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: Follin is investigating the electrostatic pendulum. His apparatus consists of an insulating Styrofoam ball with a mass of $14 \mathrm{mg}$ and radius $r=0.5 \mathrm{~cm}$ suspended on a uniform electrically-insulating string of length $1 \mathrm{~m}$ and mass per unit length density of $1.1 \cdot 10^{-5} \mathrm{~kg} / \mathrm{m}$ between two large metal plates separated by a distance $17 \mathrm{~cm}$ with a voltage drop of $10 \mathrm{kV}$ between them, such that when the ball is in equilibrium, its center of mass is exactly equidistant to the two plates. Neglect the possibility of electrical discharge throughout the next two problems.
Follin then gives the ball a charge $0.15 \mathrm{nC}$. Assuming that the charge is distributed evenly across the surface of the ball, find the subsequent horizontal deflection of the pendulum bob's center of mass from its hanging point at equilibrium.

Solution:
The force on the Styrofoam ball due to the electric field from the two plates is $F=\frac{Q V}{d}$. Since the plates are conducting, the ball also experiences an attraction toward the closer plate. This force can be neglected because it is much smaller than $\frac{Q V}{d}$ (an assumption that will be justified at the end of the solution). The mass of the string, however, needs to be considered. Consider the forces on an infinitesimal segment of the string. The horizontal component of the tension must balance out the electric force on the ball and the vertical component must balance out the weight of everything below it (string and ball). This gives us
$$
\frac{d x}{d h}=\frac{F}{m g+\lambda g h}
$$
where $h$ is the height above the ball, $x$ is the horizontal displacement from equilibrium, $F$ is the electrostatic force, $\lambda$ is the mass density of the string, and $m$ is the mass of the ball. Separating



variables, we have
$$
x=F \int_{0}^{L} \frac{d h}{m g+\lambda g h}
$$
where $L$ is the length of the string. Integrating, we get
$$
x=\frac{F}{\lambda g} \ln \left(\frac{m+\lambda L}{m}\right)=0.0475 \mathrm{~m}
$$
Now, let's justify our assumption that the attraction of the ball towards the plates is negligible. Using the method of image charges, the force due to the closer plate is equivalent to a charge $-Q$ a distance of $d-2 x$ away. Infinitely many image charges exist due to the other plate but they have a smaller effect so they will not be considered. The force due to the image charge is
$$
F^{\prime}=\frac{Q^{2}}{4 \pi \epsilon_{0}(d-2 x)^{2}}=3.59 \times 10^{-8} \mathrm{~N}
$$
which is indeed much less than $F=\frac{Q V}{d}=8.82 \times 10^{-6} \mathrm{~N}$

Final answer: $0.0475$ m