weblab-llm-competition-2025-bridge/neko-prelim-HLE_SFT_OlympiadBench huggingface dataset:
Problem: d. If Carla's stopwatch measures an elapsed time $\tau$, what does Clock B measure in CIRF?

Solution: The readings on Clock B and on Clock A are not necessarily identical once Carla moves through the circle (because her motion becomes more parallel with the displacement between the two clocks, and thus simultaneity is lost).



Suppose Carla is at $(R \cos \theta, R \sin \theta)$, so her velocity is given by $(-v \sin \theta, v \cos \theta)$. Suppose we place a light bulb between the two clocks and having it propagate a light pulse. In the lab frame, the light pulse reaches the two clocks simultaneously. In CIRF, the math is a little more complicated.



We first rotate our lab coordinates so that $\hat{\mathbf{a}}=-\sin \theta \hat{\mathbf{x}}+\cos \theta \hat{\mathbf{y}}$, and $\hat{\mathbf{b}}=\cos \theta \hat{\mathbf{x}}+\sin \theta \hat{\mathbf{y}}$. We now give the coordinates of the clocks and bulb in the rotated lab frame: Clock A, $(a, b)=(0,0)$; Clock B, $(a, b)=(-R \sin \theta, R \cos \theta)$; bulb, $(a, b)=(-R \sin \theta, R \cos \theta) / 2$. In the lab frame, a light pulse is emitted at



$$

t=0, a=-(R / 2) \sin \theta, b=(R / 2) \cos \theta

$$



The light pulse reaches Clock A at



$$

t=R / 2, a=0, b=0

$$



and Clock B at



$$

t=R / 2, a=-R \sin \theta, b=R \cos \theta .

$$



Under a Lorentz tranformation from the lab frame to CIRF, we have that the light pulse reaches Clock $A$ at $t^{\prime}=\gamma R / 2$ and Clock $B$ at $t^{\prime}=\gamma R / 2+\gamma v R \sin \theta$. Thus, Clock $B$ reads the same time as Clock $A$ with offset $\gamma v R \sin \theta$ in the reference frame moving at $v_{a}=v$, $v_{b}=0$. Note that Clock $A$ ticks slower by a factor of $\gamma$ in this frame. Therefore, the time on clock $B$ is $v R \sin \theta$ behind the time on clock $A$.



Then,



$$

t_{B}(\tau)=t_{A}(\tau)-v R \sin \theta=\frac{\tau}{\sqrt{1-v^{2} / c^{2}}}-v R \sin \theta

$$



(This is the answer we expect from the rear clock ahead effect!) Finally, we use that $\theta=\omega \tau$ and $\omega=\frac{2 \pi}{T}$, where $T$ is the period in Carla's frame. Then,



$$

t_{B}(\tau)=\frac{\tau}{\sqrt{1-v^{2} / c^{2}}}-\frac{v R}{c^{2}} \sin \left(\frac{v \tau}{R \sqrt{1-v^{2} / c^{2}}}\right)

$$

Answer: $t_{B}(\tau)=\frac{\tau}{\sqrt{1-v^{2} / c^{2}}}-\frac{v R}{c^{2}} \sin \left(\frac{v \tau}{R \sqrt{1-v^{2} / c^{2}}}\right)$
Problem: a. When a faucet is turned on, a stream of water flows down with initial speed $v_{0}$ at the spout. For this problem, we define $y$ to be the vertical coordinate with its positive direction pointing up.

Assuming the water speed is only affected by gravity as the water falls, find the speed of water $v(y)$ at height $y$. Define the zero of $y$ such that the equation for $v^{2}$ has only one term and find $y_{0}$, the height of the spout.

Solution: We can use energy conservation to answer this question. For a bit of water with mass $m$, the total energy $E$ is the sum of the kinetic and gravitational potential energies,



$$

E=\frac{1}{2} m v^{2}+m g y \text {. } \tag{B1-1}

$$



(With this sign convention, $g \approx 10 \mathrm{~m} / \mathrm{s}^{2}$ is positive. As $y$ decreases, so does the potential energy.)



As the bit of water falls, its energy remains constant, and is equal to the initial value of



$$

E=\frac{1}{2} m v_{0}^{2}+m g y_{0} \tag{B1-2}

$$



Equating eliminating $E$ from equations B1-1 and B1-2, we have



$$

\frac{1}{2} m v^{2}+m g y=\frac{1}{2} m v_{0}^{2}+m g y_{0},

$$



and solving for $v$, we get



$$

v=\sqrt{v_{0}^{2}+2 g\left(y_{0}-y\right)}

$$



The equation for $v^{2}$ has three terms, but we were asked to choose the zero of $y$ such that there is only one. Evidently, two of the terms must cancel, and these must be the two constant terms, since the final term varies with $y$.



That means we need



$$

v_{0}^{2}+2 g y_{0}=0

$$



Solving for $y_{0}$, the vertical position of the spout is



$$

y_{0}=\frac{-v_{0}^{2}}{2 g}

$$







With this choice of the zero of $y$, the equation for $v$ simplifies to



$$

v=\sqrt{-2 g y} \tag{B1-3}

$$



We note that the result of this equation is real because $y<0$ at the spout, and decreases as the water falls, so this equation shows that $v$ is real and increases as the water falls.

Answer: $y_{0}=\frac{-v_{0}^{2}}{2 g}$ ,$v=\sqrt{-2 g y}$
Problem: b. Assume that the stream of water falling from the faucet is cylindrically symmetric about a vertical axis through the center of the stream. Also assume that the volume of water per unit time exiting the spout is constant, and that the shape of the stream of water is constant over time.
In this case, the radius $r$ of the stream of water is a function of vertical position $y$. Let the radius at the faucet be $r_{0}$. Using your result from part (a), find $r(y)$.


If $r(y)$ is not constant, it implies that the water has some radial velocity during its fall, in contradiction to our assumptions in part (a) that the motion is purely vertical. You may assume throughout the problem that any such radial velocity is negligibly small.

Solution: The same volume of water must fall through any horizontal cross-section of the stream each second because water doesn't disappear during its fall, and its density if constant. That volume per unit time $Q$ is the cross-sectional area of the stream multiplied by the speed of the water in the vertical direction. As an equation,



$$

Q=v \pi r^{2} \tag{B1-4}

$$



$Q$ is the same at all $y$, and is equal to its initial value of



$$

Q=v_{0} \pi r_{0}^{2} \tag{B1-5}

$$



Eliminating $Q$ from B1-4 and B1-5 and solving for $r$ gives



$$

r=r_{0} \sqrt{\frac{v_{0}}{v}}

$$



Plugging in our equation B1-3 for $v$,



$$

r=r_{0} \sqrt[4]{\frac{v_{0}^{2}}{-2 g y}}

$$

Answer: $r=r_{0} \sqrt[4]{\frac{v_{0}^{2}}{-2 g y}}$
Problem: a. Consider an atom in the interior of this container of volume $V$. Suppose the potential energy of the interaction is given by


$$

u(r)= \begin{cases}0 & r<d \\ -\epsilon\left(\frac{d}{r}\right)^{6} & r \geq d\end{cases}

$$



where $d \ll V^{1 / 3}$ is the minimum allowed distance between two atoms. Assume the gas is uniformly distributed within the container, what is the average potential energy of this atom? Write your answer in terms of $a^{\prime}=\frac{2 \pi d^{3} \epsilon}{3}, N$, and $V$.

Solution: The density of the gas is given by $N / V$. In a spherical shell of radius $r$ and thickness $\Delta r$, there are $\left(4 \pi r^{2} \Delta r\right) N / V$ atoms. The potential energy is given by



$$

\Delta U=-\left(4 \pi r^{2} \Delta r\right) N / V \epsilon d^{6} / r^{6}

$$



Then, the total potential energy is given by



$$

U=\int_{d}^{\infty}-\left(4 \pi r^{2} \mathrm{~d} r\right) N / V \epsilon d^{6} / r^{6}=-2 a^{\prime} N / V

$$

Answer: $-2 a^{\prime} N / V$
Problem: b. What is the average potential energy of an atom near the boundary of the box? Assume that there is no interaction between atoms near the boundary and the box itself.

Solution: Now only half of the shell of radius $r$ is full of gas, and the other half is outside of the box. This mean that the potential energy is lessened by a factor of two, to $-a^{\prime} N / V$.

Answer: $-a^{\prime} N / V$
Problem: c. Using Bernoulli's law $P+U+\rho v^{2} / 2=$ constant, with pressure $P$, potential energy density $U$, mass density $\rho$ and fluid velocity $v$, what is the pressure at the boundary of the box? Assume the interior pressure is given by the ideal gas law.

Solution: The potential energy density difference is $-a^{\prime} \frac{N^{2}}{V^{2}}$. Since there is no velocity difference, this is also the pressure difference. If the pressure on the interior is $\frac{N k T}{V}$, then the pressure on the box is $\frac{N k T}{V}-a^{\prime} \frac{N^{2}}{V^{2}}$

Answer: $\frac{N k T}{V}-a^{\prime} \frac{N^{2}}{V^{2}}$
Problem: d. Assuming most atoms are in the interior of the box, what is the total energy of the atoms in the box?

Solution: The total kinetic energy is $\frac{3}{2} N k T$. The total potential energy is $-a^{\prime} N^{2} / V$ (we drop a factor of two to avoid double-counting). So the total energy is $\frac{3}{2} N k T-a^{\prime} N^{2} / V$.

Answer: $\frac{3}{2} N k T-a^{\prime} N^{2} / V$
Problem: e. What is the final temperature of the gas after the expansion?

Solution: Naively, we might say that the total potential energy of the gas is $-2 a^{\prime} N^{2} / V$, but to avoid double-counting, we divide by 2 and instead arrive at $-a^{\prime} N^{2} / V$. Then, the quantity



$$

E=\frac{3}{2} N k_{B} T-\frac{a^{\prime} N^{2}}{V}

$$



is conserved. Therefore,



$$

T^{\prime}=T-\frac{a^{\prime} N}{3 k_{B} V}

$$

Answer: $T^{\prime}=T-\frac{a^{\prime} N}{3 k_{B} V}$
Problem: f. What is the increase in the entropy of the universe as a result of the free expansion? Give your answer to first order in $\frac{a^{\prime} N}{V k_{B} T}$.

Solution: The entropy of the surroundings do not increase as a result of the free expansion (no heat is dumped to the surroundings, and the surroundings remain in thermal equilibrium). However, the entropy of the gas does increase because the gas is momentarily not in equilibrium. Therefore, we just have to compute the increase in entropy of the gas.



Because entropy is a state function, we compute this change in entropy by constructing a reversible process between the initial and final states of the expansion, and computing the change in entropy for this process. Consider constant energy reversible expansion of this gas. For this process, the work done by the gas is equal to the heat the gas takes in. Therefore,



$$

\mathrm{d} S=\frac{p \mathrm{~d} v}{t}

$$



where we use lowercase letters to denote the quantities during the reversible expansion.



Recall that



$$

p v+\frac{a^{\prime} N^{2}}{v}=N k_{B} t

$$







If the energy of the system is $E$, then,



$$

\frac{3}{2} p v+\frac{3 a^{\prime} N^{2}}{2 v}-\frac{a^{\prime} N^{2}}{v}=E

$$



Then,



$$

p=\frac{2 E}{3 v}-\frac{a^{\prime} N^{2}}{3 v^{2}}

$$



From our expression of energy,



$$

t=\frac{2}{3} \frac{E+a^{\prime} N^{2} / v}{N k_{B}}

$$



Then,



$$

\Delta S=\int_{V}^{2 V} \frac{E N k_{B}}{E v+a^{\prime} N^{2}}-\frac{a^{\prime} N^{3} k_{B}}{2\left(E v^{2}+a^{\prime} N^{2} v\right)} \mathrm{d} v

$$



Taylor expanding gives us



$$

\Delta S=\int_{V}^{2 V} \frac{N k_{B}}{v}-\frac{3 a^{\prime} N^{3} k_{B}}{2 E v^{2}} \mathrm{~d} v

$$



Integrating gives us



$$

\Delta S=N k_{B} \log 2-\frac{3 a^{\prime} N^{3} k_{B}}{4 E V}

$$



Using that $E \approx 3 / 2 N k_{B} T$, we arrive at



$$

\Delta S=N k_{B} \log 2-\frac{a^{\prime} N^{2}}{2 V T}

$$

Answer: $\Delta S=N k_{B} \log 2-\frac{a^{\prime} N^{2}}{2 V T}$
Problem: a. Find the final temperature $T_{f}$ of the two objects, and the total work $W$ done by the engine.

Solution: Since a Carnot engine is reversible, it produces no entropy,



$$

d S_{1}+d S_{2}=\frac{d Q_{1}}{T_{1}}+\frac{d Q_{2}}{T_{2}}=0

$$


By the definition of heat capacity, $d Q_{i}=C d T_{i}$, so



$$

\frac{d T_{1}}{T_{1}}=-\frac{d T_{2}}{T_{2}}

$$



Integrating this equation shows that $T_{1} T_{2}$ is constant, so the final temperature is



$$

T_{f}=\sqrt{T_{1} T_{2}}

$$



The change in thermal energy of the objects is



$$

C\left(T_{f}-T_{1}\right)+C\left(T_{f}-T_{2}\right)=C\left[2 \sqrt{T_{1} T_{2}}-T_{1}-T_{2}\right]

$$



By the First Law of Thermodynamics, the missing energy has been used to do work, so



$$

W=C\left[T_{1}+T_{2}-2 \sqrt{T_{1} T_{2}}\right]

$$

Answer: $T_{f}=\sqrt{T_{1} T_{2}}$, $W=C\left[T_{1}+T_{2}-2 \sqrt{T_{1} T_{2}}\right]$
Problem: b. Find the minimum temperature $T_{L}$ to which the first object can be lowered.

Solution: By the Second Law of Thermodynamics, we must have $T_{L}=100 \mathrm{~K}$. Otherwise, we would have a process whose sole effect was a net transfer of heat from a cold body to a warm one.

Answer: $100$
Problem: c. Find the maximum temperature $T_{H}$ to which the third object can be raised.

Solution: The entropy of an object with constant heat capacity is



$$

S=\int \frac{d Q}{T}=C \int \frac{d T}{T}=C \ln T

$$







Since the total entropy remains constant, $T_{1} T_{2} T_{3}$ is constant; this is a direct generalization of the result for $T_{f}$ found in part (a). Energy is also conserved, as it makes no sense to leave stored energy unused, so $T_{1}+T_{2}+T_{3}$ is constant.



When one object is at temperature $T_{H}$, the other two must be at the same lower temperature $T_{0}$, or else further work could be extracted from their temperature difference, so



$$

T_{1}+T_{2}+T_{3}=T_{H}+2 T_{0}, \quad T_{1} T_{2} T_{3}=T_{H} T_{0}^{2}

$$



Plugging in temperatures with values divided by 100 for convenience, and eliminating $T_{0}$ gives



$$

T_{H}\left(7-T_{H}\right)^{2}=36

$$



We know that $T_{H}=1$ is one (spurious) solution, since this is the minimum possible final temperature as found in part (b). The other roots are $T_{H}=4$ and $T_{H}=9$ by the quadratic formula. The solution $T_{H}=9$ is impossible by energy conservation, so



$$

T_{H}=400 \mathrm{~K}

$$



It is also possible to solve the problem more explicitly. For example, one can run a Carnot cycle between the first two objects until they are at the same temperature, then run a Carnot cycle in reverse between the last two objects using the stored work. At this point, the first two objects will no longer be at the same temperature, so we can repeat the procedure; this yields an infinite series for $T_{H}$. Some students did this, and took only the first term of the series. This yields a fairly good approximation of $T_{H} \approx 395 \mathrm{~K}$.



Another explicit method is to continuously switch between running one Carnot engine forward and another Carnot engine in reverse; this yields three differential equations for $T_{1}, T_{2}$, and $T_{3}$. Solving the equations and setting $T_{1}=T_{2}$ yields $T_{3}=T_{H}$.

Answer: 395
Problem: a. Derive an expression for the deviation of the compass, $\delta \theta$, from north as a function of $K_{b}$, $K_{s}$, and $\theta$.

Solution: We add the fields to get the local field. The northward component is



$$

B_{\text {north }}=B_{e}-B_{e} K_{b} \cos \theta \cos \theta-B_{e} K_{s} \sin \theta \sin \theta

$$



while the eastward component is



$$

B_{\text {east }}=-B_{e} K_{b} \sin \theta \cos \theta+B_{e} K_{s} \cos \theta \sin \theta

$$



The deviation is given by



$$

\tan \delta \theta=\left(K_{s}-K_{b}\right) \frac{\sin \theta \cos \theta}{1-K_{b} \cos ^{2} \theta-K_{s} \sin ^{2} \theta}

$$



This form is particularly nice, because as we'll see below, $K_{b}$ and $K_{s}$ are small enough to ignore in the denominator.

Answer: $\delta \theta=\arctan \left(K_{s}-K_{b}\right) \frac{\sin \theta \cos \theta}{1-K_{b} \cos ^{2} \theta-K_{s} \sin ^{2} \theta}$
Problem: b. Assuming that $K_{b}$ and $K_{s}$ are both much smaller than one, at what heading(s) $\theta$ will the deviation $\delta \theta$ be largest?

Solution: By inspection, $\theta=45^{\circ}$ will yield the largest deviation. It's also acceptable to list $45^{\circ}, 135^{\circ}$, $225^{\circ}$, and $315^{\circ}$.

Answer: $45$, $135$, $225$, $315$
Problem: a. Determine $E_{p}$ in terms of $m_{p}, m_{\Delta}$, and $E_{b}$. You may assume that $E_{b}$ is small.

Solution: We can solve the problem exactly, approximating only in the last step. This certainly isn't necessary; we do this to illustrate a useful technique. We set $c=1$ throughout, and transform to an inertial frame where the proton is initially at rest. Before the collision,



$$

E_{p}=m_{p}, \quad E_{\gamma}=\left|p_{\gamma}\right|

$$



For the $\Delta$ particle, we have the usual relativistic relation



$$

E_{\Delta}^{2}=p_{\Delta}^{2}+m_{\Delta}^{2}

$$



By energy-momentum conservation,



$$

E_{p}+E_{\gamma}=E_{\Delta}, \quad p_{\gamma}=p_{\Delta} .

$$



Combining these results gives



$$

\left(m_{p}+E_{\gamma}\right)^{2}=E_{\gamma}^{2}+m_{\Delta}^{2} \Rightarrow E_{\gamma}=\frac{m_{\Delta}^{2}-m_{p}^{2}}{2 m_{p}}

$$



Now we need to transform back to the original frame, where the energy of the photon is $E_{b}$. We can use the Lorentz transformation for this, but it's a little easier to realize that $E=h f$ for photons, and apply the Doppler shift. Then



$$

\alpha \equiv \frac{E_{b}}{E_{\gamma}}=\sqrt{\frac{1-\beta}{1+\beta}}

$$



where $\beta$ is the velocity parameter of the proton in the inertial frame where the photon has energy $E_{b}$. Solving for $\beta$ in terms of the energy ratio $\alpha$,



$$

\beta=\frac{1-\alpha^{2}}{1+\alpha^{2}}

$$



To calculate the proton energy, we need the Lorentz factor,



$$

\gamma=\frac{1}{\sqrt{1-\beta^{2}}}=\frac{1+\alpha^{2}}{2 \alpha}

$$







Then the proton energy in the original frame is



$$

E_{p}=\gamma m_{p}=\frac{m_{p}}{2}\left(\alpha+\frac{1}{\alpha}\right)=\frac{m_{p}}{2}\left(\frac{2 m_{p} E_{b}}{m_{\Delta}^{2}-m_{p}^{2}}+\frac{m_{\Delta}^{2}-m_{p}^{2}}{2 m_{p} E_{b}}\right)

$$



which is the exact answer.



At this point, we can approximate. The second term is much larger than the first, so



$$

E_{p} \approx \frac{m_{\Delta}^{2}-m_{p}^{2}}{4 E_{b}}

$$



which was the required answer for this problem.

Answer: $\frac{m_{\Delta}^{2}-m_{p}^{2}}{4 E_{b}}$
Problem: b. In this case, the photon energy $E_{b}$ is that of the cosmic background radiation, which is an EM wave with wavelength $1.06 \mathrm{~mm}$. Determine the energy of the photons, writing your answer in electron volts.

Solution: Plugging in the numbers gives



$$

E=\frac{h c}{\lambda}=0.00112 \mathrm{eV}

$$

Answer: $0.00112$
Problem: c. Assuming this value for $E_{b}$, what is the energy of the proton, in electron volts, that will allow the above reaction? This sets an upper limit on the energy of cosmic rays. The mass of the proton is given by $m_{p} c^{2}=938 \mathrm{MeV}$ and the mass of the $\Delta$ is given by $m_{\Delta} c^{2}=1232 \mathrm{MeV}$.

Solution: Restoring the factors of $c$ and plugging in the numbers gives



$$

E_{p}=1.4 \times 10^{20} \mathrm{eV}

$$



This is known as the GZK bound for cosmic rays.

Answer: $1.4 \times 10^{20}$
Problem: a. Find the solar power $P$ received by the bright side of the planet. (Hint: the Stefan-Boltzmann law states that the power emitted by a blackbody with area $A$ is $\sigma A T^{4}$.)

Solution: The intensity of solar radiation at the surface of the sun is $\sigma T_{s}^{4}$, so the intensity at the planet's orbit radius is



$$

I=\sigma T_{s}^{4} \frac{R_{s}^{2}}{R_{0}^{2}}

$$



The area subtended by the planet is $\pi R^{2}$, so



$$

P=\pi \sigma T_{s}^{4} \frac{R^{2} R_{s}^{2}}{R_{0}^{2}}

$$

Answer: $P=\pi \sigma T_{s}^{4} \frac{R^{2} R_{s}^{2}}{R_{0}^{2}}$
Problem: c. Find the wind power $P_{w}$ in terms of $P$ and the temperature ratio $T_{2} / T_{1}$.

Solution: Let heat be transferred from the bright side at a rate $Q_{1}$ and transferred to the dark side at a rate $Q_{2}$. Then by conservation of energy,



$$

Q_{1}=P_{w}+Q_{2}

$$



Since the two hemispheres have constant temperatures, energy balance for each gives



$$

P=Q_{1}+\left(2 \pi R^{2} \sigma\right) T_{1}^{4}, \quad Q_{2}=A T_{2}^{4}

$$



Finally, since the engine is reversible,



$$

\frac{Q_{1}}{T_{1}}=\frac{Q_{2}}{T_{2}}

$$



By combining the first three equations, and defining $x=T_{2} / T_{1}$, we have



$$

P_{w}=Q_{1}-Q_{2}=P-\left(2 \pi R^{2} \sigma\right)\left(T_{1}^{4}+T_{2}^{4}\right)=P-\left(2 \pi R^{2} \sigma\right) T_{1}^{4}\left(1+x^{4}\right) .

$$



This is not yet in terms of $P$ and $x$, so now we use the reversibility condition,



$$

\frac{P-A T_{1}^{4}}{T_{1}}=\frac{A T_{2}^{4}}{T_{2}}

$$



which simplifies to



$$

P=\left(2 \pi R^{2} \sigma\right)\left(T_{1}^{4}+T_{1} T_{2}^{3}\right)=\left(2 \pi R^{2} \sigma\right) T_{1}^{4}\left(1+x^{3}\right)

$$



Plugging this in above, we find



$$

P_{w}=P-\frac{P}{1+x^{3}}\left(1+x^{4}\right)=\frac{x^{3}(1-x)}{1+x^{3}} P

$$

Answer: $P_{w}=\frac{x^{3}(1-x)}{1+x^{3}} P$
Problem: d. Estimate the maximum possible value of $P_{w}$ as a fraction of $P$, to one significant figure. Briefly explain how you obtained this estimate.

Solution: There are many ways to get the required answer. For example, by sketching the function, one can see that there is a unique maximum at an intermediate value of $x$, and furthermore that this maximum is at $x>0.5$, because of the rapid rise of the $x^{3}$ factor. One could then compute $P_{w} / P$ with a calculator at a few trial values such as $x=0.5,0.7,0.9$, which are already enough to get the desired accuracy.



The optimum value is $x=0.69$, at which point



$$

P_{w}^{\max }=0.077 P

$$







Hence in this model, at most $7.7 \%$ of solar energy can be converted into wind energy. Any answer within $15 \%$ of this value was accepted.



Answer: $P_{w}^{\max }=0.077 P$
Problem: a. For large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that



$$

v=\mu E

$$



where $E$ is the local electric field and $\mu$ is the charge mobility.

i. In the steady state, there is a nonzero but time-independent density of charges between the two plates. Let the charge density at position $x$ be $\rho(x)$. Use charge conservation to find a relationship between $\rho(x), v(x)$, and their derivatives.

Solution: In the steady state, the current is the same everywhere. Consider the region $(x, x+d x)$. The time it takes for the charge in the second region to leave is $\frac{\mathrm{d} x}{v(x)}$. The amount of charge that leaves is $\rho A \mathrm{~d} x$. The current is thus given by $\rho A v$, so $\rho v$ is constant. Alternatively, one can write this as



$$

v \frac{\mathrm{d} \rho}{\mathrm{d} x}+\rho \frac{\mathrm{d} v}{\mathrm{~d} x}=0

$$



Both forms were accepted.



Answer: $v \frac{\mathrm{d} \rho}{\mathrm{d} x}+\rho \frac{\mathrm{d} v}{\mathrm{~d} x}=0$
Problem: iii. Suppose that in the steady state, conditions have been established so that $V(x)$ is proportional to $x^{b}$, where $b$ is an exponent you must find, and the current is nonzero. Derive an expression for the current in terms of $V_{0}$ and the other given parameters.

Solution: We have that



$$

\rho \frac{d v}{d x}+v \frac{d \rho}{d x}=0

$$



and now that $v=-\mu \frac{d V}{d x}$, so substituting in Poisson's equation gives us that



$$

\left(\frac{d^{2} V}{d x^{2}}\right)^{2}+\frac{d V}{d x}\left(\frac{d^{3} V}{d x^{3}}\right)=0

$$



Using $V(x)=-V_{0}(x / d)^{b}$ gives



$$

b(b-1) b(b-1)=-b b(b-1)(b-2)

$$



The solution with $b=0$ cannot satisfy the boundary conditions, while $b=1$ has zero current. Assuming $b$ is neither of these values, we have $b-1=-(b-2)$, so $b=3 / 2$. Substituting gives



$$

v=-\frac{3 V_{0} \mu x^{1 / 2}}{2 d^{3 / 2}}

$$



and



$$

\rho=-\frac{3 V_{0} \epsilon_{0}}{4 d^{3 / 2} x^{1 / 2}}

$$



so



$$

I=\rho A v=\frac{9 \epsilon_{0} \mu A V_{0}^{2}}{8 d^{3}}

$$



with the current flowing from left to right.

Answer: $I=\frac{9 \epsilon_{0} \mu A V_{0}^{2}}{8 d^{3}}$
Problem: i. Assume that in the steady state, conditions have been established so that a nonzero, steady current flows, and the electric potential again satisfies $V(x) \propto x^{b^{\prime}}$, where $b^{\prime}$ is another exponent you must find. Derive an expression for the current in terms of $V_{0}$ and the other given parameters.

Solution: We again have that $V(x)=V_{0}(x / d)^{b}$. Note that from Poisson's equation, $\frac{d \rho}{d x}=-\epsilon_{0} \frac{d^{3} V}{d x^{3}}$, so we need $b=3$ for this expression to be constant. Therefore,



$$

I=\frac{6 \mu k_{B} T A \epsilon_{0} V_{0}}{q d^{3}}

$$

Answer: $I=\frac{6 \mu k_{B} T A \epsilon_{0} V_{0}}{q d^{3}}$
Problem: ii. At roughly what voltage $V_{0}$ does the system transition from this regime to the high voltage regime of the previous part?

Solution: We find the crossover voltage by equating our answers in previous parts, to get



$$

\frac{6 \mu k_{B} T A \epsilon_{0} V_{0}}{q d^{3}}=\frac{9 \epsilon_{0} \mu A V_{0}^{2}}{8 d^{3}}

$$



or



$$

V_{0}=\frac{16 k_{B} T}{3 q}

$$

Answer: $V_{0}=\frac{16 k_{B} T}{3 q}$