di-zhang-fdu/OlympiadExtended huggingface dataset:
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.
Context 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
Context answer:
\boxed{$E_{p}=\frac{2 p}{4 \pi \varepsilon_{0} r^{3}}$}


Question: 3.2 Find the expression for the force $\vec{f}$ acting on the ion due to the polarised atom. Show that this force is attractive regardless of the sign of the charge of the ion.

Solution:
The electric field seen by the atom from the ion is

$$
\vec{E}_{\text {ion }}=-\frac{Q}{4 \pi \varepsilon_{0} r^{2}} \hat{r}
$$

The induced dipole moment is then simply

$$
\vec{p}=\alpha \vec{E}_{\text {ion }}=-\frac{\alpha Q}{4 \pi \varepsilon_{0} r^{2}} \hat{r}
$$

From eq. (2)

$$
\vec{E}_{p}=\frac{2 p}{4 \pi \varepsilon_{0} r^{3}} \hat{r}
$$

The electric field intensity $\vec{E}_{p}$ at the position of an ion at that instant is, using eq. (4),

$$
\vec{E}_{p}=\frac{1}{4 \pi \varepsilon_{0} r^{3}}\left[-\frac{2 \alpha Q}{4 \pi \varepsilon_{0} r^{2}} \hat{r}\right]=-\frac{\alpha Q}{8 \pi^{2} \varepsilon_{0}^{2} r^{5}} \hat{r}
$$

The force acting on the ion is

$$
\vec{f}=Q \vec{E}_{p}=-\frac{\alpha Q^{2}}{8 \pi^{2} \varepsilon_{0}^{2} r^{5}} \hat{r}
$$

The "-" sign implies that this force is attractive and $Q^{2}$ implies that the force is attractive regardless of the sign of $Q$.

Final answer: $\vec{f}=-\frac{\alpha Q^{2}}{8 \pi^{2} \varepsilon_{0}^{2} r^{5}} \hat{r}$
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.
Context 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
Context answer:
\boxed{$E_{p}=\frac{2 p}{4 \pi \varepsilon_{0} r^{3}}$}


Context question:
3.2 Find the expression for the force $\vec{f}$ acting on the ion due to the polarised atom. Show that this force is attractive regardless of the sign of the charge of the ion.
Context answer:
\boxed{$\vec{f}=-\frac{\alpha Q^{2}}{8 \pi^{2} \varepsilon_{0}^{2} r^{5}} \hat{r}$}


Question: 3.3 What is the electric potential energy of the ion-atom interaction in terms of $\alpha, Q$ and $r$ ?

Solution:
The potential energy of the ion-atom is given by $U=\int_{r}^{\infty} \vec{f} \cdot d \vec{r}$

Using this, $U=\int_{r}^{\infty} \vec{f} \cdot d \vec{r}=-\frac{\alpha Q^{2}}{32 \pi^{2} \varepsilon_{0}^{2} r^{4}}$.

Final answer: $-\frac{\alpha Q^{2}}{32 \pi^{2} \varepsilon_{0}^{2} r^{4}}$
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.
Context 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
Context answer:
\boxed{$E_{p}=\frac{2 p}{4 \pi \varepsilon_{0} r^{3}}$}


Context question:
3.2 Find the expression for the force $\vec{f}$ acting on the ion due to the polarised atom. Show that this force is attractive regardless of the sign of the charge of the ion.
Context answer:
\boxed{$\vec{f}=-\frac{\alpha Q^{2}}{8 \pi^{2} \varepsilon_{0}^{2} r^{5}} \hat{r}$}


Context question:
3.3 What is the electric potential energy of the ion-atom interaction in terms of $\alpha, Q$ and $r$ ?
Context answer:
\boxed{$-\frac{\alpha Q^{2}}{32 \pi^{2} \varepsilon_{0}^{2} r^{4}}$}


Question: 3.4 Find the expression for $r_{\min }$, the distance of the closest approach, as shown in Figure 1.

Solution:
At the position $r_{\min }$ we have, according to the Principle of Conservation of Angular Momentum,

$$
\begin{aligned}
m v_{\max } r_{\min } & =m v_{0} b \\
v_{\max } & =v_{0} \frac{b}{r_{\min }}
\end{aligned}
$$

And according to the Principle of Conservation of Energy:

$$
\frac{1}{2} m v_{\max }^{2}+\frac{-\alpha Q^{2}}{32 \pi^{2} \varepsilon_{0}^{2} r^{4}}=\frac{1}{2} m v_{0}^{2}
$$

Eqs.(12) \& (13):

$$
\begin{array}{r}
\left(\frac{b}{r_{\text {min }}}\right)^{2}-\frac{\alpha Q^{2} / \frac{1}{2} m v_{0}^{2}}{32 \pi^{2} \varepsilon_{0}^{2} b^{4}}\left(\frac{b}{r_{\text {min }}}\right)^{4}=1 \\
\left(\frac{r_{\min }}{b}\right)^{4}-\left(\frac{r_{\min }}{b}\right)^{2}+\frac{\alpha Q^{2}}{16 \pi^{2} \varepsilon_{0}^{2} m v_{0}^{2} b^{4}}=0
\end{array}
$$

The roots of eq. (14) are:

$$
r_{\min }=\frac{b}{\sqrt{2}}\left[1 \pm \sqrt{1-\frac{\alpha Q^{2}}{4 \pi^{2} \varepsilon_{0}^{2} m v_{0}^{2} b^{4}}}\right]^{\frac{1}{2}}
$$

[Note that the equation (14) implies that $r_{\min }$ cannot be zero, unless $b$ is itself zero.]

Since the expression has to be valid at $Q=0$, which gives

$$
r_{\min }=\frac{b}{\sqrt{2}}[1 \pm 1]^{\frac{1}{2}}
$$

We have to choose "+" sign to make $r_{\min }=b$

Hence,

$$
r_{\min }=\frac{b}{\sqrt{2}}\left[1+\sqrt{1-\frac{\alpha Q^{2}}{4 \pi^{2} \varepsilon_{0}^{2} m v_{0}^{2} b^{4}}}\right]^{\frac{1}{2}}
$$

Final answer: $r_{\min }=\frac{b}{\sqrt{2}}\left[1+\sqrt{1-\frac{\alpha Q^{2}}{4 \pi^{2} \varepsilon_{0}^{2} m v_{0}^{2} b^{4}}}\right]^{\frac{1}{2}}$
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.
Context 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
Context answer:
\boxed{$E_{p}=\frac{2 p}{4 \pi \varepsilon_{0} r^{3}}$}


Context question:
3.2 Find the expression for the force $\vec{f}$ acting on the ion due to the polarised atom. Show that this force is attractive regardless of the sign of the charge of the ion.
Context answer:
\boxed{$\vec{f}=-\frac{\alpha Q^{2}}{8 \pi^{2} \varepsilon_{0}^{2} r^{5}} \hat{r}$}


Context question:
3.3 What is the electric potential energy of the ion-atom interaction in terms of $\alpha, Q$ and $r$ ?
Context answer:
\boxed{$-\frac{\alpha Q^{2}}{32 \pi^{2} \varepsilon_{0}^{2} r^{4}}$}


Context question:
3.4 Find the expression for $r_{\min }$, the distance of the closest approach, as shown in Figure 1.
Context answer:
\boxed{$r_{\min }=\frac{b}{\sqrt{2}}\left[1+\sqrt{1-\frac{\alpha Q^{2}}{4 \pi^{2} \varepsilon_{0}^{2} m v_{0}^{2} b^{4}}}\right]^{\frac{1}{2}}$}


Question: 3.5 If the impact parameter $b$ is less than a critical value $b_{0}$, the ion will descend along a spiral to the atom. In such a case, the ion will be neutralized, and the atom is, in turn, charged. This process is known as the "charge exchange" interaction. What is the cross sectional area $A=\pi b_{0}^{2}$ of this "charge exchange" collision of the atom as seen by the ion?

Solution:
A spiral trajectory occurs when (16) is imaginary (because there is no minimum distance of approach). $r_{\min }$ is real under the condition:

$$
\begin{aligned}
& 1 \geq \frac{\alpha Q^{2}}{4 \pi^{2} \varepsilon_{0}^{2} m v_{0}^{2} b^{4}} \\
& b \geq b_{0}=\left(\frac{\alpha Q^{2}}{4 \pi^{2} \varepsilon_{0}^{2} m v_{0}^{2}}\right)^{\frac{1}{4}}
\end{aligned}
$$

For $b<b_{0}=\left(\frac{\alpha Q^{2}}{4 \pi^{2} \varepsilon_{0}^{2} m v_{0}^{2}}\right)^{\frac{1}{4}}$ the ion will collide with the atom.

Hence the atom, as seen by the ion, has a cross-sectional area $A$,

$$
A=\pi b_{0}^{2}=\pi\left(\frac{\alpha Q^{2}}{4 \pi^{2} \varepsilon_{0}^{2} m v_{0}^{2}}\right)^{\frac{1}{2}}
$$

Final answer: $\pi\left(\frac{\alpha Q^{2}}{4 \pi^{2} \varepsilon_{0}^{2} m v_{0}^{2}}\right)^{\frac{1}{2}}$
Subfield: Optics


Question: a) Consider a plane-parallel transparent plate, where the refractive index, $n$, varies with distance, $z$, from the lower surface (see figure). Show that $n_{A} \sin \alpha=n_{B} \sin \beta$. The notation is that of the figure.

<img_4366>

Solution:
a) 

<img_4495>


From the figure we get

$n_{A} \sin \alpha=n_{1} \sin \alpha_{1}=n_{2} \sin \alpha_{2}=\ldots=n_{B} \sin \beta$
Subfield: Optics


Question: a) Consider a plane-parallel transparent plate, where the refractive index, $n$, varies with distance, $z$, from the lower surface (see figure). Show that $n_{A} \sin \alpha=n_{B} \sin \beta$. The notation is that of the figure.

![](https://cdn.mathpix.com/cropped/2023_12_21_cef941636e625861824cg-1.jpg?height=605&width=937&top_left_y=917&top_left_x=545)

Solution:
a) 

![](https://cdn.mathpix.com/cropped/2023_12_21_cef941636e625861824cg-1.jpg?height=454&width=534&top_left_y=2217&top_left_x=1155)


From the figure we get

$n_{A} \sin \alpha=n_{1} \sin \alpha_{1}=n_{2} \sin \alpha_{2}=\ldots=n_{B} \sin \beta$

Final answer: 证明题
Subfield: Optics



Context question:
a) Consider a plane-parallel transparent plate, where the refractive index, $n$, varies with distance, $z$, from the lower surface (see figure). Show that $n_{A} \sin \alpha=n_{B} \sin \beta$. The notation is that of the figure.

<img_4366>
Context answer:
\boxed{证明题}


Question: b) Assume that you are standing in a large flat desert. At some distance you see what appears to be a water surface. When you approach the "water" is seems to move away such that the distance to the "water" is always constant. Explain the phenomenon.

Solution:
b) The phenomenon is due to total reflexion in a warm layer of air when $\beta=90^{\circ}$. This gives



$$
n_{A} \sin \alpha=n_{B}
$$
Subfield: Thermodynamics



Context question:
a) Consider a plane-parallel transparent plate, where the refractive index, $n$, varies with distance, $z$, from the lower surface (see figure). Show that $n_{A} \sin \alpha=n_{B} \sin \beta$. The notation is that of the figure.

<img_4366>
Context answer:
\boxed{证明题}


Context question:
b) Assume that you are standing in a large flat desert. At some distance you see what appears to be a water surface. When you approach the "water" is seems to move away such that the distance to the "water" is always constant. Explain the phenomenon.
Context answer:
\boxed{证明题}


Question: c) Compute the temperature of the air close to the ground in b) assuming that your eyes are located $1.60 \mathrm{~m}$ above the ground and that the distance to the "water" is $250 \mathrm{~m}$. The refractive index of the air at $15{ }^{\circ} \mathrm{C}$ and at normal air pressure $(101.3 \mathrm{kPa})$ is 1.000276 . The temperature of the air more than $1 \mathrm{~m}$ above the ground is assumed to be constant and equal to $30{ }^{\circ} \mathrm{C}$. The atmospheric pressure is assumed to be normal. The refractive index, $n$, is such that $n-1$ is proportional to the density of the air. Discuss the accuracy of your result.

Solution:
c) As the density, $\rho$, of the air is inversely proportional to the absolute temperature, $T$, for fixed pressure we have

$$
n(T)=1+k \cdot \rho=1+k / T
$$

The value given at $15{ }^{\circ} \mathrm{C}$ determines the value of $k=0.0795$.

In order to have total reflexion we have $n_{30} \sin \alpha=n_{T}$ or

$$
\left(1+\frac{k}{303}\right) \cdot \frac{L}{\sqrt{h^{2}+L^{2}}}=\left(1+\frac{k}{T}\right) \text { with } h=1.6 \mathrm{~m} \text { and } L=250 \mathrm{~m}
$$

As $h \ll L$ we can use a power expansion in $h L$ :

$$
T=\frac{303}{\left(\frac{303}{k}+1\right) \frac{1}{\sqrt{1+h^{2} / L^{2}}}-\frac{303}{k}} \approx 303\left(1+\frac{303 h^{2}}{2 k L^{2}}\right)=328 \mathrm{~K}
$$

Final answer: 328 K
Subfield: Thermodynamics

Mid-ocean ridge

Consider a large vessel of water that is situated in a uniform gravitational field with free-fall acceleration $g$. Two vertical rectangular plates parallel to each other are fitted into the vessel so that the vertical edges of the plates are in a tight gap-less contact with the vertical walls of the vessel. Length $h$ of each plate is immersed in water (Fig. 1). The width of the plates along the $y$-axis is $w$, water density is $\rho_{0}$.

<img_4537>

A: without oil

<img_4339>

B: with oil

Figure 1. Parallel plates in water.

Oil of density $\rho_{\text {oil }}\left(\rho_{\text {oil }}<\rho_{0}\right)$ is poured into the space between the plates until the lower level of the oil has reached the lower edges of the plates. Assume that plates and vessel edges are high enough for oil not to overflow them. Surface tension and mixing of fluids can be neglected.

The effects of the surface curvature of the planets can be neglected. You might need to use the formula

$$
(1+x)^{\varepsilon} \approx 1+\varepsilon x \text {, when }|x| \ll 1 \text {. }
$$

Question: A.1 What is the $x$-component of the net force $F_{x}$ acting on the right plate (magnitude and direction)?

Solution:
A.1

<img_4394>

Figure 1

Let $h^{\prime}$ be the height of the column of oil (see Fig. 1). Then pressure at depth $h$ below the water surface must be $p_{h}=\rho_{\text {oil }} g h=\rho_{\text {oil }} g h^{\prime}$, from where $h^{\prime}=\frac{\rho_{0}}{\rho_{\text {oil }}} h$. Horizontal force on the plate $F_{x}=F_{1}-F_{0}$, where the force due to new fluid is $F_{1}=\frac{\rho_{\mathrm{oil}} h^{\prime}}{2} \cdot h^{\prime} w$ and the force due to water is $F_{0}=\frac{\rho_{0} g h}{2} \cdot h w$.

Combining all the equation above, we get

$$
F_{x}=\left(\frac{\rho_{0}}{\rho_{\mathrm{oil}}}-1\right) \frac{\rho_{0} g h^{2} w}{2}
$$

$F_{x}=\left(\frac{\rho_{0}}{\rho_{\mathrm{oil}}}-1\right) \frac{\rho_{0} g h^{2} w}{2}$.

Final answer: $F_{x}=\left(\frac{\rho_{0}}{\rho_{\mathrm{oil}}}-1\right) \frac{\rho_{0} g h^{2} w}{2}$
Subfield: Electromagnetism

Electrostatic lens

Consider a uniformly charged metallic ring of radius $R$ and total charge $q$. The ring is a hollow toroid of thickness $2 a \ll R$. This thickness can be neglected in parts A, B, C, and E. The $x y$ plane coincides with the plane of the ring, while the $z$-axis is perpendicular to it, as shown in Figure 1. In parts A and B you might need to use the formula (Taylor expansion)

$$
(1+x)^{\varepsilon} \approx 1+\varepsilon x+\frac{1}{2} \varepsilon(\varepsilon-1) x^{2}, \text { when }|x| \ll 1 \text {. }
$$

<img_4405>

Figure 1. A charged ring of radius $R$.

Part A. Electrostatic potential on the axis of the ring (1 point)

Question: A.1 Calculate the electrostatic potential $\Phi(z)$ along the axis of the ring at a $z$ distance from its center (point A in Figure 1).

Solution:
The linear charge density of the ring is $\lambda=q /(2 \pi R)$. All the points of the ring are situated a distance $\sqrt{R^{2}+z^{2}}$ away from point A. Integrating over the whole ring we readily obtain:

$$
\Phi(z)=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\sqrt{R^{2}+z^{2}}}
$$

$\Phi(z)=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\sqrt{R^{2}+z^{2}}}$.

Final answer: $\Phi(z)=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\sqrt{R^{2}+z^{2}}}$.
Subfield: Electromagnetism

Electrostatic lens

Consider a uniformly charged metallic ring of radius $R$ and total charge $q$. The ring is a hollow toroid of thickness $2 a \ll R$. This thickness can be neglected in parts A, B, C, and E. The $x y$ plane coincides with the plane of the ring, while the $z$-axis is perpendicular to it, as shown in Figure 1. In parts A and B you might need to use the formula (Taylor expansion)

$$
(1+x)^{\varepsilon} \approx 1+\varepsilon x+\frac{1}{2} \varepsilon(\varepsilon-1) x^{2}, \text { when }|x| \ll 1 \text {. }
$$

<img_4405>

Figure 1. A charged ring of radius $R$.

Part A. Electrostatic potential on the axis of the ring (1 point)
Context question:
A.1 Calculate the electrostatic potential $\Phi(z)$ along the axis of the ring at a $z$ distance from its center (point A in Figure 1).
Context answer:
\boxed{$\Phi(z)=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\sqrt{R^{2}+z^{2}}}$.}


Question: A.2 Calculate the electrostatic potential $\Phi(z)$ to the lowest non-zero power of $z$, assuming $z \ll R$.

Solution:
Using an expansion in powers of $z$ we obtain:

$$
\Phi(z)=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\sqrt{R^{2}+z^{2}}}=\frac{q}{4 \pi \varepsilon_{0} R} \frac{1}{\sqrt{1+\left(\frac{z}{R}\right)^{2}}} \approx \frac{q}{4 \pi \varepsilon_{0} R}\left(1-\frac{z^{2}}{2 R^{2}}\right) .
$$

Final answer: $\Phi(z) \approx \frac{q}{4 \pi \varepsilon_{0} R}(1-\frac{z^{2}}{2 R^{2}})$
Subfield: Electromagnetism

Electrostatic lens

Consider a uniformly charged metallic ring of radius $R$ and total charge $q$. The ring is a hollow toroid of thickness $2 a \ll R$. This thickness can be neglected in parts A, B, C, and E. The $x y$ plane coincides with the plane of the ring, while the $z$-axis is perpendicular to it, as shown in Figure 1. In parts A and B you might need to use the formula (Taylor expansion)

$$
(1+x)^{\varepsilon} \approx 1+\varepsilon x+\frac{1}{2} \varepsilon(\varepsilon-1) x^{2}, \text { when }|x| \ll 1 \text {. }
$$

<img_4405>

Figure 1. A charged ring of radius $R$.

Part A. Electrostatic potential on the axis of the ring (1 point)
Context question:
A.1 Calculate the electrostatic potential $\Phi(z)$ along the axis of the ring at a $z$ distance from its center (point A in Figure 1).
Context answer:
\boxed{$\Phi(z)=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\sqrt{R^{2}+z^{2}}}$.}


Context question:
A.2 Calculate the electrostatic potential $\Phi(z)$ to the lowest non-zero power of $z$, assuming $z \ll R$.
Context answer:
\boxed{$\Phi(z) \approx \frac{q}{4 \pi \varepsilon_{0} R}(1-\frac{z^{2}}{2 R^{2}})$}

Question: A.3 An electron (mass $m$ and charge $-e$ ) is placed at point A (Figure 1, $z \ll R$ ). What is the force acting on the electron?

Solution:
The potential energy of the electron is $V(z)=-e \Phi(z)$. The force acting on the electron is

$$
F(z)=-\frac{\mathrm{d} V(z)}{\mathrm{d} z}=+e \frac{\mathrm{d} \Phi}{\mathrm{d} z}=-\frac{q e}{4 \pi \varepsilon_{0} R^{3}} z
$$

If this is a restoring force, it should be negative for positive $z$. Thus, $q>0$.



Final answer: $F(z)=-\frac{q e}{4 \pi \varepsilon_{0} R^{3}} z$
Subfield: Electromagnetism


Question: A.3 An electron (mass $m$ and charge $-e$ ) is placed at point A (Figure 1, $z \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to oscillations. The moving electron does not influence the charge distribution on the ring.

Solution:
The potential energy of the electron is $V(z)=-e \Phi(z)$. The force acting on the electron is

$$
F(z)=-\frac{\mathrm{d} V(z)}{\mathrm{d} z}=+e \frac{\mathrm{d} \Phi}{\mathrm{d} z}=-\frac{q e}{4 \pi \varepsilon_{0} R^{3}} z
$$

If this is a restoring force, it should be negative for positive $z$. Thus, $q>0$.

Final answer: $F(z)=-\frac{q e}{4 \pi \varepsilon_{0} R^{3}} z$, q>0
Subfield: Electromagnetism

Electrostatic lens

Consider a uniformly charged metallic ring of radius $R$ and total charge $q$. The ring is a hollow toroid of thickness $2 a \ll R$. This thickness can be neglected in parts A, B, C, and E. The $x y$ plane coincides with the plane of the ring, while the $z$-axis is perpendicular to it, as shown in Figure 1. In parts A and B you might need to use the formula (Taylor expansion)

$$
(1+x)^{\varepsilon} \approx 1+\varepsilon x+\frac{1}{2} \varepsilon(\varepsilon-1) x^{2}, \text { when }|x| \ll 1 \text {. }
$$

<img_4405>

Figure 1. A charged ring of radius $R$.

Part A. Electrostatic potential on the axis of the ring (1 point)
Context question:
A.1 Calculate the electrostatic potential $\Phi(z)$ along the axis of the ring at a $z$ distance from its center (point A in Figure 1).
Context answer:
\boxed{$\Phi(z)=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\sqrt{R^{2}+z^{2}}}$.}


Context question:
A.2 Calculate the electrostatic potential $\Phi(z)$ to the lowest non-zero power of $z$, assuming $z \ll R$.
Context answer:
\boxed{$\Phi(z) \approx \frac{q}{4 \pi \varepsilon_{0} R}(1-\frac{z^{2}}{2 R^{2}})$}


Context question:
A.3 An electron (mass $m$ and charge $-e$ ) is placed at point A (Figure 1, $z \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to oscillations. The moving electron does not influence the charge distribution on the ring.
Context answer:
$F(z)=-\frac{q e}{4 \pi \varepsilon_{0} R^{3}} z$, q>0


Question: A.4 What is the angular frequency $\omega$ of such harmonic oscillations?

Solution:
The equation of motion for an electron is

$$
m \ddot{z}+\frac{q e}{4 \pi \varepsilon_{0} R^{3}} z=0
$$

(here dots denote time derivatives). We therefore get

$$
\omega=\sqrt{\frac{q e}{4 \pi m \varepsilon_{0} R^{3}}}
$$

Final answer: $\omega=\sqrt{\frac{q e}{4 \pi m \varepsilon_{0} R^{3}}}$
Subfield: Electromagnetism


Question: B.2 An electron is placed at point $\mathrm{B}$ (Figure $1, r \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to harmonic oscillations. The moving electron does not influence the charge distribution on the ring.

Solution:
The potential of the electron is $V(r)=-e \Phi(r)$. Force acting on the electron in the $x y$ plane is

$$
F(r)=-\frac{\mathrm{d} V(r)}{\mathrm{d} r}=+e \frac{\mathrm{d} \Phi(r)}{\mathrm{d} r}=\frac{q e}{8 \pi \varepsilon_{0} R^{3}} r
$$

To have oscilations we need the force to be negative for $r>0$. Thus, $q<0$.

Final answer: $F(r)=+\frac{q e}{8 \pi \varepsilon_{0} R^{3}} r . \quad q<0$.