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.

Question: C.2 Find $c$. Express your answer in terms of the constant $\beta$ from question B.1 and other known quantities.

Solution:
<img_4378>

Figure 3: Focusing of electrons.

Let us consider an electron emitted an an angle $\gamma$ to the optical axis (Figure 3). Its initial velocity in the radial direction is:

$$
v_{r ; 0}=v \sin \gamma \approx v \gamma \approx v \frac{r}{b}
$$

where $r$ is the radial distance of the electron when it reaches the plane of the ring. The velocity in the $z$-direction is

$$
v_{z}=v \cos \gamma \approx v
$$

For small angles $\gamma$ the additional velocity in the $r$-direction acquired in the "active region" is the same as in part C.1. Thus, the radial velocity after crossing the active region is

$$
v_{r}=v \frac{r}{b}+\frac{2 e q \beta r}{m} \frac{d}{v}
$$

where the first term is positive and the second term is negative, since $q<0$. If the electrons are focused, then $v_{r}<0$ (this can be verified after obtaining the final result). The electron will reach the optical axis in time

$$
t^{\prime}=\frac{r}{\left|v_{r}\right|}=-\frac{r}{\frac{2 e q \beta r}{m} \frac{d}{v}+v \frac{r}{b}}=-\frac{1}{\frac{2 e q \beta}{m} \frac{d}{v}+\frac{v}{b}}
$$

During this time it will travel a distance

$$
c=t^{\prime} v=-\frac{1}{\frac{2 e q \beta}{m} \frac{d}{v^{2}}+\frac{1}{b}}=-\frac{1}{\frac{e q \beta d}{E}+\frac{1}{b}} .
$$

Final answer: $c=-\frac{1}{\frac{e q \beta d}{E}+\frac{1}{b}}$.