di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Electromagnetism

Electric experiments in the magnetosphere of the earth. 

In May 1991 the spaceship Atlantis will be placed in orbit around the earth. We shall assume that this orbit will be circular and that it lies in the earth's equatorial plane.

At some predetermined moment the spaceship will release a satellite $S$, which is attached to a conducting rod of length $\mathrm{L}$. We suppose that the rod is rigid, has negligible mass, and is covered by an electrical insulator. We also neglect all friction. Let $\alpha$ be the angle that the rod makes to the line between the Atlantis and the centre of the earth. (see Fig. 1).

S also lies in the equatorial plane.

Assume that the mass of the satellite is much smaller than that of the Atlantis, and that $\mathrm{L}$ is much smaller than the radius of the orbit.

<img_4434>

Figure 1 The spaceship Atlantis (A) with a satellite (S) in an orbit around the earth. The orbit lies in the earth's equatorial plane.

The magnetic field (B) is perpendicular to the diagram and is directed towards the reader.

Question: $a_{1}$ - Deduce for which value(s) of $\alpha$ the configuration of the spaceship and satellite remain unchanged (with respect to the earth)? In other words, for which value(s) of $\alpha$ is $\alpha$ constant?

Solution:
$a_{1}$ - Since $m_{2}<<m_{1}$, the Atlantis travels around the earth with a constant speed. The motion of the satellite is composed of the circular motion of the Atlantis about the earth and (possibly) a circular motion of the satellite about the Atlantis.

For $\mathrm{m}_{1}$ we have:

$$
m_{1} \cdot \Omega^{2} \cdot R=\frac{G \cdot m_{1} \cdot m_{a}}{R^{2}} \rightarrow \Omega^{2}=\frac{G \cdot m_{a}}{R^{3}}
$$

<img_4332>

For $\mathrm{m}_{2}$ we have:

$$
m_{2} \cdot L \cdot \ddot{\alpha}=-\left(F_{g}-F_{c}\right) \cdot \sin (\alpha)=-\left(\frac{G \cdot m_{2} \cdot m_{a}}{(R-L \cdot \cos (\alpha))^{2}}-m_{2} \cdot \Omega^{2} \cdot(R-L \cdot \cos (\alpha))\right) \cdot \sin (\alpha)
$$

Using the approximation:

$$
\frac{1}{(R-L \cdot \cos (\alpha))^{2}} \approx \frac{1}{R^{2}}+\frac{2 \cdot L \cdot \cos (\alpha)}{R^{3}}
$$

and equation (1), one finds:

$$
L \cdot \ddot{\alpha}=-\left(\frac{G \cdot m_{a}}{R^{2}}+\frac{2 \cdot G \cdot m_{a}}{R^{3}} \cdot L \cdot \cos (\alpha)-\frac{G \cdot m_{a}}{R^{3}} \cdot R+\frac{G \cdot m_{a}}{R^{3}} \cdot L \cdot \cos (\alpha)\right) \cdot \sin (\alpha)
$$

so:

$$
\ddot{\alpha}+3 \cdot \Omega^{2} \cdot \sin (\alpha) \cdot \cos (\alpha)=0
\tag{2}
$$

If $\alpha$ is constant: $\ddot{\alpha}=0 \quad->\quad \sin (\alpha)=0 \quad->\quad \alpha=0 ; \quad \alpha=\pi$

$$
->\quad \cos (\alpha)=0 \quad->\quad \alpha=\pi / 2 ; \quad \alpha=3 \pi / 2
$$

Final answer: $0,\pi,\frac{\pi}{2},\frac{3\pi}{2}$
Subfield: Electromagnetism

Electric experiments in the magnetosphere of the earth. 

In May 1991 the spaceship Atlantis will be placed in orbit around the earth. We shall assume that this orbit will be circular and that it lies in the earth's equatorial plane.

At some predetermined moment the spaceship will release a satellite $S$, which is attached to a conducting rod of length $\mathrm{L}$. We suppose that the rod is rigid, has negligible mass, and is covered by an electrical insulator. We also neglect all friction. Let $\alpha$ be the angle that the rod makes to the line between the Atlantis and the centre of the earth. (see Fig. 1).

S also lies in the equatorial plane.

Assume that the mass of the satellite is much smaller than that of the Atlantis, and that $\mathrm{L}$ is much smaller than the radius of the orbit.

<img_4434>

Figure 1 The spaceship Atlantis (A) with a satellite (S) in an orbit around the earth. The orbit lies in the earth's equatorial plane.

The magnetic field (B) is perpendicular to the diagram and is directed towards the reader.
Context question:
$a_{1}$ - Deduce for which value(s) of $\alpha$ the configuration of the spaceship and satellite remain unchanged (with respect to the earth)? In other words, for which value(s) of $\alpha$ is $\alpha$ constant?
Context answer:
\boxed{$0,\pi,\frac{\pi}{2},\frac{3\pi}{2}$}


Question: $a_{2}$ - Discuss the stability of the equilibrium for each case.

Solution:
$\mathrm{a}_{2}$ - The situation is stable if the moment $M=m_{2} \cdot L \cdot \ddot{\alpha} \cdot L=m_{2} \cdot L^{2} \cdot \ddot{\alpha}$ changes sign in a manner opposed to that in which the sign of $\alpha-\alpha_{0}$ changes:

$\operatorname{sign}\left(\alpha-\alpha_{0}\right) \quad-\quad+\quad-\quad+\quad-\quad+\quad-\quad+\quad-\quad+$

| $\alpha$ | 0 | $\pi / 2$ | $\pi$ | $3 \pi / 2$ |  | $2 \pi$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\operatorname{sign}(\mathrm{M})$ |  | + | + | $-\quad+$ | + |  |
| $\alpha$ | 0 | $\pi / 2$ | $\pi$ | $3 \pi / 2$ |  | $2 \pi$ |

The equilibrium about the angles 0 en $\pi$ is thus stable, whereas that around $\pi / 2$ and $3 \pi / 2$ is unstable.
Subfield: Electromagnetism

Electric experiments in the magnetosphere of the earth. 

In May 1991 the spaceship Atlantis will be placed in orbit around the earth. We shall assume that this orbit will be circular and that it lies in the earth's equatorial plane.

At some predetermined moment the spaceship will release a satellite $S$, which is attached to a conducting rod of length $\mathrm{L}$. We suppose that the rod is rigid, has negligible mass, and is covered by an electrical insulator. We also neglect all friction. Let $\alpha$ be the angle that the rod makes to the line between the Atlantis and the centre of the earth. (see Fig. 1).

S also lies in the equatorial plane.

Assume that the mass of the satellite is much smaller than that of the Atlantis, and that $\mathrm{L}$ is much smaller than the radius of the orbit.

<img_4434>

Figure 1 The spaceship Atlantis (A) with a satellite (S) in an orbit around the earth. The orbit lies in the earth's equatorial plane.

The magnetic field (B) is perpendicular to the diagram and is directed towards the reader.
Context question:
$a_{1}$ - Deduce for which value(s) of $\alpha$ the configuration of the spaceship and satellite remain unchanged (with respect to the earth)? In other words, for which value(s) of $\alpha$ is $\alpha$ constant?
Context answer:
\boxed{$0,\pi,\frac{\pi}{2},\frac{3\pi}{2}$}


Context question:
$a_{2}$ - Discuss the stability of the equilibrium for each case.
Context answer:
\boxed{证明题}


Extra Supplementary Reading Materials:

Suppose that, at a given moment, the rod deviates from the stable configuration by a small angle. The system will begin to swing like a pendulum.

Question: $\mathrm{b}$ - Express the period of the swinging in terms of the period of revolution of the system around the earth.

Solution:
$\mathrm{b}$ - For small values of $\alpha$ equation (2) becomes:

$$
\ddot{\alpha}+3 \cdot \Omega^{2} \cdot \alpha=0
$$

This is the equation of a simple harmonic motion.

The square of the angular frequency is:

$$
\omega^{2}=3 . \Omega^{2}
$$

so:

$$
\omega=\Omega \cdot \sqrt{3} \quad \rightarrow \quad T_{1}=\frac{2 \pi}{\omega}=\frac{1}{3} \sqrt{3} \cdot\left(\frac{2 \pi}{\Omega}\right) \approx 0,58 \cdot T_{0}
$$
Subfield: Mechanics

A Three-body Problem and LISA 

<img_4539>

FIGURE 1 Coplanar orbits of three bodies.

Question: 1.1 Two gravitating masses $M$ and $m$ are moving in circular orbits of radii $R$ and $r$, respectively, about their common centre of mass. Find the angular velocity $\omega_{0}$ of the line joining $M$ and $m$ in terms of $R, r, M, m$ and the universal gravitational constant $G$.

Solution:
Let $\mathrm{O}$ be their centre of mass. Hence

$$
M R-m r=0
\tag{1}
$$
$$
\begin{aligned}
& m \omega_{0}^{2} r=\frac{G M m}{(R+r)^{2}}\\
& M \omega_{0}^{2} R=\frac{G M m}{(R+r)^{2}}
\end{aligned}
\tag{2}
$$

From Eq. (2), or using reduced mass, $\omega_{0}^{2}=\frac{G(M+m)}{(R+r)^{3}}$

Hence, 
$$
\omega_{0}^{2}=\frac{G(M+m)}{(R+r)^{3}}=\frac{G M}{r(R+r)^{2}}=\frac{G m}{R(R+r)^{2}}
\tag{3}
$$

Final answer: $\omega_{0}=\sqrt{\frac{G m}{R(R+r)^{2}}}$
Subfield: Thermodynamics

An Electrified Soap Bubble

A spherical soap bubble with internal air density $\rho_{i}$, temperature $T_{i}$ and radius $R_{0}$ is surrounded by air with density $\rho_{a}$, atmospheric pressure $P_{a}$ and temperature $T_{a}$. The soap film has surface tension $\gamma$, density $\rho_{s}$ and thickness $t$. The mass and the surface tension of the soap do not change with the temperature. Assume that $R_{0} \gg t$.

The increase in energy, $d E$, that is needed to increase the surface area of a soap-air interface by $d A$, is given by $d E=\gamma d A$ where $\gamma$ is the surface tension of the film.

Question: 2.1 Find the ratio $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}$ in terms of $\gamma, P_{a}$ and $R_{0}$.

Solution:
The bubble is surrounded by air.

<img_4487>

$P_{a}, T_{a}, \rho_{a}$

Cutting the sphere in half and using the projected area to balance the forces give

$$
\begin{aligned}
P_{i} \pi R_{0}^{2} & =P_{a} \pi R_{0}^{2}+2\left(2 \pi R_{0} \gamma\right) \\
P_{i} & =P_{a}+\frac{4 \gamma}{R_{0}}
\end{aligned}
$$

The pressure and density are related by the ideal gas law:

$P V=n R T$ or $P=\frac{\rho R T}{M}$, where $M=$ the molar mass of air.

Apply the ideal gas law to the air inside and outside the bubble, we get

$$
\begin{aligned}
& \rho_{i} T_{i}=P_{i} \frac{M}{R} \\
& \rho_{a} T_{a}=P_{a} \frac{M}{R}, \\
& \frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}=\frac{P_{i}}{P_{a}}=\left[1+\frac{4 \gamma}{R_{0} P_{a}}\right]
\end{aligned}
$$

Final answer: $1+\frac{4 \gamma}{R_{0} P_{a}}$
Subfield: Thermodynamics

An Electrified Soap Bubble

A spherical soap bubble with internal air density $\rho_{i}$, temperature $T_{i}$ and radius $R_{0}$ is surrounded by air with density $\rho_{a}$, atmospheric pressure $P_{a}$ and temperature $T_{a}$. The soap film has surface tension $\gamma$, density $\rho_{s}$ and thickness $t$. The mass and the surface tension of the soap do not change with the temperature. Assume that $R_{0} \gg t$.

The increase in energy, $d E$, that is needed to increase the surface area of a soap-air interface by $d A$, is given by $d E=\gamma d A$ where $\gamma$ is the surface tension of the film.
Context question:
2.1 Find the ratio $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}$ in terms of $\gamma, P_{a}$ and $R_{0}$.
Context answer:
\boxed{$1+\frac{4 \gamma}{R_{0} P_{a}}$}


Question: 2.2 Find the numerical value of $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}-1$ using $\gamma=0.0250 \mathrm{Nm}^{-1}, R_{0}=1.00 \mathrm{~cm}$, and $P_{a}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$.

Solution:
Using $\gamma=0.025 \mathrm{Nm}^{-1}, R_{0}=1.0 \mathrm{~cm}$ and $P_{a}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$, the numerical value of the ratio is

$$
\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}=1+\frac{4 \gamma}{R_{0} P_{a}}=1+0.0001
$$

(The effect of the surface tension is very small.)

Final answer: 1.0001
Subfield: Thermodynamics

An Electrified Soap Bubble

A spherical soap bubble with internal air density $\rho_{i}$, temperature $T_{i}$ and radius $R_{0}$ is surrounded by air with density $\rho_{a}$, atmospheric pressure $P_{a}$ and temperature $T_{a}$. The soap film has surface tension $\gamma$, density $\rho_{s}$ and thickness $t$. The mass and the surface tension of the soap do not change with the temperature. Assume that $R_{0} \gg t$.

The increase in energy, $d E$, that is needed to increase the surface area of a soap-air interface by $d A$, is given by $d E=\gamma d A$ where $\gamma$ is the surface tension of the film.
Context question:
2.1 Find the ratio $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}$ in terms of $\gamma, P_{a}$ and $R_{0}$.
Context answer:
\boxed{$1+\frac{4 \gamma}{R_{0} P_{a}}$}


Context question:
2.2 Find the numerical value of $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}-1$ using $\gamma=0.0250 \mathrm{Nm}^{-1}, R_{0}=1.00 \mathrm{~cm}$, and $P_{a}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$.
Context answer:
\boxed{1.0001}


Question: 2.3 The bubble is initially formed with warmer air inside. Find the minimum numerical value of $T_{i}$ such that the bubble can float in still air. Use $T_{a}=300 \mathrm{~K}, \rho_{s}=1000 \mathrm{kgm}^{-3}$, $\rho_{a}=1.30 \mathrm{kgm}^{-3}, t=100 \mathrm{~nm}$ and $g=9.80 \mathrm{~ms}^{-2}$.

Solution:
Let $W=$ total weight of the bubble, $F=$ buoyant force due to air around the bubble

$$
\begin{aligned}
W & =(\text { mass of film }+ \text { mass of air }) g \\
& =\left(4 \pi R_{0}^{2} \rho_{s} t+\frac{4}{3} \pi R_{0}^{3} \rho_{i}\right) g \\
& =4 \pi R_{0}^{2} \rho_{s} t g+\frac{4}{3} \pi R_{0}^{3} \frac{\rho_{a} T_{a}}{T_{i}}\left[1+\frac{4 \gamma}{R_{0} P_{a}}\right] g
\end{aligned}
$$

The buoyant force due to air around the bubble is

$$
B=\frac{4}{3} \pi R_{0}^{3} \rho_{a} g
$$

If the bubble floats in still air,

$$
\begin{aligned}
B & \geq W \\
\frac{4}{3} \pi R_{0}^{3} \rho_{a} g & \geq 4 \pi R_{0}^{2} \rho_{s} \operatorname{tg}+\frac{4}{3} \pi R_{0}^{3} \frac{\rho_{a} T_{a}}{T_{i}}\left[1+\frac{4 \gamma}{R_{0} P_{a}}\right] g
\end{aligned}
$$

Rearranging to give

$$
\begin{aligned}
T_{i} & \geq \frac{R_{0} \rho_{a} T_{a}}{R_{0} \rho_{a}-3 \rho_{s} t}\left[1+\frac{4 \gamma}{R_{0} P_{a}}\right] \\
& \geq 307.1 \mathrm{~K}
\end{aligned}
$$

The air inside must be about $7.1^{\circ} \mathrm{C}$ warmer.

Final answer: 307.1 K
Subfield: Thermodynamics

An Electrified Soap Bubble

A spherical soap bubble with internal air density $\rho_{i}$, temperature $T_{i}$ and radius $R_{0}$ is surrounded by air with density $\rho_{a}$, atmospheric pressure $P_{a}$ and temperature $T_{a}$. The soap film has surface tension $\gamma$, density $\rho_{s}$ and thickness $t$. The mass and the surface tension of the soap do not change with the temperature. Assume that $R_{0} \gg t$.

The increase in energy, $d E$, that is needed to increase the surface area of a soap-air interface by $d A$, is given by $d E=\gamma d A$ where $\gamma$ is the surface tension of the film.
Context question:
2.1 Find the ratio $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}$ in terms of $\gamma, P_{a}$ and $R_{0}$.
Context answer:
\boxed{$1+\frac{4 \gamma}{R_{0} P_{a}}$}


Context question:
2.2 Find the numerical value of $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}-1$ using $\gamma=0.0250 \mathrm{Nm}^{-1}, R_{0}=1.00 \mathrm{~cm}$, and $P_{a}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$.
Context answer:
\boxed{1.0001}


Context question:
2.3 The bubble is initially formed with warmer air inside. Find the minimum numerical value of $T_{i}$ such that the bubble can float in still air. Use $T_{a}=300 \mathrm{~K}, \rho_{s}=1000 \mathrm{kgm}^{-3}$, $\rho_{a}=1.30 \mathrm{kgm}^{-3}, t=100 \mathrm{~nm}$ and $g=9.80 \mathrm{~ms}^{-2}$.
Context answer:
\boxed{307.1}


Extra Supplementary Reading Materials:

After the bubble is formed for a while, it will be in thermal equilibrium with the surrounding. This bubble in still air will naturally fall towards the ground.

Question: 2.4 Find the minimum velocity $u$ of an updraught (air flowing upwards) that will keep the bubble from falling at thermal equilibrium. Give your answer in terms of $\rho_{s}, R_{0}, g, t$ and the air's coefficient of viscosity $\eta$. You may assume that the velocity is small such that Stokes's law applies, and ignore the change in the radius when the temperature lowers to the equilibrium. The drag force from Stokes' Law is $F=6 \pi \eta R_{0} u$.

Solution:
Ignore the radius change $\rightarrow$ Radius remains $R_{0}=1.0 \mathrm{~cm}$

(The radius actually decreases by $0.8 \%$ when the temperature decreases from $307.1 \mathrm{~K}$ to $300 \mathrm{~K}$. The film itself also becomes slightly thicker.)

The drag force from Stokes' Law is $F=6 \pi \eta R_{0} u$

If the bubble floats in the updraught,

$F \geq W-B$

$6 \pi \eta R_{0} u \geq\left(4 \pi R_{0}^{2} \rho_{s} t+\frac{4}{3} \pi R_{0}^{3} \rho_{i}\right) g-\frac{4}{3} \pi R_{0}^{3} \rho_{a} g$

When the bubble is in thermal equilibrium $T_{i}=T_{a}$.

$6 \pi \eta R_{0} u \geq\left(4 \pi R_{0}^{2} \rho_{s} t+\frac{4}{3} \pi R_{0}^{3} \rho_{a}\left[1+\frac{4 \gamma}{R_{0} P_{a}}\right]\right) g-\frac{4}{3} \pi R_{0}^{3} \rho_{a} g$

Rearranging to give

$u \geq \frac{4 R_{0} \rho_{s} t g}{6 \eta}+\frac{\frac{4}{3} R_{0}^{2} \rho_{a} g\left(\frac{4 \gamma}{R_{0} P_{a}}\right)}{6 \eta}$

Final answer: $\frac{4 R_{0} \rho_{s} t g}{6 \eta}+\frac{\frac{4}{3} R_{0}^{2} \rho_{a} g\left(\frac{4 \gamma}{R_{0} P_{a}}\right)}{6 \eta}$
Subfield: Thermodynamics

An Electrified Soap Bubble

A spherical soap bubble with internal air density $\rho_{i}$, temperature $T_{i}$ and radius $R_{0}$ is surrounded by air with density $\rho_{a}$, atmospheric pressure $P_{a}$ and temperature $T_{a}$. The soap film has surface tension $\gamma$, density $\rho_{s}$ and thickness $t$. The mass and the surface tension of the soap do not change with the temperature. Assume that $R_{0} \gg t$.

The increase in energy, $d E$, that is needed to increase the surface area of a soap-air interface by $d A$, is given by $d E=\gamma d A$ where $\gamma$ is the surface tension of the film.
Context question:
2.1 Find the ratio $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}$ in terms of $\gamma, P_{a}$ and $R_{0}$.
Context answer:
\boxed{$1+\frac{4 \gamma}{R_{0} P_{a}}$}


Context question:
2.2 Find the numerical value of $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}-1$ using $\gamma=0.0250 \mathrm{Nm}^{-1}, R_{0}=1.00 \mathrm{~cm}$, and $P_{a}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$.
Context answer:
\boxed{1.0001}


Context question:
2.3 The bubble is initially formed with warmer air inside. Find the minimum numerical value of $T_{i}$ such that the bubble can float in still air. Use $T_{a}=300 \mathrm{~K}, \rho_{s}=1000 \mathrm{kgm}^{-3}$, $\rho_{a}=1.30 \mathrm{kgm}^{-3}, t=100 \mathrm{~nm}$ and $g=9.80 \mathrm{~ms}^{-2}$.
Context answer:
\boxed{307.1}


Extra Supplementary Reading Materials:

After the bubble is formed for a while, it will be in thermal equilibrium with the surrounding. This bubble in still air will naturally fall towards the ground.
Context question:
2.4 Find the minimum velocity $u$ of an updraught (air flowing upwards) that will keep the bubble from falling at thermal equilibrium. Give your answer in terms of $\rho_{s}, R_{0}, g, t$ and the air's coefficient of viscosity $\eta$. You may assume that the velocity is small such that Stokes's law applies, and ignore the change in the radius when the temperature lowers to the equilibrium. The drag force from Stokes' Law is $F=6 \pi \eta R_{0} u$.
Context answer:
\boxed{$\frac{4 R_{0} \rho_{s} t g}{6 \eta}+\frac{\frac{4}{3} R_{0}^{2} \rho_{a} g\left(\frac{4 \gamma}{R_{0} P_{a}}\right)}{6 \eta}$}


Question: 2.5 Calculate the numerical value for $u$ using $\eta=1.8 \times 10^{-5} \mathrm{kgm}^{-1} \mathrm{~s}^{-1}$.

Solution:
The numerical value is $u \geq 0.36 \mathrm{~m} / \mathrm{s}$.

Final answer: $0.36$ $ \mathrm{~m} / \mathrm{s}$
Subfield: Thermodynamics


Question: 2.8 What must be the magnitude of this charge $q$ in terms of $t, \rho_{a}, \rho_{s}, \varepsilon_{0}, R_{0}, P_{a}$ in order that the bubble will float motionlessly in still air? Calculate also the numerical value of $q$. The permittivity of free space $\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{farad} / \mathrm{m}$.

Solution:
The bubble will float if

$$
\begin{aligned}
B & \geq W \\
\frac{4}{3} \pi R_{1}^{3} \rho_{a} g & \geq 4 \pi R_{0}^{2} \rho_{s} t g+\frac{4}{3} \pi R_{0}^{3} \rho_{i} g
\end{aligned}
$$

Initially, $T_{i}=T_{a} \Rightarrow \rho_{i}=\rho_{a}$ for $\gamma \rightarrow 0$ and $R_{1}=R_{0}\left(1+\frac{\Delta R}{R_{0}}\right)$


$$
\begin{aligned}
& \frac{4}{3} \pi R_{0}^{3}\left(1+\frac{\Delta R}{R_{0}}\right)^{3} \rho_{a} g \geq 4 \pi R_{0}^{2} \rho_{s} t g+\frac{4}{3} \pi R_{0}^{3} \rho_{a} g \\
& \frac{4}{3} \pi(3 \Delta R) \rho_{a} g \geq 4 \pi R_{0}^{2} \rho_{s} t g \\
& \frac{4}{3} \pi \frac{3 q^{2}}{96 \pi^{2} \varepsilon_{0} R_{0} P_{a}} \rho_{a} g \geq 4 \pi R_{0}^{2} \rho_{s} t g \\
& q^{2} \geq \frac{96 \pi^{2} R_{0}^{3} \rho_{s} t \varepsilon_{0} P_{a}}{\rho_{a}}
\end{aligned}
$$

$$
q \approx 256 \times 10^{-9} \mathrm{C} \approx 256 \mathrm{nC}
$$

Note that if the surface tension term is retained, we get

$$
R_{1} \approx\left(1+\frac{q^{2} / 96 \pi^{2} \varepsilon_{0} R_{0}^{4} P_{a}}{\left[1+\frac{2}{3}\left(\frac{4 \gamma}{R_{0} P_{a}}\right)\right]}\right) R_{0}
$$

Final answer: $q^{2} \geq \frac{96 \pi^{2} R_{0}^{3} \rho_{s} t \varepsilon_{0} P_{a}}{\rho_{a}}$ , $q \approx 256 \times 10^{-9} \mathrm{C} \approx 256 \mathrm{nC}$
Subfield: Modern Physics


Question: 3.1 Calculate the electric field intensity $\vec{E}_{p}$ at a distance $r$ from an ideal electric dipole $\vec{p}$ at the origin $\mathrm{O}$ along the direction of $\vec{p}$ in Figure 2.

$p=2 a q, \quad r \gg a$

![](https://cdn.mathpix.com/cropped/2023_12_21_15ed4f80b426c75b5506g-1.jpg?height=433&width=735&top_left_y=1846&top_left_x=660)

FIGURE 2

Solution:
Using Coulomb's Law, we write the electric field at a distance $r$ is given by

$$
\begin{aligned}
& E_{p}=\frac{q}{4 \pi \varepsilon_{0}(r-a)^{2}}-\frac{q}{4 \pi \varepsilon_{0}(r+a)^{2}} \\
& E_{p}=\frac{q}{4 \pi \varepsilon_{0} r^{2}}\left(\frac{1}{\left(1-\frac{a}{r}\right)^{2}}-\frac{1}{\left(1+\frac{a}{r}\right)^{2}}\right)
\end{aligned}
\tag{1}
$$

Using binomial expansion for small $a$,

$$
\begin{aligned}
E_{p} & =\frac{q}{4 \pi \varepsilon_{0} r^{2}}\left(1+\frac{2 a}{r}-1+\frac{2 a}{r}\right) \\
& =+\frac{4 q a}{4 \pi \varepsilon_{0} r^{3}}=+\frac{q a}{\pi \varepsilon_{0} r^{3}} \\
& =\frac{2 p}{4 \pi \varepsilon_{0} r^{3}}
\end{aligned}
\tag{2}
$$

Final answer: $E_{p}=\frac{2 p}{4 \pi \varepsilon_{0} r^{3}}$
Subfield: Modern Physics

3. To Commemorate the Centenary of Rutherford's Atomic Nucleus: the Scattering of an Ion by a Neutral Atom 

<img_4421>

An ion of mass $m$, charge $Q$, is moving with an initial non-relativistic speed $v_{0}$ from a great distance towards the vicinity of a neutral atom of mass $M>>m$ and of electrical polarisability $\alpha$. The impact parameter is $b$ as shown in Figure 1.

The atom is instantaneously polarised by the electric field $\vec{E}$ of the in-coming (approaching) ion. The resulting electric dipole moment of the atom is $\vec{p}=\alpha \vec{E}$. Ignore any radiative losses in this problem.

Question: 3.1 Calculate the electric field intensity $\vec{E}_{p}$ at a distance $r$ from an ideal electric dipole $\vec{p}$ at the origin $\mathrm{O}$ along the direction of $\vec{p}$ in Figure 2.

$p=2 a q, \quad r \gg a$

<img_4331>

FIGURE 2

Solution:
Using Coulomb's Law, we write the electric field at a distance $r$ is given by

$$
\begin{aligned}
& E_{p}=\frac{q}{4 \pi \varepsilon_{0}(r-a)^{2}}-\frac{q}{4 \pi \varepsilon_{0}(r+a)^{2}} \\
& E_{p}=\frac{q}{4 \pi \varepsilon_{0} r^{2}}\left(\frac{1}{\left(1-\frac{a}{r}\right)^{2}}-\frac{1}{\left(1+\frac{a}{r}\right)^{2}}\right)
\end{aligned}
\tag{1}
$$

Using binomial expansion for small $a$,

$$
\begin{aligned}
E_{p} & =\frac{q}{4 \pi \varepsilon_{0} r^{2}}\left(1+\frac{2 a}{r}-1+\frac{2 a}{r}\right) \\
& =+\frac{4 q a}{4 \pi \varepsilon_{0} r^{3}}=+\frac{q a}{\pi \varepsilon_{0} r^{3}} \\
& =\frac{2 p}{4 \pi \varepsilon_{0} r^{3}}
\end{aligned}
\tag{2}
$$

Final answer: $E_{p}=\frac{2 p}{4 \pi \varepsilon_{0} r^{3}}$