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 $\Phi(r) \approx q\left(\alpha+\beta r^{2}\right)$.

Question: B.1 Find the expression for $\beta$. You might need to use the Taylor expansion formula given above.

Solution:
There are two different ways to solve this problem: (i) using direct integration; (ii) using Gauss's law and the result of part $A$.
<img_4552>

Figure 1: Calculating electrostatic potential in the plane of the ring through direct integration.

(i) Direct integration. We will follow the notations of Figure 1. Since the potential has cylindrical symmetry, let the point $\mathrm{B}$, where we calculate the potential, be on the $x$-axis. Let

$$
|\mathrm{OB}|=r ;|\mathrm{OC}|=R
$$

Thus:

$$
|\mathrm{BC}|^{2}=R^{2}+r^{2}-2 R r \cos \phi .
$$



Electrostatic potential created by ring element $\mathrm{d} \phi$ at the point $\mathrm{B}$ :

$$
\mathrm{d} \Phi=\frac{1}{4 \pi \varepsilon_{0}} \frac{\lambda R \mathrm{~d} \phi}{\sqrt{R^{2}+r^{2}-2 R r \cos \phi}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\lambda \mathrm{d} \phi}{\sqrt{1+\frac{r^{2}}{R^{2}}-2 \frac{r}{R} \cos \phi}} .
$$

Using the expansion given in the formulation of the problem for $\varepsilon=-1 / 2$ we have:

$$
\mathrm{d} \Phi \approx \frac{\lambda \mathrm{d} \phi}{4 \pi \varepsilon_{0}}\left[1-\frac{1}{2}\left(\frac{r^{2}}{R^{2}}-2 \frac{r}{R} \cos \phi\right)+\frac{3}{8}\left(\frac{r^{2}}{R^{2}}-2 \frac{r}{R} \cos \phi\right)^{2}\right]
$$

Ignoring the terms of the order $r^{3}$ and $r^{4}$ we get:

$$
\mathrm{d} \Phi \approx \frac{\lambda \mathrm{d} \phi}{4 \pi \varepsilon_{0}}\left[1+\frac{r}{R} \cos \phi+\frac{r^{2}}{R^{2}}\left(\frac{3}{2} \cos ^{2} \phi-\frac{1}{2}\right)\right] .
$$

Integrating over all angles we finally obtain:

$$
\begin{gathered}
\Phi(r)=\frac{\lambda}{4 \pi \varepsilon_{0}} \int_{0}^{2 \pi}\left[1+\frac{r}{R} \cos \phi+\frac{r^{2}}{R^{2}}\left(\frac{3}{2} \cos ^{2} \phi-\frac{1}{2}\right)\right] \mathrm{d} \phi . \\
\Phi(r)=\frac{q}{4 \pi \varepsilon_{0} R}\left(1+\frac{r^{2}}{4 R^{2}}\right) .
\end{gathered}
$$

From here, comparing with the expression $\Phi(r)=q\left(\alpha+\beta r^{2}\right)$, we obtain

$$
\beta=\frac{1}{16 \pi \varepsilon_{0} R^{3}}
$$

(ii) Gauss's law.

<img_4515>

Figure 2: Calculating electrostatic potential in the plane of the ring via Gauss's law.

Let us analyze a small cylinder of radius $r$. The center of the cylinder coincides with the center of the ring. In part A we analyzed the potential along the $z$-axis, while in this part we analyze the potential along the radius $r$. For any $z \ll R$ and $r \ll R$ the potential has an expression:

$$
\Phi(z, r)=\frac{q}{4 \pi \varepsilon_{0} R}\left(1-\frac{z^{2}}{2 R^{2}}\right)+q \beta r^{2}
$$



The lowest order terms are quadratic in $r$ and $z$. Due to reflection symmetry the potential does not contain terms of the type $r z$. This, for example, immediately gives us $\alpha=1 /\left(4 \pi \varepsilon_{0} R\right)$. Thus, for small $r$ and $z$ electric fields in the radial and axial directions are:

$$
\mathcal{E}_{z}(z, r)=+\frac{q}{4 \pi \varepsilon_{0} R^{3}} z, \quad \mathcal{E}_{r}(z, r)=-2 q \beta r .
$$

Applying Gauss's law to the cylinder we obtain:

$$
\oint \overrightarrow{\mathcal{E}} \cdot \mathrm{d} \vec{S}=0 \quad \Rightarrow \quad \int_{\text {side }} \overrightarrow{\mathcal{E}} \cdot \mathrm{d} \vec{S}+\int_{\text {base }} \overrightarrow{\mathcal{E}} \cdot \mathrm{d} \vec{S}=0
$$

The second integral is:

$$
\int_{\text {base }} \overrightarrow{\mathcal{E}} \cdot \mathrm{d} \vec{S}=2 \pi r^{2} \mathcal{E}_{z}(z, r)=\frac{q z r^{2}}{2 \varepsilon_{0} R^{3}}
$$

The first integral is:

$$
\int_{\text {side }} \overrightarrow{\mathcal{E}} \cdot \mathrm{d} \vec{S}=4 \pi r z \mathcal{E}_{r}(z, r)=-8 \pi q \beta r^{2} z
$$

Gauss's theorem thus gives:

$$
\frac{q z r^{2}}{2 \varepsilon_{0} R^{3}}-8 \pi q \beta r^{2} z=0
$$

This immediately yields

$$
\beta=\frac{1}{16 \pi \varepsilon_{0} R^{3}},
$$

which agrees with the result obtained via direct integration.

Final answer: $\beta=\frac{1}{16 \pi \varepsilon_{0} R^{3}}$