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

$$
\int \frac{\mathrm{d} x}{\sqrt{1+x^{2}}}=\ln \left|x+\sqrt{1+x^{2}}\right|+\text { const. }
$$

Question: D.1 Calculate the capacitance $C$ of the ring. Consider that the ring has a finite width $2 a$, but remember that $a \ll R$.

Solution:
<img_4561>

Figure 4: Calculation of the capacitance of the ring.

Let us sub-divide the entire ring into two parts: a part corresponding to the angle $2 \alpha \ll 1$, and the rest of the ring, as shown in Figure 4. While the angle is small in comparison to 1, let us assume that the length of the first part, $\alpha R$, is still large compared to $a(\alpha R \gg a)$. Let us calculate the electrostatic potential $\Phi$ at point K. It it a sum of two terms: the first one produced by the cut-out part with an angle $2 \alpha$ (contribution $\Phi_{1}$ ) and the second one originating from the rest of the ring (contribution $\Phi_{2}$ ).

Contribution $\Phi_{1}$. Since $\alpha \ll 1$, we can neglect the curvature of the cylinder that is cut out from the ring. The linear charge density on the ring is $\lambda=\frac{q}{2 \pi R}$. The potential at the center of the



cylinder is then given by an integral:

$$
\Phi_{1}=2 \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{2 \pi R} \int_{0}^{\alpha R} \frac{\mathrm{d} x}{\sqrt{x^{2}+a^{2}}}=\frac{q}{4 \pi^{2} \varepsilon_{0} R} \int_{0}^{\alpha R} \frac{\mathrm{d}(x / a)}{\sqrt{1+(x / a)^{2}}}=\frac{q}{4 \pi^{2} \varepsilon_{0} R} \int_{0}^{\alpha R / a} \frac{\mathrm{d} y}{\sqrt{1+y^{2}}}
$$

Using the integral provided in the description of the problem we get:

$$
\Phi_{1}=\left.\frac{q}{4 \pi^{2} \varepsilon_{0} R} \ln \left(y+\sqrt{1+y^{2}}\right)\right|_{0} ^{\alpha R / a}=\frac{q}{4 \pi^{2} \varepsilon_{0} R} \ln \left(\frac{\alpha R}{a}+\sqrt{1+\left(\frac{\alpha R}{a}\right)^{2}}\right)
$$

As $\alpha R \gg a$,

$$
\Phi_{1} \approx \frac{q}{4 \pi^{2} \varepsilon_{0} R} \ln \left(\frac{2 \alpha R}{a}\right)
$$

<img_4416>

Figure 5: Calculation of the capacitance of the ring

Contribution $\Phi_{2}$. In this case we can neglect the thickness $a$. Using the cosine theorem we can derive the distance between points $\mathrm{K}$ and $\mathrm{L}$ of Figure 5 :

$$
|\mathrm{KL}|=2 R \sin \frac{\phi}{2}
$$

The contribution $\Phi_{2}$ can then be written as an integral:

$$
\Phi_{2}=2 \frac{q}{2 \pi} \frac{1}{4 \pi \varepsilon_{0}} \int_{\alpha}^{\pi} \frac{\mathrm{d} \phi}{2 R \sin \frac{\phi}{2}}=\frac{q}{8 \pi^{2} \varepsilon_{0} R} \int_{\alpha}^{\pi} \frac{\mathrm{d} \phi}{\sin \frac{\phi}{2}}=\frac{q}{4 \pi^{2} \varepsilon_{0} R} \int_{\alpha}^{\pi} \frac{\mathrm{d}\left(\frac{\phi}{2}\right)}{\sin \frac{\phi}{2}}=\frac{q}{4 \pi^{2} \varepsilon_{0} R} \int_{\alpha / 2}^{\pi / 2} \frac{\mathrm{d} \chi}{\sin \chi} .
$$

Using the integral from the formulation of the problem, we calculate:

$$
\int_{\alpha / 2}^{\pi / 2} \frac{\mathrm{d} \chi}{\sin \chi}=-\left.\ln \left(\frac{\cos \chi+1}{\sin \chi}\right)\right|_{\alpha / 2} ^{\pi / 2}=\ln \left(\frac{\cos \alpha / 2+1}{\sin \alpha / 2}\right) \approx \ln \left(\frac{4}{\alpha}\right)
$$

for $\alpha \ll 1$. Therefore

$$
\Phi_{2} \approx \frac{q}{4 \pi^{2} \varepsilon_{0} R} \ln \left(\frac{4}{\alpha}\right)
$$



The total potential and capacitance. The total potential is the sum of $\Phi_{1}$ and $\Phi_{2}$ :

$$
\Phi=\Phi_{1}+\Phi_{2}=\frac{q}{4 \pi^{2} \varepsilon_{0} R} \ln \left(\frac{2 \alpha R}{a}\right)+\frac{q}{4 \pi^{2} \varepsilon_{0} R} \ln \left(\frac{4}{\alpha}\right)=\frac{q}{4 \pi^{2} \varepsilon_{0} R} \ln \left(\frac{8 R}{a}\right) .
$$

$\alpha$ drops out from the expression. From here we obtain the capacitance $C=q / \Phi$ :

$$
C=\frac{4 \pi^{2} \varepsilon_{0} R}{\ln \left(\frac{8 R}{a}\right)}
$$

$C \rightarrow 0$ as $a \rightarrow 0$.

Final answer: $C=\frac{4 \pi^{2} \varepsilon_{0} R}{\ln (\frac{8 R}{a})}$