weblab-llm-competition-2025-bridge/neko-prelim-HLE_SFT_OlympiadBench huggingface dataset:
Problem: In a resource-limited ecological system, a population of organisms cannot keep growing forever (such as lab bacteria growing inside culture tube). The effective growth rate $g$ (including

contributions from births and deaths) depends on the instantaneous abundance of resource $R(t)$, which in this problem we will consider the simple case of linear-dependency:

$$
\frac{\mathrm{d}}{\mathrm{d} t} N=g(R) N=\alpha R N
$$

where $N(t)$ is the population size at time $t$. The resources is consumed at a constant rate $\beta$ by each organism:

$$
\frac{\mathrm{d}}{\mathrm{d} t} R=-\beta N
$$

Initially, the total amount of resources is $R_{0}$ and the population size is $N_{0}$. Given that $\alpha=10^{-9}$ resourceunit $^{-1} \mathrm{~s}^{-1}, \beta=1$ resource-unit/s, $R_{0}=10^{6}$ resource-units and $N_{0}=1$ cell, find the total time it takes from the beginning to when all resources are depleted (in hours).

Solution: We can find the analytical solution for the following set of two ODEs describing the populationresource dynamics:

$$
\frac{d N}{d t}=\alpha R N,
\tag{1}
$$
$$
\frac{d R}{d t}=-\beta N .
\tag{2}
$$

Divide Eq.(1) by Eq.(2) on both sides, we get a direct relation between the population size $N(t)$ and the amount of resource $R(t)$ :

$$
\frac{d N}{d R}=-\frac{\alpha}{\beta} R \Longrightarrow N=N_{0}+\frac{\alpha}{2 \beta}\left(R_{0}^{2}-R^{2}\right)
$$

Plug this in Eq.(2), we obtain the total time $T$ it takes from the beginning to when the resource is depleted:

$$
\begin{aligned}
\frac{d R}{d t}=-\frac{\alpha}{2}\left[\left(\frac{2 \beta}{\alpha} N_{0}+R_{0}^{2}\right)-R^{2}\right] \Longrightarrow & \Longrightarrow \\
\left.\right|_{R=0} & =\frac{2}{\alpha} \int_{0}^{R_{0}} d R\left[\left(\frac{2 \beta}{\alpha} N_{0}+R_{0}^{2}\right)-R^{2}\right]^{-1} \\
& =\frac{2}{\alpha \sqrt{\frac{2 \beta}{\alpha} N_{0}+R_{0}^{2}}} \operatorname{arctanh}\left(\frac{R_{0}}{\sqrt{\frac{2 \beta}{\alpha} N_{0}+R_{0}^{2}}}\right) .
\end{aligned}
$$

Use the given numerical values, we arrive at $\left.t\right|_{R=0} \approx 7594.3 \mathrm{~s} \approx 2.1095 \mathrm{hrs}$.

Answer: 2.1095
Problem: The path of an asteroid that comes close to the Earth can be modeled as follows: neglect gravitational effects due to other bodies, and assume the asteroid comes in from far away with some speed $v$ and lever arm distance $r$ to Earth's center. On January 26, 2023, a small asteroid called 2023 BU came to a close distance of $3541 \mathrm{~km}$ to Earth's surface with a speed of $9300 \mathrm{~m} / \mathrm{s}$. Although BU had a very small mass estimated to be about $300,000 \mathrm{~kg}$, if it was much more massive, it could have hit the Earth. How massive would BU have had to have been to make contact with the Earth? Express your answer in scientific notation with 3 significant digits. Use $6357 \mathrm{~km}$ as the radius of the Earth. The parameters that remain constant when the asteroid mass changes are $v$ and $r$, where $v$ is the speed at infinity and $r$ is the impact parameter.

Solution: Let $v_{1}=9300 \mathrm{~m} / \mathrm{s}, d=3541 \mathrm{~km}$, and $m=300,000 \mathrm{~kg}$, and let $M$ and $R$ be the Earth's mass and radius.

First we find $v$ and $r$. We use the reference frame of the Earth, where the asteroid has reduced mass $\mu=\frac{M m}{M+m}$ and the Earth has mass $M+m$. Then by energy and angular momentum conservation, we have

$$
\mu v r=\mu v_{1}(R+d)
$$



and

$$
\frac{1}{2} \mu v^{2}=\frac{1}{2} \mu v_{1}^{2}-\frac{G M m}{R+d}
$$

We solve for

$$
v=\sqrt{2 G(M+m) \cdot \frac{R+d}{r^{2}-(R+d)^{2}}}
$$

so

$$
v_{1}=\sqrt{\frac{2 G(M+m)}{R+d} \cdot \frac{r^{2}}{r^{2}-(R+d)^{2}}},
$$

and we compute $r=37047 \mathrm{~km}$ and $v=2490 \mathrm{~m} / \mathrm{s}$.

Now we consider when the asteroid is massive enough to touch the Earth. We let $m^{\prime}$ and $\mu^{\prime}$ be the mass of the asteroid and its reduced mass, and using a similar method to above, we arrive at

$$
v=\sqrt{2 G\left(M+m^{\prime}\right) \cdot \frac{R}{r^{2}-R^{2}}}
$$

so we can solve for $m^{\prime}=3.74 \times 10^{24} \mathrm{~kg}$.

Answer: $3.74 \times 10^{24}$
Problem: Pegasi and Betelgeuse are two star systems that can undergo a supernova. Betelgeuse is 548 light-years away from Earth and IK Pegasi is 154 light-years away from Earth. Assume that the two star systems are 500 light-years away from each other.
Astronomers on Earth observe that the two star systems undergo a supernova explosion 300 years apart. A spaceship, the OPhO Galaxia Explorer which left Earth in an unknown direction before the first supernova observes both explosions occur simultaneously. Assume that this spaceship travels in a straight line at a constant speed $v$. How far are the two star systems according to the OPhO Galaxia Explorer at the moment of the simultaneous supernovae? Answer in light-years.

Note: Like standard relativity problems, we are assuming intelligent observers that know the finite speed of light and correct for it.

Solution: For any inertial observer, define the 4-distance between two events as $\Delta s^{\mu}=(c \Delta t, \Delta \mathbf{x})$, where $\Delta t$ and $\Delta \mathrm{x}$ are the temporal and spatial intervals measured by the observer. By the properties of 4 -vectors, the following quantity is Lorentz-invariant:
$$
\Delta s^{\mu} \Delta s_{\mu}=c^{2} \Delta t^{2}-\|\Delta \mathbf{x}\|^{2}
$$

In the spaceship's frame, this is equal to $\Delta s^{\mu} \Delta s_{\mu}=-\left\|\Delta \mathbf{x}^{\prime}\right\|^{2}$, since the two supernovas are simultaneous; hence

$$
\left\|\Delta \mathbf{x}^{\prime}\right\|=\sqrt{\|\Delta \mathbf{x}\|^{2}-c^{2} \Delta t^{2}}
$$

Since the observers have already taken into account the delay due to the nonzero speed of light, $c \Delta t=300 \mathrm{ly},\|\Delta \mathbf{x}\|=500 \mathrm{ly}$, and

$$
\left\|\Delta \mathbf{x}^{\prime}\right\|=\sqrt{\|\Delta \mathbf{x}\|^{2}-c^{2} \Delta t^{2}}=400 l y
$$

Answer: 400
Problem: A ball of mass $1 \mathrm{~kg}$ is thrown vertically upwards and it faces a quadratic drag with a terminal velocity of $20 \mathrm{~m} / \mathrm{s}$. It reaches a maximum height of $30 \mathrm{~m}$ and falls back to the ground. Calculate the energy dissipated until the point of impact (in J).

Solution: If we suppose that the magnitude of quadratic drag is $F_{\mathrm{drag}}=c v^{2}$ for some constant $c$, then when the ball is falling downwards at terminal velocity $v_{t}$ this upwards drag force must cancel gravity to provide zero net force:
$$
m g=c v_{t}^{2} \Longrightarrow c=\frac{m g}{v_{t}^{2}}
$$

so we can rewrite our equations of motion in terms of terminal velocity. In the ascending portion of the ball's trajectory, suppose that the ball's speed changes from $v$ to $v+d v$ after traveling an infinitesimal distance from $h$ to $h+d h$. Then:

$d($ kinetic energy $)=-d($ potential energy and energy lost to drag $) \Longrightarrow m v d v=-\frac{m g v^{2}}{v_{t}^{2}} d h-m g d h$

where we used chain rule and $d W=F \cdot d x$. This is a separable differential equation for velocity in terms of height (!), which we can solve:

$$
\int \frac{v d v}{1+\frac{v^{2}}{v_{t}^{2}}}=\int-g d h+\text { constant }=\text { constant }-g h
$$

We u-sub the entire denominator $\left(u=1+v^{2} / v_{t}^{2} \Longrightarrow d u=2 v / v_{t}^{2} d v\right)$, which nicely cancels the numerator. Our left-hand integral is

$$
\frac{v_{t}^{2}}{2} \int \frac{d u}{u}=\frac{v_{t}^{2}}{2} \ln \left|1+\frac{v^{2}}{v_{t}^{2}}\right|=\text { constant }-g h
$$

Now to find the constant! At our maximum height $h_{0}=30 \mathrm{~m}$, the speed and left-hand side are zero, so the constant must be $g h_{0}$. Then we can find the initial speed of the projectile by substituting $h=0$, from which:

$$
v_{\text {initial }}=v_{t} \sqrt{e^{\frac{2 g h_{0}}{v_{t}^{2}}}-1}
$$

We can use a similar argument for the downwards trajectory, albeit with drag pointing upwards. We should get that the final speed immediately before impact is

$$
v_{\text {final }}=v_{t} \sqrt{1-e^{\frac{-2 g h_{0}}{v_{t}^{2}}}}
$$

from which the dissipated energy is the difference in final and initial kinetic energies, or

$$
\frac{1}{2} m\left(v_{\text {final }}^{2}-v_{\text {initial }}^{2}\right) \approx 515.83 \text { joules }
$$

Answer: 515.83
Problem: Two parallel square plates of side length $1 \mathrm{~m}$ are placed a distance $30 \mathrm{~cm}$ apart whose centers are at $(-15 \mathrm{~cm}, 0,0)$ and $(15 \mathrm{~cm}, 0,0)$ have uniform charge densities $-10^{-6} \mathrm{C} / \mathrm{m}^{2}$ and $10^{-6} \mathrm{C} / \mathrm{m}^{2}$ respectively. Find the magnitude of the component of the electric field perpendicular to axis passing through the centers of the two plates at $(10 \mathrm{~cm}, 1 \mathrm{~mm}, 0)$.

Solution: By symmetry, the electric field due to the portion of the plates between $y=50 \mathrm{~cm}$ and $y=-49.8 \mathrm{~cm}$ will cancel out. We only need to consider the electric field from two $2 \mathrm{~mm}$ wide strips of charge, which are small enough to be approximated as wires with charge per unit length $\lambda=\sigma w= \pm 2 \times 10^{-9} \mathrm{C} / \mathrm{m}^{2}$ at a distance $y=50 \mathrm{~cm}$ away. The y-component of the electric field from these wires is then
$$
\begin{aligned}
& E_{y}=\frac{0.5 \lambda}{4 \pi \epsilon_{0}} \int_{-0.5}^{0.5}\left(\frac{1}{\left(z^{2}+0.5^{2}+0.05^{2}\right)^{\frac{3}{2}}}-\frac{1}{\left(z^{2}+0.5^{2}+0.25^{2}\right)^{\frac{3}{2}}}\right) \mathrm{d} z \\
& E_{y}=\frac{0.5 \lambda}{4 \pi \epsilon_{0}}\left(\frac{z}{\left(0.5^{2}+0.05^{2}\right) \sqrt{z^{2}+0.5^{2}+0.05^{2}}}-\frac{z}{\left(0.5^{2}+0.25^{2}\right) \sqrt{z^{2}+0.5^{2}+0.25^{2}}}\right) \\
& E_{y}=\frac{0.5 \lambda}{4 \pi \epsilon_{0}}\left(\frac{1}{\left(0.5^{2}+0.05^{2}\right) \sqrt{0.5^{2}+0.5^{2}+0.05^{2}}}-\frac{1}{\left(0.5^{2}+0.25^{2}\right) \sqrt{0.5^{2}+0.5^{2}+0.25^{2}}}\right)=11.9 \mathrm{~N} / \mathrm{C}
\end{aligned}
$$

Answer: $11.9$
Problem: A space elevator consists of a heavy counterweight placed near geostationary orbit, a thread that connects it to the ground (assume this is massless), and elevators that run on the threads (also massless). The mass of the counterweight is $10^{7} \mathrm{~kg}$. Mass is continuously delivered to the counterweight at a rate of $0.001 \mathrm{~kg} / \mathrm{s}$. The elevators move upwards at a rate of $20 \mathrm{~m} / \mathrm{s}$. Assume there are many elevators, so their discreteness can be neglected. The elevators are massless. The counterweight orbits the Earth.


Find the minimum possible displacement radially of the counterweight. Specify the sign.

Solution: As the orbit is geostationary, we can balance forces to write that
$$
\frac{G M_{E}}{R_{\mathrm{GS}}^{2}}=\omega^{2} R_{\mathrm{GS}} \Longrightarrow R_{\mathrm{GS}}=\left(\frac{G M_{E} T^{2}}{4 \pi}\right)^{1 / 3}
$$
The total mass of masses $=1790 \mathrm{~kg} \ll 10^{6} \mathrm{~kg}$ which means the displacements are small. The total gravity of masses is
$$
\int_{R_{E}}^{R_{E s}} \frac{G M_{E}}{x^{2}}(\lambda d x)=\operatorname{GME} \lambda\left(\frac{1}{R_{E}}-\frac{1}{R_{\cos }}\right)=2650 \mathrm{~N}
$$
where $\lambda=5 \cdot 10^{-5} \mathrm{~kg} / \mathrm{m}$. The total centrifugal of masses is
$$
\int_{R_{E}}^{R_{E}} \frac{4 \pi^{2}}{T^{2}} \times(\lambda d x)=\frac{2 \pi^{2}}{T^{2}} \lambda\left(R_{E s}^{2}-R_{E}^{2}\right)=230 \mathrm{~V}
$$
Hence, the total force is $2420 \mathrm{~N}$. The outwards force is just
$$
F_{\text {out }}=\omega^{2} a M-\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{G M_{E}}{x^{2}}\right) a M
$$
where $a$ is the horizontal displacement which is much less than $R_{E}, R_{\mathrm{GS}}$. Hence, this can be rewritten as
$$
F_{\text {out }}=\left(\frac{4 \pi^{2}}{T^{2}}-2 \frac{G M_{E}}{R_{\mathrm{GS}}^{2}}\right) a M=2420 \mathrm{~N}
$$
This implies that $a=15.21 \mathrm{~km}$.

Answer: $15.21$
Problem: A stable star of radius $R$ has a mass density profile $\rho(r)=\alpha(1-r / R)$. Here, "stable" means that the star doesn't collapse under its own gravity. If the internal pressure at the core is provided solely by the radiation of photons, calculate the temperature at the core. Assume the star is a perfect black body and treat photons as a classical ideal gas. Use $R=7 \times 10^{5} \mathrm{~km}$ and $\alpha=3 \mathrm{~g} / \mathrm{cm}^{3}$. Round your answer to the nearest kilokelvin. We treat photons as a classical gas here to neglect any relativistic effects.

Solution: For a star that doesn't collapse in on itself, there must be some source of pressure $p(r)$ that balances out the pressure due to gravity. Define the function:
$$
m(r)=\int_{0}^{r} 4 \pi r^{2} \rho(r) d r
$$
inside the star. The gravitational pressure on an infinitesimal shell of radius $r$ and thickness $d r$ is given by:
$$
d p_{g}=-\frac{G m(r) \rho(r) 4 \pi r^{2} d r}{r^{2} 4 \pi r^{2}}=-\frac{G m(r) \rho(r)}{r^{2}} d r
$$
Hence, the pressure source must provide a pressure gradient $\frac{d p}{d r}$ given by:
$$
\frac{d p}{d r}=\frac{G m(r) \rho(r)}{r^{2}}
$$
If $\rho(r)=\alpha(1-r / R)$, then $m(r)=\frac{1}{3} \pi \alpha r^{3}(3 r / R-4)$. We need the pressure gradient:
$$
\frac{d p}{d r}=\frac{1}{3} \pi G \alpha^{2} r(3 r / R-4)(r / R-1)
$$
We integrate this equation from $r=0$ to $R$ with the obvious boundary condition $p(R)=0$. We find that $p(0)=\frac{5}{36} \pi G \alpha^{2} R^{2}$. From the Stefan-Boltzmann law, along with elementary kinetic theory, the pressure due to local radiation is given by $\frac{4 \sigma}{3 c} T_{c}^{4}$. Plugging in values gives $T_{c}=26718 \mathrm{kK}$

Answer: $26718$
Problem: In this problem, we consider a simple model for a thermoacoustic device. The device uses heavily amplified sound to provide work for a pump that can then extract heat. Sound waves form standing waves in a tube of radius $0.25 \mathrm{~mm}$ that is closed on both sides, and a two-plate stack is inserted in the tube. A temperature gradient forms between the plates of the stack, and the parcel of gas trapped between the plates oscillates sinusoidally between a maximum pressure of $1.03 \mathrm{MPa}$ and a minimum of $0.97 \mathrm{MPa}$. The gas is argon, with density $1.78 \mathrm{~kg} / \mathrm{m}^{3}$ and adiabatic constant $5 / 3$. The speed of sound is $323 \mathrm{~m} / \mathrm{s}$. The heat pump itself operates as follows:
The parcel of gas starts at minimum pressure. The stack plates adiabatically compress the parcel of gas to its maximum pressure, heating the gas to a temperature higher than that of the hotter stack plate. Then, the gas is allowed to isobarically cool to the temperature of the hotter stack plate. Next, the plates adiabatically expand the gas back to its minimum pressure, cooling it to a temperature lower than that of the colder plate. Finally, the gas is allowed to isobarically heat up to the temperature of the colder stack plate.

Find the power at which the thermoacoustic heat pump emits heat.

Solution: The efficiency of the heat engine is $\epsilon=1-\left(\frac{P_{2}}{P_{1}}\right)^{\frac{\gamma-1}{\gamma}}=0.0237$. The parcel oscillates between pressures $P_{1}$ and $P_{2}$ sinusoidally with amplitude $P_{0}=\frac{P_{1}-P_{2}}{2}$. For a sound wave, the pressure amplitude is $\rho s_{0} \omega v$, where $s_{0}$ is the position amplitude and $v$ is the speed of sound.

Then the average power with which the sound wave does work on the plates is
$$
\langle P\rangle=\frac{1}{2} \rho \omega^{2} s_{0}^{2} A v=\frac{P_{0}^{2}}{2 \rho v} A
$$
where $A$ is the area of each plate. From this, the heat power generated by the engine is $\langle P\rangle / \epsilon=$ $6.47 \mathrm{~W}$.

Answer: $6.47$
Problem: The following information applies for the next two problems. For your mass spectroscopy practical you are using an apparatus consisting of a solenoid enclosed by a uniformly charged hollow cylinder of charge density $\sigma=50 \mu \mathrm{C} / \mathrm{m}^{2}$ and radius $r_{0}=7 \mathrm{~cm}$. There exists an infinitesimal slit of insular material between the cylinder and solenoid to stop any charge transfer. Also, assume that there is no interaction



between the solenoid and the cylinder, and that the magnetic field produced by the solenoid can be easily controlled to a value of $B_{0}$.

An electron is released from rest at a distance of $R=10 \mathrm{~cm}$ from the axis. Assume that it is small enough to pass through the cylinder in both directions without exchanging charge. It is observed that the electron reaches a distance $R$ at different points from the axis 7 times before returning to the original position.


Calculate $B_{0}$ under the assumption that the path of the electron does not self-intersect with itself.

Solution: By Gauss' law, we have
$$
\varepsilon_{0} \oint \vec{E} \cdot \mathrm{d} \vec{A}=q_{\mathrm{enc}} \Longrightarrow \varepsilon_{0}(2 \pi r h) E(r)=\left(2 \pi r_{0} h\right) \sigma \Longrightarrow E(r)=\frac{\sigma r_{0}}{\varepsilon_{0} r} \hat{r}
$$
We find the work done on the electron from a distance $r$ to $R$ is
$$
W=-\int_{R}^{r} q E(r) \mathrm{d} r=\int_{r}^{R} \frac{q \sigma r_{0}}{\varepsilon_{0} r} \mathrm{~d} r=\frac{q \sigma r_{0}}{\varepsilon_{0}} \ln \frac{R}{r}
$$
Therefore, by conservation of energy:
$$
\frac{1}{2} m v(r)^{2}=\frac{q \sigma r_{0}}{\varepsilon_{0}} \ln \frac{R}{r} \Longrightarrow v(r)=\sqrt{\frac{2 q \sigma r_{0}}{m} \ln \frac{R}{r}}
$$
Since the electron is moving in a magnetic field, the radius of the electron follows
$$
\frac{m v^{2}}{a}=q v B \Longrightarrow a=\frac{m v}{q B}=\sqrt{\frac{2 q \sigma r_{0}}{q B_{0}^{2}} \ln \frac{R}{r_{0}}}
$$
This means that $B=\sqrt{\frac{2 \sigma m}{q \varepsilon_{0} r_{0}} \ln \frac{R}{r_{0}}}$. The trajectory of the electron can follow the following paths

<img_4320>

As the path is specified to be non-intersecting, we analyze the first path. We can say that the radius of each "circle" the electron travels through is $s=r \tan \frac{\pi}{8}=\frac{m v}{q B}$. Hence,
$$
B=\cot \frac{\pi}{8} \sqrt{\frac{2 \sigma m}{q \varepsilon_{0} r_{0}} \ln \frac{R}{r_{0}}}
$$

Answer: $\cot \frac{\pi}{8} \sqrt{\frac{2 \sigma m}{q \varepsilon_{0} r_{0}} \ln \frac{R}{r_{0}}}$
Problem: In the far future, the Earth received an enormous amount of charge as a result of Mad Scientist ecilA's nefarious experiments. Specifically, the total charge on Earth is $Q=1.0 \times 10^{11} \mathrm{C}$. (compare this with the current $5 \times 10^{5} \mathrm{C}$ ).
Estimate the maximum height of a "mountain" on Earth that has a circular base with diameter $w=1.0$ $\mathrm{km}$, if it has the shape of a spherical sector. You may assume that $h_{\max } \ll w$. The tensile strength of rock is $10 \mathrm{MPa}$.

Solution: You can approximate the mound as a conducting sphere of radius $r$ connecting by a wire to the Earth of radius $R$. We can therefore say that
$$
\sigma_{\text {Mound }} \sim \frac{R}{r} \sigma_{\text {Earth }}
$$
as voltages are equal and proportional to $Q / R$. The electrostatic pressure can be given as $P=\frac{\sigma E}{2}$, where $E / 2$ comes from the fact that the section does not contribute a force on itself. This can be rewritten as
$$
P=\left(\frac{R}{r} \frac{Q}{4 \pi R^{2}}\right) \cdot \frac{1}{2} \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{R^{2}}=\frac{Q^{2}}{32 \pi \varepsilon_{0} R^{3} r}
$$
By using Pythagorean theorem, we can find that $r^{2}-(r-h)^{2}=w^{2}$ which means that $h=\frac{w^{2}}{2 r}$. We can hence write the tensile strength as
$$
Y=\frac{Q^{2} h}{16 \pi \varepsilon_{0} R^{3} w^{2}} \Longrightarrow h=\frac{16 \pi \varepsilon_{0} R^{3} w^{2} Y}{G^{2}}=115 \mathrm{~m}
$$

Answer: 115
Problem: 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 $F=\frac{Q V}{d}=8.82 \times 10^{-6} \mathrm{~N}$

Answer: $0.0475$