Phys 1402 - Chapter 15

Ionized Nitrogen Molecule

An ionized nitrogen molecule (N₂) at point A has a charge of +e (elementary charge).
It moves with a high velocity of 1.92 \times 10^3 meters per second in the positive x-direction.
A constant electric force acts in the negative x-direction, slowing the molecule until it stops at point B.
The distance between points A and B is 0.5856 mm.
The tasks are to calculate the x-component of the electric field and the potential difference between points A and B.

Ionization is the process of gaining or losing an electron.
In this problem, the nitrogen molecule loses one electron, resulting in a positive charge.
This means one of the nitrogen's protons is not covered by a negative charge.

Draw a coordinate system with the nitrogen molecule (N₂⁺) at point A moving in the positive x-direction.
The velocity at point A is 1.92 \times 10^3 m/s.
At point B, the velocity is zero.
The distance \Delta x between A and B is given.
The electric field is pointing in the negative x-direction.

Imagine a parallel plate capacitor to visualize the constant electric field.
Positive charges on the right plate repel the nitrogen molecule, causing it to slow down.
Negative charges are on the left plate.
The electric field points from the positive plate to the negative plate, in the negative x-direction.

Use the work-energy theorem: the work done on the molecule equals the change in its kinetic energy.
\Delta U + \Delta K = 0 where \Delta U is the change in potential energy and \Delta K is the change in kinetic energy.
\Delta K = Kf - Ki where Kf is the final kinetic energy (0 at point B) and Ki is the initial kinetic energy at point A.
K_i = \frac{1}{2} m v^2 where m is the mass of the nitrogen molecule and v is its initial velocity.
\Delta U = -qE \Delta x where q is the charge of the nitrogen molecule, E is the electric field, and \Delta x is the distance between A and B.
-qE \Delta x - \frac{1}{2} m v^2 = 0
Solve for E: E = -\frac{mv^2}{2q \Delta x}

m = 4.65 \times 10^{-26} kg
v = 1.92 \times 10^3 m/s
q = 1.6 \times 10^{-19} C
\Delta x = 0.856 \times 10^{-3} m
E = -\frac{(4.65 \times 10^{-26} kg)(1.92 \times 10^3 m/s)^2}{2(1.6 \times 10^{-19} C)(0.856 \times 10^{-3} m)}
E \approx -626 V/m

* \Delta V = -E \Delta x
* \Delta V = -(-626 V/m)(0.856 \times 10^{-3} m)
* \Delta V \approx 0.535 V

The change in potential energy plus the change in kinetic energy must equal zero.
Final potential energy plus final kinetic energy equals initial potential energy plus initial kinetic energy.

In real-world scenarios, conditions are not always constant.
We often analyze systems on a small scale where conditions appear constant, then piece together the results.
This approach allows us to handle complexities that our species is not inherently equipped to manage.

The potential energy lost by the nitrogen molecule is equal to the potential difference.
To find the potential difference, divide the energy lost by the charge: \text{Potential Difference} = \frac{\text{Energy Lost}}{q}

The energy lost by a charged particle moving between parallel plates is given by q \Delta V
\Delta V = -E \Delta x
This implies that the potential energy lost is simply the electric field strength multiplied by the distance traveled and the charge.

For 10^23 charged particles (e.g., charges on plates), \Delta V = -\frac{\sigma}{\varepsilon0} (x{\text{final}} - x_{\text{initial}})
For a single charged particle: The electric field generated by a single charge is E = k \frac{q}{r^2}, and the energy associated with this electric field is V = k \frac{q}{r}. The change in energy is \Delta V = k q \left( \frac{1}{rb} - \frac{1}{ra} \right)

Equipotential surfaces are regions where the potential energy of an electric field is constant.
These surfaces are always perpendicular to the direction of the electric field.
In a parallel plate capacitor, equipotential surfaces are vertical lines.
In a gravitational field, equipotential surfaces are surfaces of constant height.

The equipotential surfaces are circles, as all points at the same distance from the charge have the same potential.
The electric field lines are radial and perpendicular to these circular surfaces.
No work is done when a charged particle moves along an equipotential surface.

The potential energy created by a single proton (hydrogen atom) can be calculated using the same principles.
The potential energy is V = k \frac{q}{r}, where k = 8.85 \times 10^{-12}, q is the charge, and r is the distance from the proton.

The unit of electron volt (eV) is used to measure atomic energy transitions. 1 eV is the energy required to move an electron across a potential difference of 1 volt.
To convert Joules to electron volts, divide by 1.6 \times 10^{-19}