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

Image of a charge in a metallic object 

Introduction - Method of images

A point charge $q$ is placed in the vicinity of a grounded metallic sphere of radius $R$ [see Fig. 1(a)], and consequently a surface charge distribution is induced on the sphere. To calculate the electric field and potential from the distribution of the surface charge is a formidable task. However, the calculation can be considerably simplified by using the so called method of images. In this method, the electric field and potential produced by the charge distributed on the sphere can be represented as an electric field and potential of a single point charge $q^{\prime}$ placed inside the sphere (you do not have to prove it). Note: The electric field of this image charge $q$ ' reproduces the electric field and the potential only outside the sphere (including its surface).

<img_4456>

Task 1 - The image charge

The symmetry of the problem dictates that the charge $q$ ' should be placed on the line connecting the point charge $q$ and the center of the sphere [see Fig. 1(b)].
Context question:
a) What is the value of the potential on the sphere?
Context answer:
\boxed{0}


Context question:
b) Express $q^{\prime}$ and the distance $d^{\prime}$ of the charge $q^{\prime}$ from the center of the sphere, in terms of $q, d$, and $R$.
Context answer:
\boxed{$d^{\prime}=\frac{R^{2}}{d}$ , $q^{\prime}=-q \frac{R}{d}$}


Context question:
c) Find the magnitude of force acting on charge $q$. Is the force repulsive?
Context answer:
\boxed{$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R d}{\left(d^{2}-R^{2}\right)^{2}}$ , attractive}


Extra Supplementary Reading Materials:

Task 2 - Shielding of an electrostatic field

Consider a point charge $q$ placed at a distance $d$ from the center of a grounded metallic sphere of radius $R$. We are interested in how the grounded metallic sphere affects the electric field at point $A$ on the opposite side of the sphere (see Fig. 2). Point $A$ is on the line connecting charge $q$ and the center of the sphere; its distance from the point charge $q$ is $r$.


<img_4361>

Fig 2 . The electric field at point $A$ is partially screened by the grounded sphere.

Question: a) Find the vector of the electric field at point $A$.

Solution:
The electric field at the point $A$ amounts to



$$
\vec{E}_{A}=\left(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}-\frac{1}{4 \pi \varepsilon_{0}} \frac{q \frac{R}{d}}{\left(r-d+\frac{R^{2}}{d}\right)^{2}}\right) \hat{r}
\tag{11}
$$

Final answer: $\vec{E}_{A}=\left(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}-\frac{1}{4 \pi \varepsilon_{0}} \frac{q \frac{R}{d}}{\left(r-d+\frac{R^{2}}{d}\right)^{2}}\right) \hat{r}$
Subfield: Electromagnetism


Question: c) In which limit of $d$ does the grounded metallic sphere screen the field of the charge $q$ completely, such that the electric field at point $A$ is exactly zero?

Solution:
In the limit $d \rightarrow R$ the electric field at the point $A$ vanishes and the grounded metallic sphere screens the point charge completely.

Final answer: $d \rightarrow R$
Subfield: Thermodynamics

2. Chimney physics 

Introduction

Gaseous products of burning are released into the atmosphere of temperature $T_{\text {Air }}$ through a high chimney of cross-section $A$ and height $h$ (see Fig. 1). The solid matter is burned in the furnace which is at temperature $T_{\text {smoke }}$. The volume of gases produced per unit time in the furnace is $B$.

Assume that:

- the velocity of the gases in the furnace is negligibly small
- the density of the gases (smoke) does not differ from that of the air at the same temperature and pressure; while in furnace, the gases can be treated as ideal
- the pressure of the air changes with height in accordance with the hydrostatic law; the change of the density of the air with height is negligible
- the flow of gases fulfills the Bernoulli equation which states that the following quantity is conserved in all points of the flow:

$\frac{1}{2} \rho v^{2}(z)+\rho g z+p(z)=$ const,

where $\rho$ is the density of the gas, $v(z)$ is its velocity, $p(z)$ is pressure, and $z$ is the height

- the change of the density of the gas is negligible throughout the chimney

<img_4538>

Task 1
Context question:
a) What is the minimal height of the chimney needed in order that the chimney functions efficiently, so that it can release all of the produced gas into the atmosphere? Express your result in terms of $B, A, T_{\text {Air, }} g=9.81 \mathrm{~m} / \mathrm{s}^{2}, \Delta T=T_{\text {Smoke }}-T_{\text {Air }}$. Important: in all subsequent tasks assume that this minimal height is the height of the chimney.
Context answer:
\boxed{$\frac{B^{2}}{A^{2}} \frac{1}{2 g} \frac{T_{\text {Air }}}{\Delta T}$}


Question: b) Assume that two chimneys are built to serve exactly the same purpose. Their cross sections are identical, but are designed to work in different parts of the world: one in cold regions, designed to work at an average atmospheric temperature of $-30{ }^{\circ} \mathrm{C}$ and the other in warm regions, designed to work at an average atmospheric temperature of $30^{\circ} \mathrm{C}$. The temperature of the furnace is $400{ }^{\circ} \mathrm{C}$. It was calculated that the height of the chimney designed to work in cold regions is $100 \mathrm{~m}$. How high is the other chimney?

Solution:
| $\frac{h(30)}{h(-30)}=\frac{\frac{T(30)}{T_{\text {Smoke }}-T(30)}}{\frac{T(-30)}{T_{\text {Smoke }}-T(-30)}} ; h(30)=145 m$. | (9) |
| :--- | :--- |

Final answer: 145 m
Subfield: Thermodynamics

2. Chimney physics 

Introduction

Gaseous products of burning are released into the atmosphere of temperature $T_{\text {Air }}$ through a high chimney of cross-section $A$ and height $h$ (see Fig. 1). The solid matter is burned in the furnace which is at temperature $T_{\text {smoke }}$. The volume of gases produced per unit time in the furnace is $B$.

Assume that:

- the velocity of the gases in the furnace is negligibly small
- the density of the gases (smoke) does not differ from that of the air at the same temperature and pressure; while in furnace, the gases can be treated as ideal
- the pressure of the air changes with height in accordance with the hydrostatic law; the change of the density of the air with height is negligible
- the flow of gases fulfills the Bernoulli equation which states that the following quantity is conserved in all points of the flow:

$\frac{1}{2} \rho v^{2}(z)+\rho g z+p(z)=$ const,

where $\rho$ is the density of the gas, $v(z)$ is its velocity, $p(z)$ is pressure, and $z$ is the height

- the change of the density of the gas is negligible throughout the chimney

<img_4538>

Task 1
Context question:
a) What is the minimal height of the chimney needed in order that the chimney functions efficiently, so that it can release all of the produced gas into the atmosphere? Express your result in terms of $B, A, T_{\text {Air, }} g=9.81 \mathrm{~m} / \mathrm{s}^{2}, \Delta T=T_{\text {Smoke }}-T_{\text {Air }}$. Important: in all subsequent tasks assume that this minimal height is the height of the chimney.
Context answer:
\boxed{$\frac{B^{2}}{A^{2}} \frac{1}{2 g} \frac{T_{\text {Air }}}{\Delta T}$}


Context question:
b) Assume that two chimneys are built to serve exactly the same purpose. Their cross sections are identical, but are designed to work in different parts of the world: one in cold regions, designed to work at an average atmospheric temperature of $-30{ }^{\circ} \mathrm{C}$ and the other in warm regions, designed to work at an average atmospheric temperature of $30^{\circ} \mathrm{C}$. The temperature of the furnace is $400{ }^{\circ} \mathrm{C}$. It was calculated that the height of the chimney designed to work in cold regions is $100 \mathrm{~m}$. How high is the other chimney?
Context answer:
\boxed{145}


Question: c) How does the velocity of the gases vary along the height of the chimney? Find its expression.

Solution:
The velocity is constant,
$$
v=\sqrt{2gh\left(\frac{\rho_{Air}}{\rho_{Smoke}}-1 \right)}=\sqrt{2gh\left(\frac{T_{Smoke}}{T_{Air}}-1 \right)}=\sqrt{2gh\frac{\Delta T}{T_{Air}}}
\tag{10}
$$
This can be seen from the equation of continuity $A v=$ const. ( $\rho_{\text {Smoke }}$ is constant). It has a sudden jump from approximately zero velocity to this constant value when the gases enter the chimney from the furnace. In fact, since the chimney operates at minimal height this constant is equal to $B$, that is $v=B / A$.

Final answer: $v=\sqrt{2gh\frac{\Delta T}{T_{Air}}}$
Subfield: Thermodynamics

2. Chimney physics 

Introduction

Gaseous products of burning are released into the atmosphere of temperature $T_{\text {Air }}$ through a high chimney of cross-section $A$ and height $h$ (see Fig. 1). The solid matter is burned in the furnace which is at temperature $T_{\text {smoke }}$. The volume of gases produced per unit time in the furnace is $B$.

Assume that:

- the velocity of the gases in the furnace is negligibly small
- the density of the gases (smoke) does not differ from that of the air at the same temperature and pressure; while in furnace, the gases can be treated as ideal
- the pressure of the air changes with height in accordance with the hydrostatic law; the change of the density of the air with height is negligible
- the flow of gases fulfills the Bernoulli equation which states that the following quantity is conserved in all points of the flow:

$\frac{1}{2} \rho v^{2}(z)+\rho g z+p(z)=$ const,

where $\rho$ is the density of the gas, $v(z)$ is its velocity, $p(z)$ is pressure, and $z$ is the height

- the change of the density of the gas is negligible throughout the chimney

<img_4538>

Task 1
Context question:
a) What is the minimal height of the chimney needed in order that the chimney functions efficiently, so that it can release all of the produced gas into the atmosphere? Express your result in terms of $B, A, T_{\text {Air, }} g=9.81 \mathrm{~m} / \mathrm{s}^{2}, \Delta T=T_{\text {Smoke }}-T_{\text {Air }}$. Important: in all subsequent tasks assume that this minimal height is the height of the chimney.
Context answer:
\boxed{$\frac{B^{2}}{A^{2}} \frac{1}{2 g} \frac{T_{\text {Air }}}{\Delta T}$}


Context question:
b) Assume that two chimneys are built to serve exactly the same purpose. Their cross sections are identical, but are designed to work in different parts of the world: one in cold regions, designed to work at an average atmospheric temperature of $-30{ }^{\circ} \mathrm{C}$ and the other in warm regions, designed to work at an average atmospheric temperature of $30^{\circ} \mathrm{C}$. The temperature of the furnace is $400{ }^{\circ} \mathrm{C}$. It was calculated that the height of the chimney designed to work in cold regions is $100 \mathrm{~m}$. How high is the other chimney?
Context answer:
\boxed{145}


Context question:
c) How does the velocity of the gases vary along the height of the chimney? Find its expression.
Context answer:
\boxed{$v=\sqrt{2gh\frac{\Delta T}{T_{Air}}}$}


Question: d) How does the pressure of the gases vary along the height of the chimney?

Solution:
At some height $z$, from the Bernoulli equation one gets
$$
p_{\text {smoke }}(z)=p(0)-\left(\rho_{\text {Air }}-\rho_{\text {Smoke }}\right) g h-\rho_{\text {Smoke }} g z
\tag{11}
$$

Thus the pressure of smoke suddenly changes as it enters the chimney from the furnace and acquires velocity.

Final answer: $p_{\text {smoke }}(z)=p(0)-\left(\rho_{\text {Air }}-\rho_{\text {Smoke }}\right) g h-\rho_{\text {Smoke }} g z$
Subfield: Modern Physics

3. Simple model of an atomic nucleus 

Introduction

Although atomic nuclei are quantum objects, a number of phenomenological laws for their basic properties (like radius or binding energy) can be deduced from simple assumptions: (i) nuclei are built from nucleons (i.e. protons and neutrons); (ii) strong nuclear interaction holding these nucleons together has a very short range (it acts only between neighboring nucleons); (iii) the number of protons $(Z)$ in a given nucleus is approximately equal to the number of neutrons $(N)$, i.e. $Z \approx N \approx A / 2$, where $A$ is the total number of nucleons $(A \gg 1)$. Important: Use these assumptions in Tasks 1-4 below.

Task 1 - Atomic nucleus as closely packed system of nucleons

In a simple model, an atomic nucleus can be thought of as a ball consisting of closely packed nucleons [see Fig. 1(a)], where the nucleons are hard balls of radius $r_{N}=0.85 \mathrm{fm}\left(1 \mathrm{fm}=10^{-15} \mathrm{~m}\right)$. The nuclear force is present only for two nucleons in contact. The volume of the nucleus $V$ is larger than the volume of all nucleons $A V_{N}$, where $V_{N}=\frac{4}{3} r_{N}^{3} \pi$. The ratio $f=A V_{N} / V$ is called the packing factor and gives the percentage of space filled by the nuclear matter.

<img_4362>

Question: a) Calculate what would be the packing factor $f$ if nucleons were arranged in a "simple cubic" (SC) crystal system, where each nucleon is centered on a lattice point of an infinite cubic lattice [see Fig. 1(b)].

Solution:
In the SC-system, in each of 8 corners of a given cube there is one unit (atom, nucleon, etc.), but it is shared by 8 neighboring cubes - this gives a total of one nucleon per cube. If nucleons are touching, as we assume in our simplified model, then $a=2 r_{N}$ is the cube edge length $a$. The volume of one nucleon is then
$$
V_{N}=\frac{4}{3} r_{N}^{3} \pi=\frac{4}{3}\left(\frac{a}{2}\right)^{3} \pi=\frac{4 a^{3}}{3 \cdot 8} \pi=\frac{\pi}{6} a^{3}
\tag{1}
$$
from which we obtain
$$
f=\frac{V_{N}}{a^{3}}=\frac{\pi}{6} \approx 0.52
\tag{2}
$$

Final answer: $\frac{\pi}{6}$
Subfield: Modern Physics


Question: b) Assume that $d=2 R(A / 2)$ and evaluate the expression for $E_{k i n}$ obtained in part a) for $A=$ 100, 150, 200 and 250 (express the results in units of MeV). Estimate the values of $A$ for which fission is possible in the model described above?

![](https://cdn.mathpix.com/cropped/2023_12_21_75c6e1b0517dbb434e54g-1.jpg?height=591&width=1513&top_left_y=2035&top_left_x=337)

Solution:
The kinetic energy when $d=2 R(A / 2)$ is given with:
$$
\begin{aligned}
& E_{k i n}=2 E_{b}\left(\frac{A}{2}\right)-E_{b}(A)-\frac{1}{4 \pi \varepsilon_{0}} \frac{2^{1 / 3} A^{2} e^{2}}{16 \cdot 2 r_{N} A^{1 / 3} f^{-1 / 3}}= \\
& =-3 f^{1 / 3} A^{2 / 3} a_{V}\left(2^{1 / 3}-1\right)+6 f^{2 / 3} A^{1 / 3} a_{V}\left(2^{2 / 3}-1\right) \\
& -4 f a_{V}-\frac{e^{2} f^{1 / 3}}{\pi \varepsilon_{0} r_{N}}\left[\frac{3}{80}\left(2^{-2 / 3}-1\right)+\frac{2^{1 / 3}}{128}\right] A^{5 / 3}-\frac{e^{2} f^{1 / 3}}{\pi \varepsilon_{0} r_{N}}\left[\frac{3}{40}\left(2^{1 / 3}-1\right)\right] A^{2 / 3}= \\
& =\left(0.02203 A^{5 / 3}-10.0365 A^{2 / 3}+36.175 A^{1 / 3}-33.091\right) \mathrm{MeV}
\end{aligned}
\tag{17}
$$
Numerically one gets:

$A=100 \ldots E_{k i n}=-33.95 \mathrm{MeV}$,

$A=150 \ldots E_{k i n}=-30.93 \mathrm{MeV}$,



$A=200 \ldots E_{k i n}=-14.10 \mathrm{MeV}$,

$A=250 \ldots E_{\text {kin }}=+15.06 \mathrm{MeV}$.

In our model, fission is possible when $E_{k i n}(d=2 R(A / 2)) \geq 0$. From the numerical evaluations given above, one sees that this happens approximately halfway between $A=200$ and $A=250$ - a rough estimate would be $A \approx 225$. Precise numerical evaluation of the equation:
$$
E_{\text {kin }}=\left(0.02203 A^{5 / 3}-10.0365 A^{2 / 3}+36.175 A^{1 / 3}-33.091\right) \mathrm{MeV} \geq 0
\tag{18}
$$
gives that for $A \geq 227$ fission is possible.

Final answer: Numerically one gets:
$A=100 \ldots E_{k i n}=-33.95 \mathrm{MeV}$,
$A=150 \ldots E_{k i n}=-30.93 \mathrm{MeV}$,
$A=200 \ldots E_{k i n}=-14.10 \mathrm{MeV}$,
$A=250 \ldots E_{\text {kin }}=+15.06 \mathrm{MeV}$.

and $A \geq 227$
Subfield: Optics

Figure 1.1

<img_4357>

A plane monochromatic light wave, wavelength $\lambda$ and frequency $f$, is incident normally on two identical narrow slits, separated by a distance $d$, as indicated in Figure 1.1. The light wave emerging at each slit is given, at a distance $x$ in a direction $\theta$ at time $t$, by

$$
y=a \cos [2 \pi(f t-x / \lambda)]
$$

where the amplitude $a$ is the same for both waves. (Assume $x$ is much larger than $d$ ).

Question: (i) Show that the two waves observed at an angle $\theta$ to a normal to the slits, have a resultant amplitude A which can be obtained by adding two vectors, each having magnitude $a$, and each with an associated direction determined by the phase of the light wave.

Verify geometrically, from the vector diagram, that

$$
A=2 a \cos \theta
$$

where

$$
\beta=\frac{\pi}{\lambda} d \sin \theta
$$

Solution:
(i) Vector Diagram
<img_4350>

If the phase of the light from the first slit is zero, the phase from second slit is

$$
\phi=\frac{2 \pi}{\lambda} d \sin \theta
$$

Adding the two waves with phase difference $\phi$ where $\xi=2 \pi\left(f t-\frac{x}{\lambda}\right)$,

$$
\begin{aligned}
& a \cos (\xi+\phi)+a \cos (\xi)=2 a \cos (\phi / 2)(\xi+\phi / 2) \\
& a \cos (\xi+\phi)+a \cos (\xi)=2 a \cos \beta\{\cos (\xi+\beta)\}
\end{aligned}
$$

This is a wave of amplitude $A=2 a \cos \beta$ and phase $\beta$. From vector diagram, in isosceles triangle $\mathrm{OPQ}$,

$$
\beta=\frac{1}{2} \phi=\frac{\pi}{\lambda} d \sin \theta \quad(N B \quad \phi=2 \beta)
$$

and

$$
A=2 a \cos \beta
$$

Thus the sum of the two waves can be obtained by the addition of two vectors of amplitude a and angular directions 0 and $\phi$.
Subfield: Optics

Figure 1.1

<img_4357>

A plane monochromatic light wave, wavelength $\lambda$ and frequency $f$, is incident normally on two identical narrow slits, separated by a distance $d$, as indicated in Figure 1.1. The light wave emerging at each slit is given, at a distance $x$ in a direction $\theta$ at time $t$, by

$$
y=a \cos [2 \pi(f t-x / \lambda)]
$$

where the amplitude $a$ is the same for both waves. (Assume $x$ is much larger than $d$ ).
Context question:
(i) Show that the two waves observed at an angle $\theta$ to a normal to the slits, have a resultant amplitude A which can be obtained by adding two vectors, each having magnitude $a$, and each with an associated direction determined by the phase of the light wave.

Verify geometrically, from the vector diagram, that

$$
A=2 a \cos \theta
$$

where

$$
\beta=\frac{\pi}{\lambda} d \sin \theta
$$
Context answer:
\boxed{证明题}


Question: (ii) The double slit is replaced by a diffraction grating with $N$ equally spaced slits, adjacent slits being separated by a distance $d$. Use the vector method of adding amplitudes to show that the vector amplitudes, each of magnitude $a$, form a part of a regular polygon with vertices on a circle of radius $R$ given by

$$
R=\frac{a}{2 \sin \beta}
$$

Deduce that the resultant amplitude is

$$
\frac{a \sin N \beta}{\sin \beta}
$$



and obtain the resultant phase difference relative to that of the light from the slit at the edge of the grating.

Solution:
(ii) Each slit in diffraction grating produces a wave of amplitude a with phase $2 \beta$ relative to previous slit wave. The vector diagram consists of a 'regular' polygon with sides of constant length $a$ and with constant angles between adjacent sides.

Let $\mathrm{O}$ be the centre of circumscribing circle passing through the vertices of the polygon. Then radial lines such as OS have length $\mathrm{R}$ and bisect the internal angles of the polygon. Figure 1.2.

Figure 1.2

<img_4428>





$$
\begin{aligned}
& O\hat{ S }T=O\hat{ T }S=\frac{1}{2}(180-\phi) \\
& \text { and } T \hat{O }S=\phi
\end{aligned}
$$
In the triangle $T O S$, for example
$$
\begin{gathered}
a=2 R \sin (\phi / 2)=2 R \sin \beta \text { as }(\phi=2 \beta) \\
\therefore R=\frac{a}{2 \sin \beta} \quad \text { (1) }
\end{gathered}
$$

As the polygon has $N$ faces then:

$$
T \hat{O} Z=N(T \hat{O} Z)=N \phi=2 N \beta
$$

Therefore in isosceles triangle TOZ, the amplitude of the resultant wave, TZ, is given by

$$
2 R \sin N \beta \text {. }
$$

Hence form (1) this amplitude is

$$
\frac{a \sin N \beta}{\sin \beta}
$$

Resultant phase is



$$
\begin{aligned}
& =Z \hat{T }S \\
& =O \hat{T} S-O \hat{T} Z \\
& \left(90-\frac{\phi}{2}\right)-\frac{1}{2}(180-N \phi) \\
& -\frac{1}{2}(N-1) \phi \\
& =(N-1) \beta
\end{aligned}
$$
Subfield: Optics

Figure 1.1

<img_4357>

A plane monochromatic light wave, wavelength $\lambda$ and frequency $f$, is incident normally on two identical narrow slits, separated by a distance $d$, as indicated in Figure 1.1. The light wave emerging at each slit is given, at a distance $x$ in a direction $\theta$ at time $t$, by

$$
y=a \cos [2 \pi(f t-x / \lambda)]
$$

where the amplitude $a$ is the same for both waves. (Assume $x$ is much larger than $d$ ).
Context question:
(i) Show that the two waves observed at an angle $\theta$ to a normal to the slits, have a resultant amplitude A which can be obtained by adding two vectors, each having magnitude $a$, and each with an associated direction determined by the phase of the light wave.

Verify geometrically, from the vector diagram, that

$$
A=2 a \cos \theta
$$

where

$$
\beta=\frac{\pi}{\lambda} d \sin \theta
$$
Context answer:
\boxed{证明题}


Context question:
(ii) The double slit is replaced by a diffraction grating with $N$ equally spaced slits, adjacent slits being separated by a distance $d$. Use the vector method of adding amplitudes to show that the vector amplitudes, each of magnitude $a$, form a part of a regular polygon with vertices on a circle of radius $R$ given by

$$
R=\frac{a}{2 \sin \beta}
$$

Deduce that the resultant amplitude is

$$
\frac{a \sin N \beta}{\sin \beta}
$$



and obtain the resultant phase difference relative to that of the light from the slit at the edge of the grating.
Context answer:
\boxed{证明题}


Question: (iv) Determine the intensities of the principal intensity maxima.

Solution:
(iv) For the principle maxima $\beta=\pi p \quad$ where $p=0 \pm 1 \pm 2 \ldots \ldots$.

$$
I_{\max }=a^{2}\left(\frac{N \beta^{\prime}}{\beta^{\prime}}\right)=N^{2} a^{2} \quad \beta^{\prime}=0 \text { and } \beta=\pi p+\beta^{\prime}
$$

Final answer: $N^{2} a^{2}$