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


Context question:
A.4 What is the angular frequency $\omega$ of such harmonic oscillations?
Context answer:
\boxed{$\omega=\sqrt{\frac{q e}{4 \pi m \varepsilon_{0} R^{3}}}$}


Extra Supplementary Reading Materials:

Part B. Electrostatic potential in the plane of the ring

In this part of the problem you will have to analyze the potential $\Phi(r)$ in the plane of the ring $(z=0)$ for $r \ll R$ (point $\mathrm{B}$ in Figure 1). To the lowest non-zero power of $r$ the electrostatic potential is given by $\Phi(r) \approx q\left(\alpha+\beta r^{2}\right)$.
Context question:
B.1 Find the expression for $\beta$. You might need to use the Taylor expansion formula given above.
Context answer:
\boxed{$\beta=\frac{1}{16 \pi \varepsilon_{0} R^{3}}$}


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


Extra Supplementary Reading Materials:

Part C. The focal length of the idealized electrostatic lens: instantaneous charging

One wants to build a device to focus electrons-an electrostatic lens. Let us consider the following construction. The ring is situated perpendicularly to the $z$-axis, as shown in Figure 2. We have a source that produces on-demand packets of non-relativistic electrons. Kinetic energy of these electrons is $E=m v^{2} / 2$ ( $v$ is velocity) and they leave the source at precisely controlled moments. The system is programmed so that the ring is charge-neutral most of the time, but its charge becomes $q$ when electrons are closer than a distance $d / 2(d \ll R)$ from the plane of the ring (shaded region in Figure 2, called "active region"). In part $\mathrm{C}$ assume that charging and de-charging processes are instantaneous and the electric field "fills the space" instantaneously as well. One can neglect magnetic fields and assume that the velocity of electrons in the $z$-direction is constant. Moving electrons do not perturb the charge distribution on the ring.

<img_4381>

Figure 2. A model of an electrostatic lens.
Context question:
C.1 Determine the focal length $f$ of this lens. Assume that $f \gg d$. Express your answer in terms of the constant $\beta$ from question B. 1 and other known quantities. Assume that before reaching the "active region" the electron packet is parallel to the $z$-axis and $r \ll R$. The sign of $q$ is such so that the lens is focusing.
Context answer:
\boxed{$f=-\frac{E}{e q d \beta}$}


Extra Supplementary Reading Materials:

In reality the electron source is placed on the $z$-axis at a distance $b>f$ from the center of the ring. Consider that electrons are no longer parallel to the $z$-axis before reaching the "active region", but are emitted from a point source at a range of different angles $\gamma \ll 1$ rad to the $z$-axis. Electrons are focused in a point situated at a distance $c$ from the center of the ring.
Context question:
C.2 Find $c$. Express your answer in terms of the constant $\beta$ from question B.1 and other known quantities.
Context answer:
\boxed{$c=-\frac{1}{\frac{e q \beta d}{E}+\frac{1}{b}}$.}


Context question:
C.3 Is the equation of a thin optical lens

$$
\frac{1}{b}+\frac{1}{c}=\frac{1}{f}
$$

fulfilled for the electrostatic lens? Show it by explicitly calculating $1 / b+1 / c$.
Context answer:
证明题


Extra Supplementary Reading Materials:

Part D. The ring as a capacitor

The model of considered above was idealized and we assumed that the ring charged instantaneously. In reality charging is non-instantaneous, as the ring is a capacitor with a finite capacitance $C$. In this part we will analyze the properties of this capacitor. You might need the following integrals:

$$
\int \frac{\mathrm{d} x}{\sin x}=-\ln \left|\frac{\cos x+1}{\sin x}\right|+\text { const }
$$

and

$$
\int \frac{\mathrm{d} x}{\sqrt{1+x^{2}}}=\ln \left|x+\sqrt{1+x^{2}}\right|+\text { const. }
$$
Context question:
D.1 Calculate the capacitance $C$ of the ring. Consider that the ring has a finite width $2 a$, but remember that $a \ll R$.
Context answer:
\boxed{$C=\frac{4 \pi^{2} \varepsilon_{0} R}{\ln (\frac{8 R}{a})}$}


Extra Supplementary Reading Materials:

When electrons reach the "active region", the ring is connected to a source of voltage $V_{0}$ (Figure 3). When electrons pass the "active region", the ring is connected to the ground. The resistance of contacts is $R_{0}$ and the resistance of the ring itself can be neglected.

<img_4401>

Figure 3. Charging of the electrostatic lens.
Context question:
D.2 Determine the dependence of the charge on the ring as a function of time, $q(t)$. $t=0$ corresponds to a time moment when electrons are in the plane of the ring. What is the charge on the ring $q_{0}$ when the absolute value $q(t)$ is maximum? The capacitance of the ring is $C$ (i.e., you do not have to use the actual expression found in D.1).

Remark: the drawn polarity in Figure 3 is for indicative purposes only. The sign should be chosen so that the lens is focusing.
Context answer:
For $-\frac{d}{2 v}<t<\frac{d}{2 v}, \quad q(t)=C V_{0}\left(1-\mathrm{e}^{-\frac{d}{2 v R_{0} C}} \mathrm{e}^{-\frac{t}{R_{0} C}}\right)$.

For $t>\frac{d}{2 v}, \quad q(t)=C V_{0}\left(\mathrm{e}^{\frac{d}{2 v R_{0} C}}-\mathrm{e}^{-\frac{d}{2 v R_{0} C}}\right) \mathrm{e}^{-\frac{t}{R_{0} C}}$.

$q_{0}=C V_{0}\left(1-\mathrm{e}^{-\frac{d}{v R_{0} C}}\right) . \quad$


Extra Supplementary Reading Materials:

Part E. Focal length of a more realistic lens: non-instantaneous charging

In this part of the problem, we will consider the action of this more realistic lens. Here we will again neglect the width of the ring $2 a$ and will assume that electrons travel parallel to the $z$-axis before reaching the "active region". However, the charging of the ring is no longer instantaneous.

Question: E.1 Find the focal length $f$ of the lens. Assume that $f / v \gg R_{0} C$, but $d / v$ and $R_{0} C$ are of the same order of magnitude. Express your answer in terms of the constant $\beta$ from part $\mathrm{B}$ and other known quantities.

Solution:
Like in part $C$, the radial equation of motion of an electron is:

$$
m \ddot{r}=2 e q(t) \beta r,
$$

where in this case $q(t)$ depends on time. Using the notation $\eta=2 e \beta / m$, we obtain:

$$
\ddot{r}-\eta q(t) r=0 .
$$

As $f / v \gg R_{0} C$, then during charging-decharging the electron does not substantially change its radial position $r$, and we can assume $r$ to be constant during the entire charging-decharging process. In this case the acquired vertical velocity is

$$
v_{r}=\eta r \int_{-d /(2 v)}^{\infty} q(t) \mathrm{d} t
$$



We can use the derived equations for $q(t)$ and find the integrals. The integral $\int_{-d /(2 v)}^{d /(2 v)} q(t) \mathrm{d} t$ is (using the notation $d / v=t_{0}, R_{0} C=\tau, C V_{0}=Q_{0}$ ):

$$
\int_{-t_{0} / 2}^{t_{0} / 2} q(t) \mathrm{d} t=\int_{-t_{0} / 2}^{t_{0} / 2} Q_{0}\left(1-\mathrm{e}^{-\frac{t_{0}}{2 \tau}} \mathrm{e}^{-\frac{t}{\tau}}\right) \mathrm{d} t=Q_{0}\left(t_{0}-\tau\left[1-\mathrm{e}^{-t_{0} / \tau}\right]\right)
$$

The integral $\int_{d /(2 v)}^{\infty} q(t) \mathrm{d} t$ is

$$
\int_{t_{0} / 2}^{\infty} Q_{0}\left(e^{\frac{t_{0}}{2 \tau}}-e^{-\frac{t_{0}}{2 \tau}}\right) \mathrm{e}^{-\frac{t}{\tau}} \mathrm{d} t=Q_{0} \tau\left[1-\mathrm{e}^{-t_{0} / \tau}\right]
$$

Adding the two integrals we obtain for the final integral:

$$
\int_{-t_{0} / 2}^{\infty} q(t) d t=Q_{0} t_{0}
$$

Interestingly, it does not depend on $\tau=R_{0} C$. Therefore, the acquired vertical velocity of the electron is

$$
v_{r}=\eta r \frac{C V_{0} d}{v}=\frac{2 e \beta C V_{0} d r}{m v}
$$

Following the logic similar to part $\mathrm{C}$, we derive the focal length

$$
f=-\frac{E}{e C V_{0} d \beta}
$$

Final answer: $f=-\frac{E}{e C V_{0} d \beta}$