Question
Two electrons and two protons form a square and fly off to infinity due to Coulomb's repulsion. Determine the ratio of the final velocities of the protons and electrons.
Answers
GPT-4o mini
To determine the ratio of the final velocities of the protons and electrons when they form a square and fly off to infinity due to Coulomb's repulsion, we can apply the laws of conservation of energy and the principles of electrostatics.
We start with the initial configuration: two electrons and two protons positioned at the corners of a square. The side length of this square is denoted by \( a \). The charges are arranged in a way that the electrons and protons are diagonally opposite each other in the square.
**Step 1: Calculate the initial potential energy.**
The initial potential energy \( U \) of the system due to Coulomb's repulsion can be calculated by considering the pairwise interactions:
1. **Between the two electrons:**
\[
U_{ee} = k \frac{e^2}{a}
\]
2. **Between the two protons:**
\[
U_{pp} = k \frac{e^2}{a}
\]
3. **Between the electron and the proton:**
The distance between each electron and proton in the square is \( a \) for side pairs and \( \sqrt{2}a \) for diagonal pairs. There are two side pairs (each electron is near a proton), and one diagonal pair.
\[
U_{ep} = k \frac{e^2}{\sqrt{2}a}
\]
There are two ep interactions:
\[
U_{ep, \text{total}} = 2 \cdot k \frac{e^2}{\sqrt{2} a}
\]
So, the total potential energy \( U \):
\[
U = U_{ee} + U_{pp} + U_{ep, \text{total}} = k \frac{e^2}{a} + k \frac{e^2}{a} + 2 k \frac{e^2}{\sqrt{2} a}
\]
\[
U = 2 k \frac{e^2}{a} + 2 k \frac{e^2}{\sqrt{2} a}
\]
Factoring out common terms, we find:
\[
U = k \frac{e^2}{a} \left( 2 + \frac{2}{\sqrt{2}} \right) = k \frac{e^2}{a} (2 + \sqrt{2})
\]
**Step 2: Calculate the final kinetic energy.**
As the particles move to infinite separation, the potential energy is converted entirely to kinetic energy. By conservation of energy, we have:
\[
U = K
\]
Where \( K \) is the total kinetic energy:
\[
K = \frac{1}{2} m_p v_p^2 + \frac{1}{2} m_e v_e^2
\]
Where \( m_p \) is the mass of the proton (\( m_p \approx 1836 m_e \)) and \( m_e \) is the mass of the electron.
**Step 3: Setting the equation.**
We now write down our expression for the total kinetic energy:
\[
k \frac{e^2}{a} (2 + \sqrt{2}) = \frac{1}{2} (1836 m_e) v_p^2 + \frac{1}{2} m_e v_e^2
\]
**Step 4: Finding the ratio of velocities.**
By rearranging for velocity terms:
\[
k \frac{e^2}{a} (2 + \sqrt{2}) = \frac{1}{2} m_e (1836 v_p^2 + v_e^2)
\]
From energy conservation and considering that momentum is also conserved, we can state:
\[
m_p v_p + m_e v_e = 0 \Rightarrow v_e = -\frac{m_p}{m_e} v_p = -1836 v_p
\]
Now assuming positive values for speed, we convert this to velocity ratios:
Take magnitudes:
\[
Implying \quad \frac{v_e}{v_p} = \frac{m_p}{m_e} \quad \text{(due to conservation of momentum)}
\]
This implies:
\[
\frac{v_p}{v_e} = \frac{m_e}{m_p} = \frac{1}{1836}
\]
The ratio of the final velocities of the protons to the electrons is thus:
\[
\frac{v_p}{v_e} = \frac{1}{1836}.
\]
This gives us a complete picture of understanding how these charged particles behave under Coulomb's forces and motion dynamics away from each other after mutual repulsion.
We start with the initial configuration: two electrons and two protons positioned at the corners of a square. The side length of this square is denoted by \( a \). The charges are arranged in a way that the electrons and protons are diagonally opposite each other in the square.
**Step 1: Calculate the initial potential energy.**
The initial potential energy \( U \) of the system due to Coulomb's repulsion can be calculated by considering the pairwise interactions:
1. **Between the two electrons:**
\[
U_{ee} = k \frac{e^2}{a}
\]
2. **Between the two protons:**
\[
U_{pp} = k \frac{e^2}{a}
\]
3. **Between the electron and the proton:**
The distance between each electron and proton in the square is \( a \) for side pairs and \( \sqrt{2}a \) for diagonal pairs. There are two side pairs (each electron is near a proton), and one diagonal pair.
\[
U_{ep} = k \frac{e^2}{\sqrt{2}a}
\]
There are two ep interactions:
\[
U_{ep, \text{total}} = 2 \cdot k \frac{e^2}{\sqrt{2} a}
\]
So, the total potential energy \( U \):
\[
U = U_{ee} + U_{pp} + U_{ep, \text{total}} = k \frac{e^2}{a} + k \frac{e^2}{a} + 2 k \frac{e^2}{\sqrt{2} a}
\]
\[
U = 2 k \frac{e^2}{a} + 2 k \frac{e^2}{\sqrt{2} a}
\]
Factoring out common terms, we find:
\[
U = k \frac{e^2}{a} \left( 2 + \frac{2}{\sqrt{2}} \right) = k \frac{e^2}{a} (2 + \sqrt{2})
\]
**Step 2: Calculate the final kinetic energy.**
As the particles move to infinite separation, the potential energy is converted entirely to kinetic energy. By conservation of energy, we have:
\[
U = K
\]
Where \( K \) is the total kinetic energy:
\[
K = \frac{1}{2} m_p v_p^2 + \frac{1}{2} m_e v_e^2
\]
Where \( m_p \) is the mass of the proton (\( m_p \approx 1836 m_e \)) and \( m_e \) is the mass of the electron.
**Step 3: Setting the equation.**
We now write down our expression for the total kinetic energy:
\[
k \frac{e^2}{a} (2 + \sqrt{2}) = \frac{1}{2} (1836 m_e) v_p^2 + \frac{1}{2} m_e v_e^2
\]
**Step 4: Finding the ratio of velocities.**
By rearranging for velocity terms:
\[
k \frac{e^2}{a} (2 + \sqrt{2}) = \frac{1}{2} m_e (1836 v_p^2 + v_e^2)
\]
From energy conservation and considering that momentum is also conserved, we can state:
\[
m_p v_p + m_e v_e = 0 \Rightarrow v_e = -\frac{m_p}{m_e} v_p = -1836 v_p
\]
Now assuming positive values for speed, we convert this to velocity ratios:
Take magnitudes:
\[
Implying \quad \frac{v_e}{v_p} = \frac{m_p}{m_e} \quad \text{(due to conservation of momentum)}
\]
This implies:
\[
\frac{v_p}{v_e} = \frac{m_e}{m_p} = \frac{1}{1836}
\]
The ratio of the final velocities of the protons to the electrons is thus:
\[
\frac{v_p}{v_e} = \frac{1}{1836}.
\]
This gives us a complete picture of understanding how these charged particles behave under Coulomb's forces and motion dynamics away from each other after mutual repulsion.