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 1.92×1031.92 \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.

Understanding Ionization

  • 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.

Setting up the Problem

  • Draw a coordinate system with the nitrogen molecule (N₂⁺) at point A moving in the positive x-direction.
  • The velocity at point A is 1.92×103m/s1.92 \times 10^3 m/s.
  • At point B, the velocity is zero.
  • The distance Δx\Delta x between A and B is given.
  • The electric field is pointing in the negative x-direction.

Parallel Plate Capacitor Analogy

  • 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.

Calculating the Electric Field

  • Use the work-energy theorem: the work done on the molecule equals the change in its kinetic energy.
  • ΔU+ΔK=0\Delta U + \Delta K = 0 where ΔU\Delta U is the change in potential energy and ΔK\Delta K is the change in kinetic energy.
  • ΔK=K<em>f−K</em>i\Delta K = K<em>f - K</em>i where K<em>fK<em>f is the final kinetic energy (0 at point B) and K</em>iK</em>i is the initial kinetic energy at point A.
  • Ki=12mv2K_i = \frac{1}{2} m v^2 where m is the mass of the nitrogen molecule and v is its initial velocity.
  • ΔU=−qEΔx\Delta U = -qE \Delta x where q is the charge of the nitrogen molecule, E is the electric field, and Δx\Delta x is the distance between A and B.
  • −qEΔx−12mv2=0-qE \Delta x - \frac{1}{2} m v^2 = 0
  • Solve for E: E=−mv22qΔxE = -\frac{mv^2}{2q \Delta x}

Plugging in the Values

  • m=4.65×10−26kgm = 4.65 \times 10^{-26} kg
  • v=1.92×103m/sv = 1.92 \times 10^3 m/s
  • q=1.6×10−19Cq = 1.6 \times 10^{-19} C
  • Δx=0.856×10−3m\Delta x = 0.856 \times 10^{-3} m
  • E=−(4.65×10−26kg)(1.92×103m/s)22(1.6×10−19C)(0.856×10−3m)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≈−626V/mE \approx -626 V/m

Calculating the Potential Difference

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

Discussion of Conservation of Energy

  • 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.

Practical Applications and Approximations

  • 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.

Potential Energy and Potential Difference

  • 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: Potential Difference=Energy Lostq\text{Potential Difference} = \frac{\text{Energy Lost}}{q}

Generalization with Parallel Plate Capacitor

  • The energy lost by a charged particle moving between parallel plates is given by qΔVq \Delta V
  • ΔV=−EΔx\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.

Single Charge Particle

  • For 10^23 charged particles (e.g., charges on plates), ΔV=−σε<em>0(x</em>final−xinitial)\Delta V = -\frac{\sigma}{\varepsilon<em>0} (x</em>{\text{final}} - x_{\text{initial}})
  • For a single charged particle: The electric field generated by a single charge is E=kqr2E = k \frac{q}{r^2}, and the energy associated with this electric field is V=kqrV = k \frac{q}{r}. The change in energy is ΔV=kq(1r<em>b−1r</em>a)\Delta V = k q \left( \frac{1}{r<em>b} - \frac{1}{r</em>a} \right)

Equipotential Surfaces

  • 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.

Equipotential Surfaces for a Single Charge

  • 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.

Energy Units and the Hydrogen Atom

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

Electron Volt

  • 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 1.6×10−191.6 \times 10^{-19}