Answer :
To compare the gravitational and electrical forces between a proton and a neutron, let's start by reviewing the masses and charges of both particles:
- Mass of proton: [tex]\(1.673 \times 10^{-27}\)[/tex] kg
- Mass of neutron: [tex]\(1.675 \times 10^{-27}\)[/tex] kg
- Charge of proton: [tex]\(1.61 \times 10^{-19}\)[/tex] C
- Charge of neutron: [tex]\(0\)[/tex] C
Next, we will use these constants to calculate the respective forces:
### Gravitational Force
The formula to calculate the gravitational force [tex]\(F_{\text{gravity}}\)[/tex] between two masses [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex] separated by a distance [tex]\(r\)[/tex] is given by Newton's law of gravitation:
[tex]\[ F_{\text{gravity}} = G \frac{m_1 m_2}{r^2} \][/tex]
where [tex]\(G\)[/tex] is the gravitational constant, [tex]\(G = 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}\)[/tex].
Using the provided values:
[tex]\[ F_{\text{gravity}} = 6.67430 \times 10^{-11} \frac{(1.673 \times 10^{-27}) (1.675 \times 10^{-27})}{r^2} \][/tex]
Calculation result shows that for [tex]\(r = 1\)[/tex] meter, the gravitational force is:
[tex]\[ F_{\text{gravity}} \approx 1.8703224032499998 \times 10^{-64} \, \text{N} \][/tex]
### Electrical Force
The formula to calculate the electrical force [tex]\(F_{\text{electric}}\)[/tex] between two charges [tex]\(q_1\)[/tex] and [tex]\(q_2\)[/tex] separated by a distance [tex]\(r\)[/tex] is given by Coulomb's law:
[tex]\[ F_{\text{electric}} = k \frac{q_1 q_2}{r^2} \][/tex]
where [tex]\(k\)[/tex] is Coulomb's constant, [tex]\(k = 8.98755 \times 10^9 \, \text{N m}^2 \text{C}^{-2}\)[/tex].
Using the provided values:
[tex]\[ F_{\text{electric}} = 8.98755 \times 10^9 \frac{(1.61 \times 10^{-19}) (0)}{r^2} \][/tex]
Since the charge of the neutron is [tex]\(0\)[/tex] C, the electrical force is:
[tex]\[ F_{\text{electric}} = 0 \, \text{N} \][/tex]
### Comparison
Given the calculations:
- Gravitational force between the proton and the neutron: [tex]\(1.8703224032499998 \times 10^{-64} \, \text{N}\)[/tex]
- Electrical force between the proton and the neutron: [tex]\(0 \, \text{N}\)[/tex]
Thus, the gravitational force, although extremely small, is the only force acting between the proton and the neutron in this scenario with the given distances.
Conclusively:
- The gravitational force is much larger than the electrical force for any distance between the particles because the electrical force is always zero due to the zero charge of the neutron.
Therefore, the correct statement is:
The gravitational force is much larger than the electrical force for any distance between the particles.
- Mass of proton: [tex]\(1.673 \times 10^{-27}\)[/tex] kg
- Mass of neutron: [tex]\(1.675 \times 10^{-27}\)[/tex] kg
- Charge of proton: [tex]\(1.61 \times 10^{-19}\)[/tex] C
- Charge of neutron: [tex]\(0\)[/tex] C
Next, we will use these constants to calculate the respective forces:
### Gravitational Force
The formula to calculate the gravitational force [tex]\(F_{\text{gravity}}\)[/tex] between two masses [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex] separated by a distance [tex]\(r\)[/tex] is given by Newton's law of gravitation:
[tex]\[ F_{\text{gravity}} = G \frac{m_1 m_2}{r^2} \][/tex]
where [tex]\(G\)[/tex] is the gravitational constant, [tex]\(G = 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}\)[/tex].
Using the provided values:
[tex]\[ F_{\text{gravity}} = 6.67430 \times 10^{-11} \frac{(1.673 \times 10^{-27}) (1.675 \times 10^{-27})}{r^2} \][/tex]
Calculation result shows that for [tex]\(r = 1\)[/tex] meter, the gravitational force is:
[tex]\[ F_{\text{gravity}} \approx 1.8703224032499998 \times 10^{-64} \, \text{N} \][/tex]
### Electrical Force
The formula to calculate the electrical force [tex]\(F_{\text{electric}}\)[/tex] between two charges [tex]\(q_1\)[/tex] and [tex]\(q_2\)[/tex] separated by a distance [tex]\(r\)[/tex] is given by Coulomb's law:
[tex]\[ F_{\text{electric}} = k \frac{q_1 q_2}{r^2} \][/tex]
where [tex]\(k\)[/tex] is Coulomb's constant, [tex]\(k = 8.98755 \times 10^9 \, \text{N m}^2 \text{C}^{-2}\)[/tex].
Using the provided values:
[tex]\[ F_{\text{electric}} = 8.98755 \times 10^9 \frac{(1.61 \times 10^{-19}) (0)}{r^2} \][/tex]
Since the charge of the neutron is [tex]\(0\)[/tex] C, the electrical force is:
[tex]\[ F_{\text{electric}} = 0 \, \text{N} \][/tex]
### Comparison
Given the calculations:
- Gravitational force between the proton and the neutron: [tex]\(1.8703224032499998 \times 10^{-64} \, \text{N}\)[/tex]
- Electrical force between the proton and the neutron: [tex]\(0 \, \text{N}\)[/tex]
Thus, the gravitational force, although extremely small, is the only force acting between the proton and the neutron in this scenario with the given distances.
Conclusively:
- The gravitational force is much larger than the electrical force for any distance between the particles because the electrical force is always zero due to the zero charge of the neutron.
Therefore, the correct statement is:
The gravitational force is much larger than the electrical force for any distance between the particles.
