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: Hoping to get a larger deflection, Follin replaces the insulating Styrofoam ball with a conducting pith ball of mass $250 \mathrm{mg}$ and $2 \mathrm{~cm}$ and daisy chains 4 additional $10 \mathrm{kV}$ High Voltage Power Supplies to increase the voltage drop across the plates to $50 \mathrm{kV}$. Leaving the plate separation and the string unchanged, he repeats the same experiment as before, but forgets to measure the charge on the ball. Nonetheless, once the ball reaches equilibrium, he measures the deflection from the hanging point to be $5.6 \mathrm{~cm}$. Find the charge on the ball.

Solution:
When a conducting ball is placed in a uniform electric field, the charges separate so that the sphere becomes a dipole with electric dipole moment
$$
p=4 \pi \epsilon_{0} r^{3} E_{0}
$$
Let $Q$ be the charge on the ball. The ball can be approximated as a point dipole with dipole moment $p$ and a point charge of magnitude $Q$ at the center of the ball. Since the plates are conducting and must be equipotential, the charge and dipole will experience an attractive force to the plates. Using the method of image charges, the force caused by the closer plate is equivalent to that of a point charge $-Q$ and a dipole moment $p$ at a distance $d-2 x$ from the center of the ball, where $d=17 \mathrm{~cm}$ is the separation between the plates and $x=5.6 \mathrm{~cm}$ is the deflection of the ball from the hanging point. The forces from the farther plate are negligible compared to the force due to the electric field, $Q E_{0}$. The force between a point charge $Q$ and a point dipole moment $p$ a distance $r$ apart is
$$
F=\frac{p Q}{2 \pi \epsilon_{0} r^{3}}
$$
The force between two dipoles with dipole moments of magnitude $p$ in the same direction as the vector from one dipole to the other is
$$
F=\frac{3 p^{2}}{2 \pi \epsilon_{0} r^{4}}
$$
Adding up the charge-image charge, charge-image dipole, dipole-image charge, and dipole-image dipole interactions, we get that the total electrostatic force on the ball is
$$
F=Q E_{0}+\frac{Q^{2}}{4 \pi \epsilon_{0}(d-2 x)^{2}}+\frac{p Q}{\pi \epsilon_{0}(d-2 x)^{3}}+\frac{3 p^{2}}{2 \pi \epsilon_{0}(d-2 x)^{4}}
$$
From the previous problem, we know that the displacement $x$ of the ball from equilibrium as a function of the electrostatic force $F$ is
$$
x=\frac{F}{l g} \ln \left(\frac{m+l}{m}\right)
$$

Rearranging this as a quadratic in $Q$ and solving, we have
$$
\begin{gathered}
\frac{1}{4 \pi \epsilon_{0}(d-2 x)^{2}} Q^{2}+\left(\frac{p}{\pi \epsilon_{0}(d-2 x)^{3}}+E_{0}\right) Q+\left(\frac{3 p^{2}}{2 \pi \epsilon_{0}(d-2 x)^{4}}-\frac{\lg x}{\ln \left(\frac{m+l}{m}\right)}\right)=0 \\
Q=\frac{-\left(\frac{p}{\pi \epsilon_{0}(d-2 x)^{3}}+E_{0}\right)+\sqrt{\left(\frac{p}{\pi \epsilon_{0}(d-2 x)^{3}}+E_{0}\right)^{2}-4\left(\frac{1}{4 \pi \epsilon_{0}(d-2 x)^{2}}\right)\left(\frac{3 p^{2}}{2 \pi \epsilon_{0}(d-2 x)^{4}}-\frac{\lg x}{\ln \left(\frac{m+l}{m}\right)}\right)}}{2 \frac{1}{4 \pi \epsilon_{0}(d-2 x)^{2}}} .
\end{gathered}
$$
Plugging gives $Q=4.48 \times 10^{-10} \mathrm{C}$.

Final answer: $4.48 \times 10^{-10}$ C