IPhO 2021 · Theory problem 2

Electrostatic lens

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Consider a uniformly charged metallic ring of radius \(R\) and total charge \(q\). The ring is a hollow toroid of thickness \(2a\ll R\). This thickness can be neglected in parts A, B, C, and E. The \(xy\) 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)

\[ (1+x)^\varepsilon\approx 1+\varepsilon x+\frac12\varepsilon(\varepsilon-1)x^2, \quad\text{when }|x|\ll1. \]

Figure 1

Figure 1. A charged ring of radius \(R\).

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

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). 0.3pt
A.2 Calculate the electrostatic potential \(\Phi(z)\) to the lowest non-zero power of \(z\), assuming \(z\ll R\). 0.4pt
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. 0.2pt
A.4 What is the angular frequency \(\omega\) of such harmonic oscillations? 0.1pt

Part B. Electrostatic potential in the plane of the ring (1.7 points)

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 B in Figure 1). To the lowest non-zero power of \(r\) the electrostatic potential is given by \(\Phi(r)\approx q(\alpha+\beta r^2)\).

B.1 Find the expression for \(\beta\). You might need to use the Taylor expansion formula given above. 1.5pt
B.2 An electron is placed at point 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. 0.2pt

Part C. The focal length of the idealized electrostatic lens: instantaneous charging (2.3 points)

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 \(E=mv^2/2\) (\(v\) 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 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.

Figure 2

Figure 2. A model of an electrostatic lens.

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. 1.3pt

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 \(\gamma\ll1\) rad to the \(z\)-axis. Electrons are focused in a point situated at a distance \(c\) from the center of the ring.

C.2 Find \(c\). Express your answer in terms of the constant \(\beta\) from question B.1 and other known quantities. 0.8pt
C.3 Is the equation of a thin optical lens \(\displaystyle \frac1b+\frac1c=\frac1f\) fulfilled for the electrostatic lens? Show it by explicitly calculating \(1/b+1/c\). 0.2pt

Part D. The ring as a capacitor (3 points)

The model 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:

\[ \int\frac{\mathrm dx}{\sin x} =-\ln\left|\frac{\cos x+1}{\sin x}\right|+\mathrm{const} \]

and

\[ \int\frac{\mathrm dx}{\sqrt{1+x^2}} =\ln\left|x+\sqrt{1+x^2}\right|+\mathrm{const}. \]

D.1 Calculate the capacitance \(C\) of the ring. Consider that the ring has a finite width \(2a\), but remember that \(a\ll R\). 2.0pt

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.

Figure 3

Figure 3. Charging of the electrostatic lens.

D.2 Determine the dependence of the charge on the ring as a function of time, \(q(t)\), and make a schematic plot of this dependence. \(t=0\) 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 maximal? 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. 1.0pt

Part E. Focal length of a more realistic lens: non-instantaneous charging (2 points)

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 \(2a\) 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.

E.1 Find the focal length \(f\) of the lens. Assume that \(f/v\gg R_0C\), but \(d/v\) and \(R_0C\) are of the same order of magnitude. Express your answer in terms of the constant \(\beta\) from part B and other known quantities. 1.7pt
E.2 You will see, that the result for \(f\) is similar to that obtained in part C, whereby the value \(q\) is substituted with \(q_{\mathrm{eff}}\). Find the expression for \(q_{\mathrm{eff}}\) in terms of quantities given in formulation of the problem. 0.3pt

Official English text of the problem.

Answers

Use \[ k=\frac{1}{4\pi\varepsilon_0},\qquad T=\frac dv,\qquad \tau=R_0C,\qquad Q=CV_0. \] Here \(e>0\), the potential vanishes at infinity, and \(V_0\) is the signed applied voltage: focusing requires \(V_0<0\), hence \(Q<0\). All approximate results have the accuracy stipulated in the problem.

  • A.1: \(\displaystyle \Phi(z)=\frac{kq}{\sqrt{R^2+z^2}}\).
  • A.2: \(\displaystyle \Phi(z)\approx\frac{kq}{R}\left(1-\frac{z^2}{2R^2}\right)\).
  • A.3: \(\displaystyle \mathbf F\approx-\frac{keq}{R^3}z\,\hat{\mathbf z}\). Axial oscillations require \(q>0\).
  • A.4: \(\displaystyle \omega=\sqrt{\frac{keq}{mR^3}}\), for \(q>0\).
  • B.1: \(\displaystyle \beta=\frac{k}{4R^3}=\frac{1}{16\pi\varepsilon_0R^3}\).
  • B.2: \(\displaystyle \mathbf F_\perp=2e\beta q\,\mathbf r_\perp=\frac{keq}{2R^3}\mathbf r_\perp\). Oscillations in the plane require \(q<0\).
  • C.1: \(\displaystyle f=-\frac{E}{e\beta qd}=\frac{E}{e\beta|q|d}\).
  • C.2: \(\displaystyle c=\frac{bE}{e\beta|q|bd-E}=\frac{bf}{b-f}\).
  • C.3: Yes: \(\displaystyle \frac1b+\frac1c=\frac{e\beta|q|d}{E}=\frac1f\).
  • D.1: \(\displaystyle C\approx\frac{4\pi^2\varepsilon_0R}{\ln(8R/a)}\).
  • D.2: \[ q(t)= \begin{cases} 0, & t<-T/2,\\[2pt] Q\left[1-\exp\!\left(-\dfrac{t+T/2}{\tau}\right)\right], & -T/2\le t\le T/2,\\[5pt] q_0\exp\!\left(-\dfrac{t-T/2}{\tau}\right), & t>T/2, \end{cases} \qquad q_0=Q\left(1-e^{-T/\tau}\right)<0. \] The graph falls from zero to its minimum \(q_0\) at \(t=T/2\), then approaches zero exponentially from below.
  • E.1: \(\displaystyle f=-\frac{E}{e\beta CV_0d}=\frac{E}{e\beta C|V_0|d}\).
  • E.2: \(\displaystyle q_{\mathrm{eff}}=CV_0=\frac{q_0}{1-e^{-d/(vR_0C)}}\).

Solution

For an electron, the potential energy is \(U=-e\Phi\). Therefore \[ \mathbf F=-\boldsymbol\nabla U=e\boldsymbol\nabla\Phi. \] This fixes the force signs throughout the solution.

A.1

Every charge element on the ring is at the same distance \(\sqrt{R^2+z^2}\) from the observation point on its axis. Superposition gives \[ \Phi(z)=k\int\frac{\mathrm dq}{\sqrt{R^2+z^2}} =\boxed{\frac{kq}{\sqrt{R^2+z^2}}}. \]

A.2

For \(|z|\ll R\), \[ \Phi(z)=\frac{kq}{R} \left(1+\frac{z^2}{R^2}\right)^{-1/2}. \] Using the supplied expansion with \(\varepsilon=-1/2\), \[ \boxed{\Phi(z)\approx\frac{kq}{R}-\frac{kq}{2R^3}z^2}. \] The first nonconstant term is quadratic because the potential is unchanged by \(z\mapsto-z\).

A.3

Differentiating the axial potential, \[ F_z=e\frac{\mathrm d\Phi}{\mathrm dz} =-\frac{keqz}{(R^2+z^2)^{3/2}} \approx-\frac{keq}{R^3}z. \] Thus \[ \boxed{\mathbf F\approx-\frac{keq}{R^3}z\,\hat{\mathbf z}}. \]

A restoring force must point towards \(z=0\). Consequently, \[ \boxed{q>0} \] is required for axial oscillations.

A.4

Newton’s equation becomes \[ m\ddot z=-\frac{keq}{R^3}z. \] Comparison with \(\ddot z+\omega^2z=0\) gives \[ \boxed{\omega=\sqrt{\frac{keq}{mR^3}} =\sqrt{\frac{eq}{4\pi\varepsilon_0mR^3}}}. \]

B.1

Place the observation point at \((r,0,0)\). A ring element at polar angle \(\theta\) has charge \[ \mathrm dq=\frac{q}{2\pi}\,\mathrm d\theta \] and distance \[ s=\sqrt{R^2+r^2-2Rr\cos\theta} \] from that point. Hence, with \(\rho=r/R\), \[ \Phi(r)=\frac{kq}{2\pi R} \int_0^{2\pi} \frac{\mathrm d\theta}{\sqrt{1+\rho^2-2\rho\cos\theta}}. \]

Expanding through second order in \(\rho\), \[ \begin{aligned} (1+\rho^2-2\rho\cos\theta)^{-1/2} &\approx 1-\frac12(\rho^2-2\rho\cos\theta) +\frac38(4\rho^2\cos^2\theta)\\ &= 1+\rho\cos\theta +\rho^2\left(-\frac12+\frac32\cos^2\theta\right). \end{aligned} \] The angular averages are \[ \langle\cos\theta\rangle=0,\qquad \langle\cos^2\theta\rangle=\frac12. \] Therefore \[ \Phi(r)\approx\frac{kq}{R} \left(1+\frac{r^2}{4R^2}\right). \] Comparing with \(q(\alpha+\beta r^2)\), \[ \alpha=\frac{k}{R},\qquad \boxed{\beta=\frac{k}{4R^3} =\frac{1}{16\pi\varepsilon_0R^3}}. \]

B.2

Let \(\mathbf r_\perp=x\hat{\mathbf x}+y\hat{\mathbf y}\). Since \(r^2=x^2+y^2\), \[ \boxed{\mathbf F_\perp =e\boldsymbol\nabla_\perp\Phi =2e\beta q\,\mathbf r_\perp =\frac{keq}{2R^3}\mathbf r_\perp}. \] The force is restoring in the ring’s plane precisely when \[ \boxed{q<0}. \] For this sign, each transverse coordinate satisfies a harmonic-oscillator equation. Thus axial and transverse stability require opposite signs of ring charge.

C.1

The electron spends \[ T=\frac dv \] in the active region. Since \(d\ll R\), the transverse force there can be evaluated using B.2. Also, \(f\gg d\) allows us to neglect the change of transverse position during the interaction.

For an incident electron at transverse distance \(r\), its transverse impulse is \[ m\Delta v_r =F_rT =2e\beta qr\,\frac dv. \] The incident transverse velocity is zero, so \[ v_{r,\mathrm{out}}=\frac{2e\beta qrd}{mv}<0 \] for the focusing choice \(q<0\).

After leaving the active region, the electron moves along a straight line. To thin-lens accuracy, the distance to the axis is \[ f=-\frac{vr}{v_{r,\mathrm{out}}}. \] Consequently, \[ \boxed{f=-\frac{mv^2}{2e\beta qd} =-\frac{E}{e\beta qd} =\frac{E}{e\beta|q|d}}. \] The cancellation of \(r\) shows that all sufficiently near-axis parallel rays have the same focal length.

C.2

An electron emitted at angle \(\gamma\ll1\) from a source at distance \(b\) reaches the lens at \[ r\approx b\gamma. \] Its incoming transverse velocity satisfies \[ \frac{v_{r,\mathrm{in}}}{v}\approx\gamma\approx\frac rb. \]

The lens produces the same transverse velocity change as in C.1. Thus \[ \frac{v_{r,\mathrm{out}}}{v} =\frac rb+\frac{2e\beta qrd}{mv^2} =r\left(\frac1b+\frac{e\beta qd}{E}\right). \] For the outgoing trajectory to cross the axis a distance \(c\) beyond the ring, \[ \frac{v_{r,\mathrm{out}}}{v}=-\frac rc. \] It follows that \[ \frac1c=-\frac1b-\frac{e\beta qd}{E} =\frac{e\beta|q|d}{E}-\frac1b. \] Therefore \[ \boxed{c=\frac{bE}{e\beta|q|bd-E} =\frac{bf}{b-f}}. \] The specified condition \(b>f\) ensures \(c>0\).

C.3

Using the result just obtained, \[ \frac1b+\frac1c =\frac1b+\left(\frac{e\beta|q|d}{E}-\frac1b\right) =\frac{e\beta|q|d}{E}. \] By C.1 this equals \(1/f\), so \[ \boxed{\frac1b+\frac1c=\frac1f}. \] The thin optical lens equation is therefore fulfilled within the stated approximations.

D.1

The isolated ring’s capacitance is \[ C=\frac{q}{\Phi_s}, \] where \(\Phi_s\) is its surface potential relative to infinity.

For a thin torus, introduce the charge per unit length \[ \lambda=\frac{q}{2\pi R}. \] Choose an intermediate length \(\ell\) satisfying \[ a\ll\ell\ll R. \] We calculate the surface potential by combining a nearby, approximately straight portion of the ring with its distant curved portion.

Locally the conductor resembles a cylinder of radius \(a\). Its exterior potential can be represented, to the required accuracy, by charge on its axis. The contribution from the segment \(-\ell<s<\ell\) is therefore \[ \begin{aligned} \Phi_{\mathrm{near}} &\approx k\lambda\int_{-\ell}^{\ell} \frac{\mathrm ds}{\sqrt{s^2+a^2}}\\ &=2k\lambda \ln\left(\frac{\ell+\sqrt{\ell^2+a^2}}{a}\right)\\ &\approx2k\lambda\ln\frac{2\ell}{a}. \end{aligned} \]

For the distant portion, the thickness can be neglected. Let \[ \delta=\frac{\ell}{R}\ll1. \] The distance between two points on the ring separated by angle \(\theta\) is \(2R\sin(\theta/2)\). Hence \[ \begin{aligned} \Phi_{\mathrm{far}} &\approx k\lambda \int_\delta^{2\pi-\delta} \frac{R\,\mathrm d\theta}{2R\sin(\theta/2)}\\ &=k\lambda\int_{\delta/2}^{\pi-\delta/2} \frac{\mathrm du}{\sin u}\\ &=2k\lambda\ln\cot\frac{\delta}{4}\\ &\approx2k\lambda\ln\frac{4R}{\ell}. \end{aligned} \]

Adding the contributions eliminates the arbitrary intermediate length: \[ \begin{aligned} \Phi_s &\approx2k\lambda \left(\ln\frac{2\ell}{a}+\ln\frac{4R}{\ell}\right)\\ &=2k\lambda\ln\frac{8R}{a}\\ &=\frac{q}{4\pi^2\varepsilon_0R}\ln\frac{8R}{a}. \end{aligned} \] Thus \[ \boxed{C\approx\frac{4\pi^2\varepsilon_0R}{\ln(8R/a)}}. \] The finite radius \(a\) supplies the short-distance cutoff that makes the surface potential finite.

D.2

The electrons enter and leave the active region at \[ t_-=-\frac{T}{2},\qquad t_+=\frac{T}{2},\qquad T=\frac dv. \] The ring is initially neutral. Define \[ \tau=R_0C,\qquad Q=CV_0<0. \]

While connected to the source, the circuit equation is \[ R_0\frac{\mathrm dq}{\mathrm dt}+\frac qC=V_0, \] or \[ \tau\dot q+q=Q. \] With \(q(t_-)=0\), \[ q(t)=Q\left[1-e^{-(t+T/2)/\tau}\right], \qquad -T/2\le t\le T/2. \]

When the ring is grounded, \[ \tau\dot q+q=0. \] Charge is continuous at switching, so \[ q(t)=q_0e^{-(t-T/2)/\tau}, \qquad t>T/2, \] where \[ \boxed{q_0=Q(1-e^{-T/\tau}) =CV_0\left[1-\exp\left(-\frac{d}{vR_0C}\right)\right]}. \]

Combining the intervals, \[ \boxed{ q(t)= \begin{cases} 0, & t<-T/2,\\[2pt] Q\left[1-e^{-(t+T/2)/\tau}\right], & -T/2\le t\le T/2,\\[2pt] Q(1-e^{-T/\tau})e^{-(t-T/2)/\tau}, & t>T/2. \end{cases}} \]

For the requested sketch, draw \(q(t)\) below the time axis after charging begins. It is initially zero, falls continuously and with decreasing steepness during charging, reaches its most negative value at \(t=T/2\), and then rises exponentially towards zero. Its slope changes discontinuously at both switching times, while \(q(t)\) remains continuous.

Useful points are:

Time Charge
\(-T/2\) \(0\)
\(0\) \(Q(1-e^{-T/(2\tau)})\)
\(T/2\) \(q_0=Q(1-e^{-T/\tau})\)
\(T/2+\tau\) \(q_0\exp(-1)\)
\(t\to+\infty\) \(0\), approached from below

Thus \(|q(t)|\) is maximal when the electrons leave the active region.

E.1

Here \(\tau\) and \(T=d/v\) are comparable. Therefore the charging and appreciable discharge occur while the electrons remain within a longitudinal distance of order \(d\) from the ring. Since \(d\ll R\), the transverse force from B.2 remains applicable: \[ F_r(t)\approx2e\beta q(t)r. \]

Also, \(f/v\gg\tau\) means that focusing occurs long after the charge has substantially decayed. During the force pulse, the transverse displacement is small compared with \(r\), so \(r\) can be held constant when calculating the impulse: \[ m\Delta v_r =2e\beta r\int_{-T/2}^{\infty}q(t)\,\mathrm dt. \]

The charging contribution is \[ \begin{aligned} I_{\mathrm{ch}} &=Q\int_{-T/2}^{T/2} \left[1-e^{-(t+T/2)/\tau}\right]\mathrm dt\\ &=Q\left[T-\tau(1-e^{-T/\tau})\right]. \end{aligned} \] The discharge contribution is \[ \begin{aligned} I_{\mathrm{dis}} &=q_0\int_{T/2}^{\infty}e^{-(t-T/2)/\tau}\,\mathrm dt\\ &=q_0\tau =Q\tau(1-e^{-T/\tau}). \end{aligned} \] Their sum is \[ \boxed{\int_{-T/2}^{\infty}q(t)\,\mathrm dt=QT =CV_0\frac dv}. \]

Consequently, \[ \Delta v_r =\frac{2e\beta rQd}{mv}. \] Using the same thin-lens geometry as in C.1, \[ \boxed{ f=-\frac{vr}{\Delta v_r} =-\frac{mv^2}{2e\beta CV_0d} =-\frac{E}{e\beta CV_0d} =\frac{E}{e\beta C|V_0|d}}. \]

The resistance changes the shape and duration of the charge pulse, but its contributions to the total impulse cancel between charging and discharging. The finite extent of the interaction produces corrections beyond the assumed thin-lens accuracy.

E.2

Comparing \[ f=-\frac{E}{e\beta qd} \quad\text{and}\quad f=-\frac{E}{e\beta CV_0d}, \] we obtain \[ \boxed{q_{\mathrm{eff}}=CV_0}. \] Equivalently, using D.2, \[ \boxed{q_{\mathrm{eff}} =\frac{q_0}{1-\exp[-d/(vR_0C)]}}. \]

This effective charge equals the charge the ring would reach if held at the applied voltage indefinitely. Its magnitude exceeds the actual peak charge because the discharge tail also contributes to focusing.

The solution was checked against the official solution.