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


Question: C.2 Recall the problem from part A where the particle traverses from $\mathrm{O}$ to $\mathrm{P}$ (see Fig. 4). Let an opaque partition be placed at the boundary $\mathrm{AB}$ between the two regions. There is a small opening $\mathrm{CD}$ of width $d$ in $\mathrm{AB}$ such that $d \ll\left(x_{0}-x_{1}\right)$ and $d \ll x_{1}$.

Consider two extreme paths OCP and ODP such that OCP lies on the classical trajectory discussed in part A. Obtain the phase difference $\Delta \phi_{C D}$ between the two paths to first order.

![](https://cdn.mathpix.com/cropped/2023_12_21_1329e5e361501978a8a6g-1.jpg?height=337&width=574&top_left_y=340&top_left_x=1249)

Figure 4

Solution:
![](https://cdn.mathpix.com/cropped/2023_12_21_1329e5e361501978a8a6g-1.jpg?height=677&width=785&top_left_y=875&top_left_x=681)

Consider the extreme trajectories $O C P$ and $O D P$ of $(\mathrm{C} 1)$

The geometrical path difference is $E D$ in region I and $C F$ in region II.

This implies (note: $d \ll\left(x_{0}-x_{1}\right)$ and $d \ll x_{1}$ )

$$
\begin{gathered}
\Delta \phi_{C D}=\frac{2 \pi d \sin \theta_{1}}{\lambda_{1}}-\frac{2 \pi d \sin \theta_{2}}{\lambda_{2}} \\
\Delta \phi_{C D}=\frac{2 \pi m v_{1} d \sin \theta_{1}}{h}-\frac{2 \pi m v_{2} d \sin \theta_{2}}{h} \\
=2 \pi \frac{m d}{h}\left(v_{1} \sin \theta_{1}-v_{2} \sin \theta_{2}\right) \\
=0 \quad(\text { from } \mathrm{A} 2 \text { or } \mathrm{B} 1)
\end{gathered}
$$

Thus near the clasical path there is invariably constructive interference.

Final answer: $\Delta \phi_{C D}=0$
Subfield: Thermodynamics

A bubble of radius $r=5.00 \mathrm{~cm}$, containing a diatomic ideal gas, has the soap film of thickness $h=$ $10.0 \mu \mathrm{m}$ and is placed in vacuum. The soap film has the surface tension $\sigma=4.00 \cdot 10^{-2} \frac{\mathrm{N}}{\mathrm{m}}$ and the density $\rho=1.10 \frac{\mathrm{g}}{\mathrm{cm}^{3}} .

Question: 1 Find formula for the molar heat capacity of the gas in the bubble for such a process when the gas is heated so slowly that the bubble remains in a mechanical equilibrium and evaluate it; 

Hint: Laplace showed that there is pressure difference between inside and outside of a curved surface, caused by surface tension of the interface between liquid and gas, so that $\Delta p=\frac{2 \sigma}{r}$.

Solution:
1) According to the first law of thermodynamics, the amount of heat transmitted $\delta Q$ to the gas in the bubble is found as

$$
\delta Q=v C_{V} d T+p d V
\tag{B.1}
$$

where the molar heat capacity at arbitrary process is as follows

$$
C=\frac{1}{v} \frac{\delta Q}{d T}=C_{V}+\frac{p}{v} \frac{d V}{d T}
\tag{B.2}
$$

Here $C_{V}$ stands for the molar heat capacity of the gas at constant volume, $p$ designates its pressure, $v$ is the total amount of moles of gas in the bubble, $V$ and $T$ denote the volume and temperature of the gas, respectively.

Evaluate the derivative standing on the right hand side of (B.2). According to the Laplace formula, the gas pressure inside the bubble is defined by

$$
p=\frac{4 \sigma}{r}
\tag{B.3}
$$

thus, the equation of any equilibrium process with the gas in the bubble is a polytrope of the form

$$
p^{3} V=\text { const. }
\tag{B.4}
$$

The equation of state of an ideal gas has the form

$$
p V=v R T
\tag{B.5}
$$

and hence equation (B.4) can be rewritten as

$$
T^{3} V^{-2}=\text { const. }
\tag{B.6}
$$

Differentiating (B.6) the derivative with respect to temperature sought is found as

$$
\frac{d V}{d T}=\frac{3 V}{2 T}
\tag{B.7}
$$

Taking into account that the molar heat capacity of a diatomic gas at constant volume is

$$
C_{V}=\frac{5}{2} R
\tag{B.8}
$$

and using (B.5) it is finally obtained that

$$
C=C_{V}+\frac{3}{2} R=4 R=33.2 \frac{\mathrm{J}}{\mathrm{mole} \cdot \mathrm{K}}
\tag{B.9}
$$

Final answer: $33.2 $ $\frac{\mathrm{J}}{\mathrm{mole} \cdot \mathrm{K}}$
Subfield: Thermodynamics

A bubble of radius $r=5.00 \mathrm{~cm}$, containing a diatomic ideal gas, has the soap film of thickness $h=$ $10.0 \mu \mathrm{m}$ and is placed in vacuum. The soap film has the surface tension $\sigma=4.00 \cdot 10^{-2} \frac{\mathrm{N}}{\mathrm{m}}$ and the density $\rho=1.10 \frac{\mathrm{g}}{\mathrm{cm}^{3}} .
Context question:
1 Find formula for the molar heat capacity of the gas in the bubble for such a process when the gas is heated so slowly that the bubble remains in a mechanical equilibrium and evaluate it; 

Hint: Laplace showed that there is pressure difference between inside and outside of a curved surface, caused by surface tension of the interface between liquid and gas, so that $\Delta p=\frac{2 \sigma}{r}$.
Context answer:
\boxed{$33.2 $}


Question: 2 Find formula for the frequency $\omega$ of the small radial oscillations of the bubble and evaluate it under the assumption that the heat capacity of the soap film is much greater than the heat capacity of the gas in the bubble. Assume that the thermal equilibrium inside the bubble is reached much faster than the period of oscillations.

Hint: Laplace showed that there is pressure difference between inside and outside of a curved surface, caused by surface tension of the interface between liquid and gas, so that $\Delta p=\frac{2 \sigma}{r}$.

Solution:
Since the heat capacity of the gas is much smaller than the heat capacity of the soap film, and there is heat exchange between them, the gas can be considered as isothermal since the soap film plays the role of thermostat. Consider the fragment of soap film, limited by the angle $\alpha$ as shown in the figure. It's area is found as

$$
S=\pi(\alpha r)^{2}
\tag{B.10}
$$

and the corresponding mass is obtained as

$$
m=\rho S h \text {. }
\tag{B.11}
$$

Let $x$ be an increase in the radius of the bubble, then the Newton second law for the fragment of the soap film mentioned above takes the form

$$
m \ddot{x}=p^{\prime} S^{\prime}-F_{\text {surf }},
\tag{B.12}
$$

where $F_{\text {surf }}$ denotes the projection of the resultant surface tension force acting in the radial direction, $p^{\prime}$ stands for the gas pressure beneath the surface of the soap film and

$$
S^{\prime}=S\left(1+2 \frac{x}{r}\right)
$$

$F_{\text {surf }}$ is easily found as

$$
F_{\text {surf }}=F_{S T} \alpha=\sigma \cdot 2 \cdot 2 \pi[(r+x) \alpha] \cdot \alpha .
\tag{B.13}
$$

Since the gaseous process can be considered isothermal, it is

<img_4410>
written that

$$
p^{\prime} V^{\prime}=p V \text {. }
\tag{B.14}
$$

Assuming that the volume increase is quite small, (B.14) yields

$$
p^{\prime}=p \frac{1}{\left(1+\frac{x}{r}\right)^{3}} \approx p \frac{1}{\left(1+\frac{3 x}{r}\right)} \approx p\left(1-\frac{3 x}{r}\right)
\tag{B.15}
$$

Thus, from (B.10) - (B.16) and (B.3) the equation of small oscillations of the soap film is derived as

$$
\rho h \ddot{x}=-\frac{8 \sigma}{r^{2}} x
\tag{B.16}
$$



with the frequency

$$
\omega=\sqrt{\frac{8 \sigma}{\rho h r^{2}}}=108 \mathrm{~s}^{-1}
\tag{B.17}
$$

Final answer: $108 $ $\mathrm{~s}^{-1}$
Subfield: Thermodynamics

Van der Waals equation of state

In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered.

Part A. Non-ideal gas equation of state

Taking into account the finite size of the molecules, the gaseous equation of state takes the form

$$
P(V-b)=R T
\tag{1}
$$

where $P, V, T$ stands for the gas pressure, its volume per mole and temperature, respectively, $R$ denotes the universal gas constant, and $b$ is a specific constant extracting some volume.

Question: A1 Estimate $b$ and express it in terms of the diameter of the molecules $d$.

Solution:
If $V=b$ is substituted into the equation of state, then the gas pressure turns infinite. It is obvious that this is the moment when all the molecules are tightly packed. Therefore, the parameter $b$ is approximately equal to the volume of all molecules, i.e.

$$
b=N_{A} d^{3}
\tag{A1.1}
$$

Final answer: $b=N_{A} d^{3}$
Subfield: Thermodynamics

Van der Waals equation of state

In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered.

Part A. Non-ideal gas equation of state

Taking into account the finite size of the molecules, the gaseous equation of state takes the form

$$
P(V-b)=R T
\tag{1}
$$

where $P, V, T$ stands for the gas pressure, its volume per mole and temperature, respectively, $R$ denotes the universal gas constant, and $b$ is a specific constant extracting some volume.
Context question:
A1 Estimate $b$ and express it in terms of the diameter of the molecules $d$.
Context answer:
\boxed{$b=N_{A} d^{3}$}


Extra Supplementary Reading Materials:

With account of intermolecular attraction forces, van der Waals proposed the following equation of state that neatly describes both the gaseous and liquid states of matter

$$
\left(P+\frac{a}{V^{2}}\right)(V-b)=R T
\tag{2}
$$

where $a$ is another specific constant.

At temperatures $T$ below a certain critical value $T_{c}$ the isotherm of equation (2) is well represented by a non-monotonic curve 1 shown in Figure 1 which is then called van der Waals isotherm. In the same figure curve 2 shows the isotherm of an ideal gas at the same temperature. A real isotherm differs from the van der Waals isotherm by a straight segment $\boldsymbol{A B}$ drawn at some constant pressure $P_{L G}$. This straight segment is located between the volumes $V_{L}$ and $V_{G}$, and corresponds to the equilibrium of the liquid phase (indicated by $L$ ) and the gaseous phase (referred to by $G$ ). From the second law of thermodynamics J. Maxwell showed that the pressure $P_{L G}$ must be chosen such that the areas $I$ and $I I$ shown in Figure 1 must be equal.

<img_4406>

Figure 1. Van der Waals isotherm of gas/liquid (curve 1) and the isotherm of an ideal gas (curve 2).

<img_4483>

Figure 2. Several isotherms for van der Waals equation of state.

With increasing temperature the straight segment $A B$ on the isotherm shrinks to a single point when the temperature and the pressure reaches some values $T_{c}$ and $P_{L G}=P_{c}$, respectively. The parameters $P_{c}$ and $T_{c}$ are called critical and can be measured experimentally with high degree of accuracy.
Context question:
A2 Express the van der Waals constants $a$ and $b$ in terms of $T_{c}$ and $P_{c}$.
Context answer:
\boxed{$a=\frac{27 R^{2} T_{c}^{2}}{64 P_{c}}$ , $b=\frac{R T_{c}}{8 P_{c}}$}


Question: A3 For water $T_{c}=647 \mathrm{~K}$ and $P_{c}=2.2 \cdot 10^{7} \mathrm{~Pa}$. Calculate $a_{w}$ and $b_{w}$ for water.

Solution:
Numericalcalculationsforwaterproduce the following result

$$
a_{w}=0.56 \frac{\mathrm{m}^{6} \cdot \mathrm{Pa}}{\mathrm{mole}^{2}}
\tag{A3.1}
$$
$$
b_{w}=3.1 \cdot 10^{-5} \frac{\mathrm{m}^{3}}{\mathrm{~mole}}
\tag{A3.2}
$$

Final answer: $a_{w}=0.56$ , $b_{w}=3.1 \cdot 10^{-5} $ $ \frac{\mathrm{m}^{6} \cdot \mathrm{Pa}}{\mathrm{mole}^{2}}$, $\frac{\mathrm{m}^{3}}{\mathrm{~mole}}$
Subfield: Thermodynamics

Van der Waals equation of state

In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered.

Part A. Non-ideal gas equation of state

Taking into account the finite size of the molecules, the gaseous equation of state takes the form

$$
P(V-b)=R T
\tag{1}
$$

where $P, V, T$ stands for the gas pressure, its volume per mole and temperature, respectively, $R$ denotes the universal gas constant, and $b$ is a specific constant extracting some volume.
Context question:
A1 Estimate $b$ and express it in terms of the diameter of the molecules $d$.
Context answer:
\boxed{$b=N_{A} d^{3}$}


Extra Supplementary Reading Materials:

With account of intermolecular attraction forces, van der Waals proposed the following equation of state that neatly describes both the gaseous and liquid states of matter

$$
\left(P+\frac{a}{V^{2}}\right)(V-b)=R T
\tag{2}
$$

where $a$ is another specific constant.

At temperatures $T$ below a certain critical value $T_{c}$ the isotherm of equation (2) is well represented by a non-monotonic curve 1 shown in Figure 1 which is then called van der Waals isotherm. In the same figure curve 2 shows the isotherm of an ideal gas at the same temperature. A real isotherm differs from the van der Waals isotherm by a straight segment $\boldsymbol{A B}$ drawn at some constant pressure $P_{L G}$. This straight segment is located between the volumes $V_{L}$ and $V_{G}$, and corresponds to the equilibrium of the liquid phase (indicated by $L$ ) and the gaseous phase (referred to by $G$ ). From the second law of thermodynamics J. Maxwell showed that the pressure $P_{L G}$ must be chosen such that the areas $I$ and $I I$ shown in Figure 1 must be equal.

<img_4406>

Figure 1. Van der Waals isotherm of gas/liquid (curve 1) and the isotherm of an ideal gas (curve 2).

<img_4483>

Figure 2. Several isotherms for van der Waals equation of state.

With increasing temperature the straight segment $A B$ on the isotherm shrinks to a single point when the temperature and the pressure reaches some values $T_{c}$ and $P_{L G}=P_{c}$, respectively. The parameters $P_{c}$ and $T_{c}$ are called critical and can be measured experimentally with high degree of accuracy.
Context question:
A2 Express the van der Waals constants $a$ and $b$ in terms of $T_{c}$ and $P_{c}$.
Context answer:
\boxed{$a=\frac{27 R^{2} T_{c}^{2}}{64 P_{c}}$ , $b=\frac{R T_{c}}{8 P_{c}}$}


Context question:
A3 For water $T_{c}=647 \mathrm{~K}$ and $P_{c}=2.2 \cdot 10^{7} \mathrm{~Pa}$. Calculate $a_{w}$ and $b_{w}$ for water.
Context answer:
\boxed{$a_{w}=0.56$ , $b_{w}=3.1 \cdot 10^{-5} $}


Question: A4 Estimate the diameter of water molecules $d_{w}

Solution:
From equations (A1.4) and (A3.2) it is found that

$$
d_{w}=\sqrt[3]{\frac{b}{N_{A}}}=3.7 \cdot 10^{-10} \mathrm{~m} \approx 4 \cdot 10^{-10} \mathrm{~m}
\tag{A4.1}
$$

Final answer: $4 \cdot 10^{-10}$ m
Subfield: Electromagnetism

Problem3. Simplest model of gas discharge

An electric current flowing through a gas is called a gas discharge. There are many types of gas discharges including glow discharge in lighting lamps, arc discharge inwelding and the well known spark discharge that occurs between the clouds and the earth in the form of lightning.

Part A. Non-self-sustained gas discharge 

In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a permanent operationan external ionizer is needed, which creates $Z_{\text {ext }}$ pairs of singly ionized ions and free electrons per unit volume and per unit time uniformly in the volume.

When an external ionizer is switched on, the number of electrons and ions starts to grow. Unlimited increase in the number densities of electrons and ions in the gas is prevented by the recombination process in which a free electron recombineswith an ion to form a neutral atom. The number of recombining events $Z_{\text {rec }}$ that occurs in the gas per unit volume and per unit time is given by

$$
Z_{\mathrm{rec}}=r n_{e} n_{i},
$$

where $r$ is a constant called the recombination coefficient, and $n_{e}, n_{i}$ denotethe electron and ion number densities, respectively.

Suppose that at time $t=0$ the external ionizer is switched on and the initial number densities of electrons and ions in the gas are both equal to zero. Then, the electron number density $n_{e}(t)$ depends on timetas follows:

$$
n_{e}(t)=n_{0}+a \tanh b t
$$

where $n_{0}, a$ and $b$ are some constants, and $\tanh x$ stands for the hyperbolic tangent.

Question: A1 Find $_{0}, a, b$ and express them in terms of $Z_{\text {ext }}$ and $r$.

Solution:
Let us derive an equation describing the change of the electron number density with time. It is determined by the two processes; the generation of ion pairs by external ionizer and the recombination of electrons with ions. At ionization process electrons and ions are generated in pairs, and at recombination processthey disappear in pairs as well.Thus, their concentrations are alwaysequal at any given time, i.e.

$$
n(t)=n_{e}(t)=n_{i}(t)
\tag{A1.1}
$$

Then the equation describing the numberdensityevolution of electrons and ions in time can be written as

$$
\frac{d n(t)}{d t}=Z_{e x t}-r n(t)^{2}
\tag{A1.2}
$$

It is easy to show that at $t \rightarrow 0$ the function $\tanh b t \rightarrow 0$, therefore, by virtue of the initial condition $n(0)=0$,one finds

$$
n_{0}=0
\tag{A1.3}
$$

Substituting $n_{e}(t)=a \tanh b t$ in (A1.2) and separating it in the independent functions (hyperbolic, or 1 and $\left.e^{x}\right)$, one gets

$$
a=\sqrt{\frac{Z_{e x t}}{r}}
\tag{A1.4}
$$
$$
b=\sqrt{r Z_{e x t}}
\tag{A1.5}
$$

Final answer: $n_{0}=0$ , $a=\sqrt{\frac{Z_{e x t}}{r}}$ , $b=\sqrt{r Z_{e x t}}$
Subfield: Electromagnetism

Problem3. Simplest model of gas discharge

An electric current flowing through a gas is called a gas discharge. There are many types of gas discharges including glow discharge in lighting lamps, arc discharge inwelding and the well known spark discharge that occurs between the clouds and the earth in the form of lightning.

Part A. Non-self-sustained gas discharge 

In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a permanent operationan external ionizer is needed, which creates $Z_{\text {ext }}$ pairs of singly ionized ions and free electrons per unit volume and per unit time uniformly in the volume.

When an external ionizer is switched on, the number of electrons and ions starts to grow. Unlimited increase in the number densities of electrons and ions in the gas is prevented by the recombination process in which a free electron recombineswith an ion to form a neutral atom. The number of recombining events $Z_{\text {rec }}$ that occurs in the gas per unit volume and per unit time is given by

$$
Z_{\mathrm{rec}}=r n_{e} n_{i},
$$

where $r$ is a constant called the recombination coefficient, and $n_{e}, n_{i}$ denotethe electron and ion number densities, respectively.

Suppose that at time $t=0$ the external ionizer is switched on and the initial number densities of electrons and ions in the gas are both equal to zero. Then, the electron number density $n_{e}(t)$ depends on timetas follows:

$$
n_{e}(t)=n_{0}+a \tanh b t
$$

where $n_{0}, a$ and $b$ are some constants, and $\tanh x$ stands for the hyperbolic tangent.
Context question:
A1 Find $_{0}, a, b$ and express them in terms of $Z_{\text {ext }}$ and $r$.
Context answer:
\boxed{$n_{0}=0$ , $a=\sqrt{\frac{Z_{e x t}}{r}}$ , $b=\sqrt{r Z_{e x t}}$}


Extra Supplementary Reading Materials:

Assume that there are two external ionizers available. When the first one is switched on, the electron number density in the gas reaches its equilibrium value of be $n_{e 1}=12 \cdot 10^{10} \mathrm{~cm}^{-3}$. When the second external ionizer is switched on, the electron number density reaches its equilibrium value of $n_{e 2}=16$. $10^{10} \mathrm{~cm}^{-3}$.

Question: A2 Find the electron number density $n_{e}$ at equilibrium when both external ionizers are switched on simultaneously.

Solution:
According to equation (A1.4) the number density of electronsat steady-state is expressed in terms of the external ionizer activity as

$$
n_{e 1}=\sqrt{\frac{Z_{e x t 1}}{r}}
\tag{A2.1}
$$
$$
n_{e 2}=\sqrt{\frac{Z_{e x t 2}}{r}}
\tag{A2.2}
$$
$$
n_{e}=\sqrt{\frac{Z_{e x t 1}+Z_{e x t 2}}{r}}
\tag{A2.3}
$$

Thus,the following analogue of the Pythagorean theorem is obtained as

$$
n_{e}=\sqrt{n_{e 1}^{2}+n_{e 2}^{2}}=20.0 \cdot 10^{10} \mathrm{~cm}^{-3}
\tag{A2.4}
$$

Final answer: $20.0 \cdot 10^{10}$ $ \mathrm{~cm}^{-3}$
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)].

Question: a) What is the value of the potential on the sphere?

Solution:
As the metallic sphere is grounded, its potential vanishes, $\mathrm{V}=0$.

Final answer: 0
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}


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 $R$.

Solution:
Let us consider an arbitrary point B on the surface of the sphere as depicted in Fig. 1.

<img_4544>

Fig 1. The potential at point $B$ is zero.

The distance of point $B$ from the charge $\mathrm{q}^{\prime}$ is

$$
r_{1}=\sqrt{R^{2}+d^{\prime 2}-2 R d^{\prime} \cos \alpha}
\tag{1}
$$

whereas the distance of the point $B$ from the charge $q$ is given with the expression

$$
r_{2}=\sqrt{R^{2}+d^{2}-2 R d \cos \alpha}
\tag{2}
$$

The electric potential at the point $B$ is

$$
V=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q}{r_{2}}+\frac{q^{\prime}}{r_{1}}\right)
\tag{3}
$$

This potential must vanish,

$$
\frac{q}{r_{2}}+\frac{q^{\prime}}{r_{1}}=0
\tag{4}
$$

i.e. its numerical value is $0 \mathrm{~V}$.



Combining (1), (2) and (3) we obtain

$$
R^{2}+d^{2}-2 R d \cos \alpha=\left(\frac{q}{q^{\prime}}\right)^{2}\left(R^{2}+d^{\prime 2}-2 R d^{\prime} \cos \alpha\right)
\tag{5}
$$

As the surface of the sphere must be equipotential, the condition (5) must be satisfied for every angle $\alpha$ what leads to the following results

$$
d^{2}+R^{2}=\left(\frac{q}{q^{\prime}}\right)^{2}\left(R^{2}+d^{\prime 2}\right)
\tag{6}
$$

and

$$
d R=\left(\frac{q}{q^{\prime}}\right)^{2}\left(d^{\prime} R\right)
\tag{7}
$$

By solving of (6) and (7) we obtain the expression for the distance d' of the charge $q^{\prime}$ from the center of the sphere

$$
d^{\prime}=\frac{R^{2}}{d}
\tag{8}
$$

and the size of the charge $\mathrm{q}^{\prime}$

$$
q^{\prime}=-q \frac{R}{d}
\tag{9}
$$

Final answer: $d^{\prime}=\frac{R^{2}}{d}$ , $q^{\prime}=-q \frac{R}{d}$