Subfield: Electromagnetism

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: Two electrons are in a uniform electric field $\mathbf{E}=E_{0} \hat{\mathbf{z}}$ where $E_{0}=10^{-11} \mathrm{~N} / \mathrm{C}$. One electron is at the origin, and another is $10 \mathrm{~m}$ above the first electron. The electron at the origin is moving at $u=10 \mathrm{~m} / \mathrm{s}$ at an angle of $30^{\circ}$ from the line connecting the electrons at $t=0$, while the other electron is at rest at $t=0$. Find the minimum distance between the electrons. You may neglect relativistic effects.

<img_4321>

Solution:
Let $\ell=10 \mathrm{~m}$. First, switch into the reference frame accelerating at $-\frac{E q}{m} \hat{z}$. In this frame, the electrons are not affected by the electric field. Now, switch into the center of mass reference frame from here. In this frame, we have both conservation of angular momentum and conservation of energy. Both electrons in this frame are moving at $\frac{u}{2}$ initially at an angle of $\theta=30^{\circ}$. At the smallest distance, both electrons will be moving perpendicular to the line connecting them. Suppose that they both move with speed $v$ and are a distance $r$ from the center of mass. By conservation of angular momentum,
$$
\begin{gathered}
2 m \cdot \frac{u}{2} \cdot \frac{\ell}{2} \sin \theta=2 m v r \\
v r=\frac{u \ell}{4} \sin \theta
\end{gathered}
$$



Now, by conservation of energy,

$$
m v^{2}+\frac{k e^{2}}{2 r}=\frac{1}{4} m u^{2}+\frac{k e^{2}}{\ell}
$$

Now, we just solve this system of equations to determine the value of $r$. Substituting $v=\frac{u l}{4 r} \sin \theta$ into the conservation of energy equation, we can solve the ensuing quadratic to find:

$$
r=\frac{\frac{k e^{2}}{2}+\sqrt{\left(\frac{k e^{2}}{2}\right)^{2}+\left(m u^{2}+\frac{4 k e^{2}}{\ell}\right)\left(\frac{m u^{2} \ell^{2}}{16} \sin ^{2}(\theta)\right)}}{\frac{1}{2} m u^{2}+\frac{2 k e^{2}}{\ell}}
$$

Finally, remembering that the distance between the electrons is actually $2 r$, we obtain $2 r=6.84 \mathrm{~m}$ as the final answer.

Final answer: 6.84 m