weblab-llm-competition-2025-bridge/neko-prelim-HLE_SFT_OlympiadBench huggingface dataset:
Problem: 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 $Q=4.48 \times 10^{-10} \mathrm{C}$.

Answer: $4.48 \times 10^{-10}$
Problem: (a) Where should a pin be placed on the optical axis such that its image is formed at the same place?

Solution: Image of object will coincide with it if ray of light after refraction from the concave surface fall normally on concave mirror so formed by silvering the convex surface. Or image after refraction from concave surface should form at centre of curvature of concave mirror or at a distance of $20 \mathrm{~cm}$ on same side of the combination. Let $x$ be the distance of pin from the given optical system.

Using,

$$
\frac{\mu_{2}}{\nu}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}
\tag{8}
$$

With proper signs,

$$
\frac{1.5}{-20}-\frac{1}{-x}=\frac{1.5-1}{-60}
\tag{9}
$$

Solve to get $x=15 \mathrm{~cm}$

Answer: 15
Problem: (b) Equation 4 is a version of the damped harmonic oscillator, and can be solved by guessing a solution $\eta=\alpha e^{\lambda t}$.

Plugging in this guess, what must $\lambda$ be?

Solution: We can plug in $\eta=\alpha e^{\lambda t}$ :

$$
0=\lambda^{2} \alpha e^{\lambda t}-2 i \lambda \Omega \alpha e^{\lambda t}-\Omega^{2} \alpha e^{\lambda t}
\tag{23}
$$

Then we can cancel common factors and find that

$$
0=\lambda^{2}-2 i \lambda \Omega-\Omega^{2}=(\lambda-i \Omega)^{2}
\tag{24}
$$

We see that $\lambda=i \Omega$.

Answer: $\lambda=i \Omega$
Problem: (c) Using your answer to part (b), and defining $\alpha=A e^{i \phi}$ where $A$ and $\phi$ are real, find $\mathbf{x}(\mathbf{t})$ and $\mathbf{y}(\mathbf{t})$.

This is the trajectory for a particle which is stationary with respect to the symmetry axis. While not required for this problem, an additional guess would reveal that $\eta=\beta t e^{\lambda t}$ is also a solution.

Solution: Using our answer to part (b) and $\alpha=A e^{i \phi}$, we have

$$
\eta(t)=A e^{i(\Omega t+\phi)}
\tag{25}
$$

Using the Euler identity, we have

$$
x(t)+i y(t)=A \cos (\Omega t+\phi)+i A \sin (\Omega t+\phi)
\tag{26}
$$

The real and imaginary parts become

$$
\begin{aligned}
& x(t)=A \cos (\Omega t+\phi) \\
& y(t)=A \sin (\Omega t+\phi)
\end{aligned}
\tag{27}
$$

Answer: $x(t)=A \cos (\Omega t+\phi)$ , $y(t)=A \sin (\Omega t+\phi)$
Problem: (d) The one-dimensional diffusion equation (also called the "heat equation") is given (for a free particle) by

$$
\frac{\partial \psi}{\partial t}=a \frac{\partial^{2} \psi}{\partial x^{2}}
\tag{5}
$$

A spatial wave can be written as $\sim e^{i k x}$ (larger $k$ 's correspond to waves oscillating on smaller length scales). Guessing a solution $\psi(x, t)=A e^{i k x-i \omega t}$, find $\omega$ in terms of k. A relationship of this time is called a "dispersion relation."

Solution: Consider the given differential equation:

$$
\frac{\partial \psi}{\partial t}=a \frac{\partial^{2} \psi}{\partial x^{2}}
\tag{28}
$$

We can plug in $\psi(x, t)=A e^{i k x-i \omega t}$ to find

$$
-i \omega A e^{i k x-i \omega t}=-k^{2} a A e^{i k x-i \omega t}
\tag{29}
$$

so that

$$
\omega=-i k^{2} a
\tag{30}
$$

Answer: $\omega=-i k^{2} a$
Problem: (e) The most important equation of non-relativistic quantum mechanics is the Schrödinger equation, which is given by

$$
i \hbar \frac{\partial \psi}{\partial t}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \psi}{\partial x^{2}}
\tag{6}
$$

Using your answer to part (d), what is the dispersion relation of the Schrödinger equation?

Solution: We see that the free Schrödinger equation takes the form of Equation 28, but with

$$
a=\frac{i \hbar}{2 m}
\tag{31}
$$

Then, using our answer from part (d), we have

$$
\omega=\frac{\hbar k^{2}}{2 m}
\tag{32}
$$

Answer: $\omega=\frac{\hbar k^{2}}{2 m}$
Problem: (g) The theory of relativity instead posits that the energy of a particle is given by $E=\sqrt{p^{2} c^{2}+m^{2} c^{4}}$. In accordance with this, we can try to guess a relativistic version of the Schrödinger equation:

$$
\frac{1}{c^{2}} \frac{\partial^{2} \phi}{\partial t^{2}}-\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{m^{2} c^{2}}{\hbar^{2}} \phi=0
\tag{7}
$$

This is called the Klein-Gordon equation. Using the same guess as before, find $\omega$ in terms of $\mathrm{k}$.

Hint: If you are careful, you should find that there is an infinite continuum of energy states extending down to negative infinity. This apparently mathematical issue hints at the existence of antimatter, and ultimately demonstrates to us that we must formulate quantum field theory to properly describe relativistic quantum physics.

Solution: We consider the Klein-Gordon equation:

$$
\frac{1}{c^{2}} \frac{\partial^{2} \phi}{\partial t^{2}}-\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{m^{2} c^{2}}{\hbar^{2}} \phi=0
\tag{35}
$$

We can, as before, plug in a guess $\phi(x, t)=A e^{i k x-i \omega t}$. This yields

$$
-\frac{\omega^{2}}{c^{2}} A e^{i k x-i \omega t}+k^{2} A e^{i k x-i \omega t}+\frac{m^{2} c^{2}}{\hbar^{2}} A e^{i k x-i \omega t}=0
\tag{36}
$$

Cancelling out common terms, we see that

$$
\omega^{2}=k^{2} c^{2}+\frac{m^{2} c^{4}}{\hbar^{2}}
\tag{37}
$$

or

$$
\omega= \pm \sqrt{k^{2} c^{2}+\frac{m^{2} c^{4}}{\hbar^{2}}}
\tag{38}
$$

Note that both positive and negative $\omega$ solve the Klein-Gordon equation.

Answer: $\sqrt{k^{2} c^{2}+\frac{m^{2} c^{4}}{\hbar^{2}}}$, $-\sqrt{k^{2} c^{2}+\frac{m^{2} c^{4}}{\hbar^{2}}}$
Problem: (c) The uniqueness theorem allows us to use "image" charges in certain settings to describe the system. Consider one such example: There is a point-like charge $q$ at a distance $L$ from a metallic sphere of radius $R$ attached to the ground. As you argued in part (a), sphere will be polarized to make sure the electric potential is constant throughout its body. Since it is attached to the ground, the constant potential will be zero. Place an image charge inside the sphere to counter the non-uniform potential of the outer charge $q$ on the surface. Where should this charge be placed, and what is its value?

Solution: Take the origin of our coordinate system at the center of the sphere and align the axes so that the coordinates of our charge is $(x, y)=(L, 0)$. Now suppose we place an image charge $q_{0}$ at $(x, 0)$ where $0<r<R$. Then the potential at the point $(R \cos \theta, R \sin \theta)$ on the surface becomes

$$
V(\theta)=\frac{q_{0}}{x_{\theta}}+\frac{q}{L_{\theta}}
$$

where

$$
\begin{aligned}
x_{\theta} & =\sqrt{x^{2}+R^{2}-2 x R \cos \theta} \\
L_{\theta} & =\sqrt{L^{2}+R^{2}-2 L R \cos \theta}
\end{aligned}
$$

We want this image charge to satisfy $V(\theta)=0 \forall \theta \in[0,2 \pi]$. Then we have,

$$
\begin{aligned}
\frac{q}{q_{0}} & =-\frac{\sqrt{L^{2}+R^{2}-2 L R \cos \theta}}{\sqrt{x^{2}+R^{2}-2 x R \cos \theta}} \\
& =-\sqrt{\frac{L}{x} \frac{\sqrt{\frac{L^{2}+R^{2}}{L}-2 R \cos \theta}}{\sqrt{\frac{x^{2}+R^{2}}{x}-2 R \cos \theta}}}
\end{aligned}
$$

Consequently,

$$
\begin{aligned}
& \frac{L^{2}+R^{2}}{L}=\frac{x^{2}+R^{2}}{x} \text { and } \\
& \frac{q}{q_{0}}=-\sqrt{\frac{L}{x}}
\end{aligned}
$$

Solution to the first equation with $x<R$ condition gives $x=\frac{R^{2}}{L}$, and from the second equation, $q_{0}$ can be found as $q_{0}=-q \frac{R}{L}$. By azimuthal symmetry, we conclude that if the



potential is zero on surface crossing the $x-y$ plane with the image charge $q_{0}=-q \frac{R}{L}$ at $x=\frac{R^{2}}{L}$, then it is zero everywhere throughout the surface.

Answer: $x=\frac{R^{2}}{L}$ , $q_{0}=-q \frac{R}{L}$
Problem: (e) Find the force of attraction between the charge and the sphere.

Solution: From part (d) and (c), this is simply

$$
F=\frac{1}{4 \pi \epsilon_{0}} \frac{q^{2} \frac{R}{L}}{\left(L-\frac{R^{2}}{L}\right)^{2}}
$$

Answer: $F=\frac{1}{4 \pi \epsilon_{0}} \frac{q^{2} \frac{R}{L}}{\left(L-\frac{R^{2}}{L}\right)^{2}}$
Problem: (a) What value of $\theta$ maximizes the range?

Solution: From the equation of motion $\frac{d \mathbf{v}}{d t}=\mathbf{g}$ we find that $\mathbf{v}=\mathbf{v}_{0}+\mathbf{g} t$. Then the position of the mass after time $t$ is given by $\mathbf{r}=\int_{0}^{t} \mathbf{v} d t=\int_{0}^{t}\left(\mathbf{v}_{0}+\mathbf{g} t\right) d t=\mathbf{v}_{0} t+\frac{1}{2} \mathbf{g} t^{2}$. Denoting the final time as $T$ and the position as $\mathbf{R}$, we find that

$$
\mathbf{R e}_{\mathbf{y}}=0=v_{0} \sin \theta T-\frac{1}{2} g T^{2}
\tag{1}
$$
$$
\mathbf{R e}_{\mathbf{x}}=L=v_{0} \cos \theta T
\tag{2}
$$

where $L$ is the range of the mass. From equations (1) and (2), we get

$$
L=\frac{2 v_{0}^{2}}{g} \sin \theta \cos \theta=\frac{v_{0}^{2}}{g} \sin (2 \theta)
$$

Maximum $L$ is achieved when $\sin (2 \theta)=1$. Therefore, $\theta=\frac{\pi}{4}$

Answer: $\frac{\pi}{4}$
Problem: (b) What value of $\theta$ maximizes the surface area under the trajectory curve?

Solution: The surface area covered by the trajectory can be found by

$$
\mathbf{A}=\frac{1}{2} \int \mathbf{r} \times \mathbf{d} \mathbf{r}=\frac{1}{2} \int(\mathbf{r} \times \mathbf{v}) d t
$$

Since the motion occurs in a plane, the surface area is always orthogonal to the plane. Therefore, the area covered by the trajectory can be found by calculating the magnitude of the vector $\mathbf{A}$. Now, inserting the equations for the position vector and the velocity as found in (1), we get

$$
\begin{aligned}
\mathbf{A}=\frac{1}{2} \int(\mathbf{r} \times \mathbf{v}) d t & =\frac{1}{4} \int\left(\mathbf{v}_{\mathbf{0}} \times \mathbf{g}\right) t^{2} d t \\
& =\frac{1}{4}\left(\mathbf{v}_{\mathbf{0}} \times \mathbf{g}\right) \int t^{2} d t \\
& \left.=\frac{1}{12} \mathbf{v}_{\mathbf{0}} \times \mathbf{g}\right) T^{3}
\end{aligned}
$$



As found in (1), $T$ is proportional to $\sin \theta$, and $\mathbf{v}_{\mathbf{0}} \times \mathbf{g}$ is proportional to $\cos \theta$. Therefore, the angle that maximizes the area maximizes the function

$$
f(\theta)=\cos \theta \sin ^{3} \theta
$$

Taking derivative of $f(\theta)$, we get the equation

$$
\frac{d f}{d \theta}=\sin ^{2} \theta\left(3 \cos ^{2} \theta-\sin ^{2} \theta\right)
$$

which gives zero at $\theta=0$ or $\theta=\arctan \sqrt{3}=\pi / 3$. Since $\theta=0$ corresponds to the minimal area (no trajectory, $\mathrm{A}=0$ ), the answer is $\theta=\pi / 3$

Answer: $\pi / 3$
Problem: (c) What is the answer for (a), if the point mass is thrown from an apartment of height $h$ ?


Now assume that we have a frictional force that is proportional to the velocity vector, such that the equation of motion is as follows

$$
\frac{d \vec{v}}{d t}=\vec{g}-\beta \vec{v}
$$

Solution: If the initial position of the mass is $\mathbf{r}_{\mathbf{0}}=h \mathbf{e}_{\mathbf{y}}$, equation (1) and (2) becomes

$$
\mathbf{R e}_{\mathbf{y}}=-h=v_{0} \sin \theta T-\frac{1}{2} g T^{2}
\tag{3}
$$
$$
\mathbf{R e}_{\mathbf{x}}=L=v_{0} \cos \theta T
\tag{4}
$$

Maximum $L$ occurs when $\frac{d L}{d \theta}=0$. Therefore, from (4), we obtain

$$
\frac{d L}{d \theta}=-v_{0} \sin \theta+v_{0} \cos \theta \frac{d T}{d \theta}=0
\tag{5}
$$

Taking derivative of both sides of (3) to find $\frac{d T}{d \theta}$ and inserting it into (5), we get

$$
\begin{aligned}
0 & =v_{0} \cos \theta T+\frac{v_{0} \sin \theta-g T}{\cos \theta} \sin \theta T \\
& \Longrightarrow v_{0}=g T \sin \theta
\end{aligned}
$$

Inserting this equation to (3), we obtain

$$
-h=\frac{v_{0}^{2}}{g}-\frac{1}{2} \frac{v_{0}^{2}}{g \sin \theta}
$$

Therefore, the answer is $\theta=\arcsin \left(\frac{1}{\sqrt{2}} \frac{v_{0}}{\sqrt{v_{0}^{2}+g h}}\right)$

Answer: $\arcsin \left(\frac{1}{\sqrt{2}} \frac{v_{0}}{\sqrt{v_{0}^{2}+g h}}\right)$
Problem: (d) Supposing that $\beta<<\frac{g}{v_{0}}$, find the duration of the motion $T$.

Solution: Under the presence of friction, equation of motion for $y$ component becomes

$$
\frac{d v_{y}}{d t}=-g-\beta v_{y}
\tag{6}
$$

Solving (6), we get

$$
v_{y}=v_{0 y}-\left(v_{0 y}+\frac{g}{\beta}\right)\left(1-e^{-\beta t}\right)
\tag{7}
$$



$y$ component of the final position vector can be found by integrating $v_{y}$,

$$
y=0=-\frac{g T}{\beta}+\frac{\left(v_{0 y}+\frac{g}{\beta}\right)}{\beta}\left(1-e^{-\beta T}\right)
\tag{8}
$$

Since $\beta$ is small compared to the dimensions of the system, we can use Taylor expansion for (8) to find an analytic equation for $T$.

$$
\begin{aligned}
& g T=\left(v_{0 y}+\frac{g}{\beta}\right)\left(1-e^{-\beta T}\right) \approx\left(v_{0 y}+\frac{g}{\beta}\right)\left(\beta T-\frac{1}{2} \beta^{2} T^{2}\right) \\
& \Longrightarrow T=\frac{2 v_{0 y}}{\beta v_{0 y}+g}
\end{aligned}
$$

Therefore, the answer is

$$
T=\frac{2 v_{0} \sin \theta}{g} \frac{1}{1+\frac{\beta v_{0} \sin \theta}{g}}
$$

which is less than the value found for frictionless trajectory, as expected.

Answer: $T=\frac{2 v_{0} \sin \theta}{g} \frac{1}{1+\frac{\beta v_{0} \sin \theta}{g}}$
Problem: (a) Compute the electric potential inside and outside the sphere.

Solution: From Gauss's theorem:

$$
\frac{Q}{\varepsilon_{0}}=4 \pi R^{2} E
$$

So:

$$
E=\frac{Q}{4 \pi \varepsilon_{0}}
$$

But:

$$
\mathbf{E}=-\nabla \Phi
$$

In this case $E=-\frac{\partial \Phi}{\partial r}$. To obtain a zero potential at infinity we get:

$$
\Phi_{+}=\frac{Q}{4 \pi \varepsilon_{0} r}
$$

The electric field inside the sphere is 0 (no charges), so the potential is constant. Because it is continuous on the surface of the sphere (no jumps):

$$
\Phi_{-}=\frac{Q}{4 \pi \varepsilon_{0} R}
$$

Answer: $\Phi_{-}=\frac{Q}{4 \pi \varepsilon_{0} R}$,$\Phi_{+}=\frac{Q}{4 \pi \varepsilon_{0} r}$
Problem: (a) Knowing that Legendre polynomials are orthogonal $\left(\left\langle P_{m}(x), P_{n}(x)\right\rangle=0\right.$ if $m \neq n)$ and $\operatorname{deg} P_{n}(x)=n$, obtain $P_{2}(x)$ and $P_{3}(x)$. For reaching the usual and most convenient form of these polynomials, divide your results by the norm: $\left\|P_{n}(x)\right\|=\frac{2}{2 n+1}$.

Let us now consider a sphere of radius $R$ centered at the origin. Suppose a point charge $q$ is put at the origin and that this is the only charge inside or outside the sphere. Furthermore, the potential is $\Phi=V_{0} \cos \theta$ on the surface of the sphere.

We know that we can write the potential induced by the charge on the sphere (without taking into account $q$ ) in the following way:

$$
\begin{array}{ll}
\Phi_{-}=\sum_{n=0}^{\infty} A_{n} r^{n} P_{n}(\cos \theta), & r<R \\
\Phi_{+}=\sum_{n=0}^{\infty} \frac{B_{n}}{r^{n+1}} P_{n}(\cos \theta), & r>R
\end{array}
$$

Solution: $$
P_{2}(x)=C_{2}\left(x^{2}+\lambda_{1} x+\lambda_{0}\right)
$$

Because $\left\langle P_{2}(x), P_{1}(x)\right\rangle=\int_{-1}^{1} P_{2}(x) P_{1}(x) d x=0$ we get $\lambda_{1}=0$. Because $\int_{-1}^{1} x^{2} d x=2 / 3$ and $\int_{-1}^{1} d x=$ 2 , we get $\lambda_{0}=-1 / 3$. The condition for the norm implies $P_{2}(x)=\frac{1}{2}\left(3 x^{2}-1\right)$. Similarly:

$$
P_{3}(x)=C_{3}\left(x^{3}+\lambda_{2} x^{2}+\lambda_{1} x+\lambda_{0}\right)
$$

Through an identical reasoning:

$$
P_{3}=\frac{1}{2}\left(5 x^{3}-3 x\right)
$$

Answer: $P_{2}(x)=C_{2}\left(x^{2}+\lambda_{1} x+\lambda_{0}\right)$ , $P_{3}=\frac{1}{2}\left(5 x^{3}-3 x\right)$
Problem: (b) Compute the electric potential both inside and outside the sphere.

Solution: The potential is given by either Poisson's or Laplace's equation:

$$
\begin{array}{ll}
\nabla^{2} \Phi_{-}=-\frac{q}{\varepsilon_{0}} \delta(r), & r<R \\
\nabla^{2} \Phi_{+}=0, & r>R
\end{array}
$$



The general solutions finite in the respective regions, taking account of the symmetry, are

$$
\begin{aligned}
& \Phi_{-}=\frac{q}{4 \pi \varepsilon_{0} r}+\sum_{n=0}^{\infty} A_{n} r^{n} P_{n}(\cos \theta), \quad r<R \\
& \Phi_{+}=\sum_{n=0}^{\infty} \frac{B_{n}}{r^{n+1}} P_{n}(\cos \theta), \quad r>R
\end{aligned}
$$

Then from the condition $\Phi_{-}=\Phi_{+}=V_{0} \cos \theta$ at $r=R$ we obtain $A_{0}=-\frac{q}{4 \pi \varepsilon_{0} R}, A_{1}=\frac{V_{0}}{R}, B_{1}=$ $V_{0} R^{2}, B_{0}=0, A_{n}=B_{n}=0$ for $n \neq 0,1$, and hence

$$
\begin{aligned}
& \Phi_{-}=\frac{q}{4 \pi \varepsilon_{0} r}-\frac{q}{4 \pi \varepsilon_{0} R}+\frac{V_{0} \cos \theta}{R} r, \quad r<R ; \\
& \Phi_{+}=\frac{V_{0} R^{2}}{r^{2}} \cos \theta, \quad r>R .
\end{aligned}
$$

Answer: $\Phi_{-}=\frac{q}{4 \pi \varepsilon_{0} r}-\frac{q}{4 \pi \varepsilon_{0} R}+\frac{V_{0} \cos \theta}{R} r$,$\Phi_{+}=\frac{V_{0} R^{2}}{r^{2}} \cos \theta$
Problem: (b) Any real inductor has undesired, or parasitic, resistance. We can model the real inductor as an ideal inductor $L$ in series with a parasitic resistance $R$.

Due to the thermal noise $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$ of its parasitic resistance, this (real) inductor will support a nonzero per-frequency mean-squared current, $\frac{d\left\langle I^{2}\right\rangle}{d f}$, even when both sides of the inductor are grounded. Compute $\frac{d\left\langle I^{2}\right\rangle}{d f}$ as a function of $f, L, T$ and $R$.

Solution: The total impedance of the circuit is

$$
Z=R+i \omega L=R+2 \pi i f L \Rightarrow|Z|=\sqrt{R^{2}+4 \pi^{2} f^{2} L^{2}}
\tag{9}
$$

Therefore, the circuit equation $V=Z I \Rightarrow\left\langle V^{2}\right\rangle=\left|Z^{2}\right|\left\langle I^{2}\right\rangle$ implies

$$
\frac{d\left\langle I^{2}\right\rangle}{d f}=\frac{1}{\left|Z^{2}\right|} \frac{d\left\langle V^{2}\right\rangle}{d f}=\frac{4 k T R}{R^{2}+4 \pi^{2} f^{2} L^{2}}
\tag{10}
$$

Answer: $\frac{4 k T R}{R^{2}+4 \pi^{2} f^{2} L^{2}}$
Problem: (b) Find the period $T$ and angular frequency $\omega$ of the orbital motion.

Solution: $$
\frac{d^{2}}{d t^{2}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right)=-\frac{G\left(M_{1}+M_{2}\right)}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right)
$$

Using polar coordinate, the time derivative of a position vector $\vec{r}$ with constant magnitude,

$$
\begin{aligned}
& \frac{d}{d t} \vec{r}=\frac{d}{d t} r \hat{r}=r \frac{d}{d t} \hat{r}=\vec{\omega} \times \vec{r} \\
& \frac{d 2}{d t^{2}} \vec{r}=\vec{\omega} \times(\vec{\omega} \times \vec{r})=-\omega^{2} \vec{r}
\end{aligned}
$$

Since $\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|$ is constant,

$$
\begin{gathered}
\omega=\sqrt{\frac{G\left(M_{1}+M_{2}\right)}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}} \\
T=2 \pi / \omega=2 \pi \sqrt{\frac{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}{G\left(M_{1}+M_{2}\right)}}
\end{gathered}
$$

Answer: $\omega=\sqrt{\frac{G\left(M_{1}+M_{2}\right)}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}}$ , $T=2 \pi \sqrt{\frac{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}{G\left(M_{1}+M_{2}\right)}}$
Problem: (a) Express $\alpha$ in terms of $M_{1}$ and $M_{2}$.

Solution: $$
\begin{gathered}
M_{1}(-\alpha R)+M_{2}((1-\alpha) R)=0 \\
\alpha=\frac{M_{2}}{M_{1}+M_{2}}
\end{gathered}
$$

Answer: $\alpha=\frac{M_{2}}{M_{1}+M_{2}}$
Problem: (b) Let $\rho_{1}(t)$ and $\rho_{2}(t)$ be the distances from $m$ to $M_{1}$ and $M_{2}$ respectively. Express $\rho_{1}(t)$ and $\rho_{2}(t)$ in terms of the coordinates and constants given.

Solution: $$
\begin{gathered}
\rho_{1}(t)=\sqrt{(x(t)+\alpha R)^{2}+(y(t))^{2}} \\
\rho_{2}(t)=\sqrt{(x(t)-(1-\alpha) R)^{2}+(y(t))^{2}}
\end{gathered}
$$

Answer: $\rho_{1}(t)=\sqrt{(x(t)+\alpha R)^{2}+(y(t))^{2}}$ , $\rho_{2}(t)=\sqrt{(x(t)-(1-\alpha) R)^{2}+(y(t))^{2}}$
Problem: (c) By considering the centrifugal acceleration $\omega^{2} \vec{r}$ and Coriolis acceleration $-2 \omega \times$ $\vec{v}$, find the acceleration $\frac{d^{2}}{d t^{2}} \vec{r}$ of the third mass in terms of the coordinates and constants given, including $\rho_{1}$ and $\rho_{2}$.

Solution: $$
\begin{gathered}
m \frac{d^{2}}{d t^{2}} \vec{r}=-G m M_{1} \frac{\vec{r}-\overrightarrow{r_{1}}}{\rho_{1}^{3}}-G m M_{2} \frac{\vec{r}-\overrightarrow{r_{2}}}{\rho_{2}^{3}}+m \omega^{2} \vec{r}-2 m \omega \times \vec{r} \\
\frac{d^{2}}{d t^{2}} \vec{r}=-G M_{1} \frac{\vec{r}-\overrightarrow{r_{1}}}{\rho_{1}^{3}}-G M_{2} \frac{\vec{r}-\overrightarrow{r_{2}}}{\rho_{2}^{3}}+\omega^{2} \vec{r}-2 \omega \times \vec{r}
\end{gathered}
$$

Answer: $\frac{d^{2}}{d t^{2}} \vec{r}=-G M_{1} \frac{\vec{r}-\vec{r_{1}}}{\rho_{1}^{3}}-G M_{2} \frac{\vec{r}-\vec{r_{2}}}{\rho_{2}^{3}}+\omega^{2} \vec{r}-2 \omega \times \vec{r}$
Problem: (d) Express $\frac{d^{2} x}{d t^{2}}$ and $\frac{d^{2} y}{d t^{2}}$ in terms of $U$, where $U=-\frac{G M_{1}}{\rho_{1}}-\frac{G M_{2}}{\rho_{2}}-\frac{\omega^{2}}{2}\left(x^{2}+y^{2}\right)$.

Solution: $$
\begin{gathered}
\ddot{x}=-\frac{G M_{1}(x+\alpha R)}{\rho_{1}^{3}}-\frac{G M_{2}(x-(1-\alpha) R)}{\rho_{2}^{3}}+\omega^{2} x+2 \omega \dot{y} \\
\ddot{y}=-\frac{G M_{1} y}{\rho_{1}^{3}}-\frac{G M_{2} y}{\rho_{2}^{3}}+\omega^{2} y-2 \omega \dot{x}
\end{gathered}
$$

Then,

$$
\begin{aligned}
\ddot{x} & =2 \omega \dot{y}-\frac{\partial U}{\partial x} \\
\ddot{y} & =-2 \omega \dot{x}-\frac{\partial U}{\partial y}
\end{aligned}
$$

Answer: $\ddot{x} =2 \omega \dot{y}-\frac{\partial U}{\partial x}$ , $\ddot{y} =-2 \omega \dot{x}-\frac{\partial U}{\partial y}$