let's step through the solution to find the electric force exerted between two protons located [tex]\(1.00 \times 10^{-15}\)[/tex] meters apart.
We will use Coulomb's law, which states that the electric force ([tex]\( F \)[/tex]) between two point charges is given by:
[tex]\[ F = k \frac{|q_1 q_2|}{r^2} \][/tex]
Where:
- [tex]\( q_1 \)[/tex] is the charge of the first proton,
- [tex]\( q_2 \)[/tex] is the charge of the second proton,
- [tex]\( r \)[/tex] is the distance between the protons,
- [tex]\( k \)[/tex] is Coulomb's constant ([tex]\(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2\)[/tex]).
Now, let's plug in the given values:
1. The charge of a proton [tex]\( q = 1.60 \times 10^{-19} \)[/tex] Coulombs.
2. The distance between the protons [tex]\( r = 1.00 \times 10^{-15} \)[/tex] meters.
3. Coulomb's constant [tex]\( k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)[/tex].
Step-by-Step Calculation:
1. Identify the charges:
[tex]\( q_1 = 1.60 \times 10^{-19} \)[/tex] C and [tex]\( q_2 = 1.60 \times 10^{-19} \)[/tex] C.
2. Identify the distance between the charges:
[tex]\( r = 1.00 \times 10^{-15} \)[/tex] meters.
3. Apply Coulomb's law:
[tex]\[ F = k \frac{|q_1 q_2|}{r^2} \][/tex]
4. Plug in the values for [tex]\( q_1, q_2, \)[/tex], and [tex]\( r \)[/tex]:
[tex]\[ F = 8.99 \times 10^9 \frac{(1.60 \times 10^{-19})^2}{(1.00 \times 10^{-15})^2} \][/tex]
5. Compute the squared terms:
[tex]\[ q_1 \times q_2 = (1.60 \times 10^{-19}) \times (1.60 \times 10^{-19}) = 2.56 \times 10^{-38} \, \text{C}^2 \][/tex]
[tex]\[ r^2 = (1.00 \times 10^{-15})^2 = 1.00 \times 10^{-30} \, \text{m}^2 \][/tex]
6. Now, compute the electric force:
[tex]\[ F = 8.99 \times 10^9 \frac{2.56 \times 10^{-38}}{1.00 \times 10^{-30}} \][/tex]
7. Simplify the fraction:
[tex]\[ \frac{2.56 \times 10^{-38}}{1.00 \times 10^{-30}} = 2.56 \times 10^{-8} \][/tex]
8. Multiply by Coulomb's constant:
[tex]\[ F = 8.99 \times 10^9 \times 2.56 \times 10^{-8} \][/tex]
[tex]\[ F = 8.99 \times 2.56 \times 10^1 \][/tex]
[tex]\[ F = 230.14399999999995 \, \text{N} \][/tex]
So, the electric force that the two protons exert on each other is approximately:
[tex]\[ F \approx 230.14 \, \text{N} \][/tex]
Therefore, the electric force between the two protons is [tex]\( \boxed{230.14 \, \text{N}} \)[/tex].
We will use Coulomb's law, which states that the electric force ([tex]\( F \)[/tex]) between two point charges is given by:
[tex]\[ F = k \frac{|q_1 q_2|}{r^2} \][/tex]
Where:
- [tex]\( q_1 \)[/tex] is the charge of the first proton,
- [tex]\( q_2 \)[/tex] is the charge of the second proton,
- [tex]\( r \)[/tex] is the distance between the protons,
- [tex]\( k \)[/tex] is Coulomb's constant ([tex]\(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2\)[/tex]).
Now, let's plug in the given values:
1. The charge of a proton [tex]\( q = 1.60 \times 10^{-19} \)[/tex] Coulombs.
2. The distance between the protons [tex]\( r = 1.00 \times 10^{-15} \)[/tex] meters.
3. Coulomb's constant [tex]\( k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)[/tex].
Step-by-Step Calculation:
1. Identify the charges:
[tex]\( q_1 = 1.60 \times 10^{-19} \)[/tex] C and [tex]\( q_2 = 1.60 \times 10^{-19} \)[/tex] C.
2. Identify the distance between the charges:
[tex]\( r = 1.00 \times 10^{-15} \)[/tex] meters.
3. Apply Coulomb's law:
[tex]\[ F = k \frac{|q_1 q_2|}{r^2} \][/tex]
4. Plug in the values for [tex]\( q_1, q_2, \)[/tex], and [tex]\( r \)[/tex]:
[tex]\[ F = 8.99 \times 10^9 \frac{(1.60 \times 10^{-19})^2}{(1.00 \times 10^{-15})^2} \][/tex]
5. Compute the squared terms:
[tex]\[ q_1 \times q_2 = (1.60 \times 10^{-19}) \times (1.60 \times 10^{-19}) = 2.56 \times 10^{-38} \, \text{C}^2 \][/tex]
[tex]\[ r^2 = (1.00 \times 10^{-15})^2 = 1.00 \times 10^{-30} \, \text{m}^2 \][/tex]
6. Now, compute the electric force:
[tex]\[ F = 8.99 \times 10^9 \frac{2.56 \times 10^{-38}}{1.00 \times 10^{-30}} \][/tex]
7. Simplify the fraction:
[tex]\[ \frac{2.56 \times 10^{-38}}{1.00 \times 10^{-30}} = 2.56 \times 10^{-8} \][/tex]
8. Multiply by Coulomb's constant:
[tex]\[ F = 8.99 \times 10^9 \times 2.56 \times 10^{-8} \][/tex]
[tex]\[ F = 8.99 \times 2.56 \times 10^1 \][/tex]
[tex]\[ F = 230.14399999999995 \, \text{N} \][/tex]
So, the electric force that the two protons exert on each other is approximately:
[tex]\[ F \approx 230.14 \, \text{N} \][/tex]
Therefore, the electric force between the two protons is [tex]\( \boxed{230.14 \, \text{N}} \)[/tex].
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