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IPhO 2021, Теория — Задача 2. Electrostatic lens

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IPhO Lithuania 2021 — Theory Q2: Electrostatic lens (10 points) · 10 т.

Условие

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 + \frac{1}{2}\varepsilon(\varepsilon-1)x^2, \ \text{when } |x| \ll 1.

A thick ring of radius R lying in the xy plane with total charge q; axes x, y, z drawn at top left. Point A lies on the axis of the ring at a height z above the centre; point B lies in the plane of the ring at a distance r from the centre. The ring thickness is marked 2a at the right edge. — 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). [0,3 т.]

A.2 Calculate the electrostatic potential $\Phi(z)$ to the lowest non-zero power of $z$ , assuming $z \ll R$ . [0,4 т.]

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. [0,2 т.]

A.4 What is the angular frequency $\omega$ of such harmonic oscillations? [0,1 т.]

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. [1,5 т.]

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. [0,2 т.]

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.

Side view of the ring (height 2R, charge q) perpendicular to the dashed z-axis, surrounded by a shaded active region of width d. An electron path parallel to the axis at distance r bends at the ring and reaches the focal point F on the axis. Axes x, y (out of page), z at bottom left. — 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. [1,3 т.]

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.

C.2 Find $c$ . Express your answer in terms of the constant $\beta$ from question B.1 and other known quantities. [0,8 т.]

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 $1/b + 1/c$ . [0,2 т.]

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

and

\int \frac{\mathrm{d}x}{\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 т.]

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.

Circuit: a switch connects the ring (circle labelled C) through a resistor R_0 either to a grounded voltage source V_0 or directly to ground. — 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 т.]

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. [1,7 т.]

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. [0,3 т.]

Отговори

Решение

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

A.1 (0.3 points)

The linear charge density of the ring is $\lambda = q/(2\pi R)$ . All the points of the ring are situated a distance $\sqrt{R^2+z^2}$ away from point A. Integrating over the whole ring we readily obtain:

\Phi(z) = \frac{q}{4\pi\varepsilon_0}\frac{1}{\sqrt{R^2+z^2}}.

A.2 (0.4 points)

Using an expansion in powers of $z$ we obtain:

\Phi(z) = \frac{q}{4\pi\varepsilon_0}\frac{1}{\sqrt{R^2+z^2}} = \frac{q}{4\pi\varepsilon_0 R}\frac{1}{\sqrt{1+\left(\frac{z}{R}\right)^2}} \approx \frac{q}{4\pi\varepsilon_0 R}\left(1-\frac{z^2}{2R^2}\right).

A.3 (0.2 points)

The potential energy of the electron is $V(z) = -e\Phi(z)$ . The force acting on the electron is

F(z) = -\frac{\mathrm{d}V(z)}{\mathrm{d}z} = +e\frac{\mathrm{d}\Phi}{\mathrm{d}z} = -\frac{qe}{4\pi\varepsilon_0 R^3}z.

If this is a restoring force, it should be negative for positive $z$ . Thus, $q > 0$ .

A.4 (0.1 points)

The equation of motion for an electron is

m\ddot{z} + \frac{qe}{4\pi\varepsilon_0 R^3}z = 0

(here dots denote time derivatives). We therefore get

\omega = \sqrt{\frac{qe}{4\pi m\varepsilon_0 R^3}}.

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

B.1 (1.5 points)

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

Top view of the ring with centre O, a point B inside at distance r from O, and a point C on the ring; dashed segments OC and BC with angle phi at O. Axes x and y, with z out of the page. — 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 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 - 2Rr\cos\phi.

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

\mathrm{d}\Phi = \frac{1}{4\pi\varepsilon_0}\frac{\lambda R\,\mathrm{d}\phi}{\sqrt{R^2+r^2-2Rr\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:

\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}{4R^2}\right).

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

\beta = \frac{1}{16\pi\varepsilon_0 R^3}.

The ring in perspective with axes x, y, z; a small coaxial cylinder of radius r and height 2z centred at the centre of the ring. — Figure 2: Calculating electrostatic potential in the plane of the ring via Gauss’s law.

(ii) 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}{2R^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 $rz$ . This, for example, immediately gives us $\alpha = 1/(4\pi\varepsilon_0 R)$ . 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, \qquad \mathcal{E}_r(z,r) = -2q\beta r.

Applying Gauss’s law to the cylinder we obtain:

\oint \vec{\mathcal{E}}\cdot\mathrm{d}\vec{S} = 0 \qquad \Rightarrow \qquad \int\limits_{\mathrm{side}}\vec{\mathcal{E}}\cdot\mathrm{d}\vec{S} + \int\limits_{\mathrm{base}}\vec{\mathcal{E}}\cdot\mathrm{d}\vec{S} = 0.

The second integral is:

\int\limits_{\mathrm{base}}\vec{\mathcal{E}}\cdot\mathrm{d}\vec{S} = 2\pi r^2\mathcal{E}_z(z,r) = \frac{qzr^2}{2\varepsilon_0 R^3}.

The first integral is:

\int\limits_{\mathrm{side}}\vec{\mathcal{E}}\cdot\mathrm{d}\vec{S} = 4\pi rz\mathcal{E}_r(z,r) = -8\pi q\beta r^2 z.

Gauss’s theorem thus gives:

\frac{qzr^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.

B.2 (0.2 points)

The potential of the electron is $V(r) = -e\Phi(r)$ . Force acting on the electron in the $xy$ plane is

F(r) = -\frac{\mathrm{d}V(r)}{\mathrm{d}r} = +e\frac{\mathrm{d}\Phi(r)}{\mathrm{d}r} = \frac{qe}{8\pi\varepsilon_0 R^3}r.

To have oscillations we need the force to be negative for $r > 0$ . Thus, $q < 0$ .

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

C.1 (1.3 points)

Let us consider an electron with the velocity $v = \sqrt{2E/m}$ at a distance $r$ from the “optical” axis (Figure 2 of the problem). The electron crosses the “active region” of the lens in time

t = \frac{d}{v}.

The equation of motion in the $r$ direction:

m\ddot{r} = 2eq\beta r.

During the time the electron crosses the active region of the lens, the electron acquires radial velocity:

v_r = \frac{2eq\beta r}{m}\frac{d}{v} < 0.

The lens will be focusing if $q < 0$ . The time it takes for an electron to reach the “optical” axis is:

t' = \frac{r}{|v_r|} = -\frac{mv}{2eq\beta d}.

During this time the electron travels in the $z$ -direction a distance

\Delta z = t'v = -\frac{mv^2}{2eq\beta d} = -\frac{E}{eqd\beta}.

$\Delta z$ does not depend on the radial distance $r$ , therefore all electron will cross the “optical” axis (will be focused) in the same spot. Thus,

f = -\frac{E}{eqd\beta}.

C.2 (0.8 points)

Side view of the ring (height 2R, charge q) in a shaded active region of width d on the dashed optical axis. An electron leaves a point on the axis at distance b from the ring at angle gamma, is deflected at the ring and reaches the axis at distance c beyond the ring. Axes y, z with x into the page. — 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{2eq\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' = \frac{r}{|v_r|} = -\frac{r}{\frac{2eq\beta r}{m}\frac{d}{v}+v\frac{r}{b}} = -\frac{1}{\frac{2eq\beta}{m}\frac{d}{v}+\frac{v}{b}}.

During this time it will travel a distance

c = t'v = -\frac{1}{\frac{2eq\beta}{m}\frac{d}{v^2}+\frac{1}{b}} = -\frac{1}{\frac{eq\beta d}{E}+\frac{1}{b}}.

C.3 (0.2 pt)

From the previous answer we obtain:

\frac{1}{b} + \frac{1}{c} = -\frac{eq\beta d}{E}.

Comparing with the answer of C.1 we immediately obtain

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

i.e. the equation of a thin optical lens is valid for an electrostatic lens as well.

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

The ring with a narrow sector of angle 2alpha cut out at the top; point K marked on the ring at the middle of the sector, dashed lines from the centre. — Figure 4: Calculation of the capacitance of the ring.

D.1 (2.0 points)

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 = \frac{q}{4\pi^2\varepsilon_0 R}\ln\left(y+\sqrt{1+y^2}\right)\Big|_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).

The ring with points K at the top and L on the right; dashed radii from the centre to K and L enclose the angle phi. — 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 K and L of Figure 5:

|\mathrm{KL}| = 2R\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}{2R\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} = -\ln\left(\frac{\cos\chi+1}{\sin\chi}\right)\Big|_{\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{8R}{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{8R}{a}\right)}.

$C \to 0$ as $a \to 0$ .

D.2 (1.0 point)

Let $q(t)$ be the charge on the ring at a time $t$ . Potential of the disk is thus $q(t)/C$ . Voltage drop of the resistor is $R_0 I(t) = R_0\,\mathrm{d}q/\mathrm{d}t$ . Therefore for time $-\frac{d}{2v} < t < \frac{d}{2v}$ :

\frac{q(t)}{C} + R_0\frac{\mathrm{d}q}{\mathrm{d}t} = V_0.

Integrating this equation and keeping in mind that $q(t) = 0$ at $t = -d/(2v)$ , we get:

q(t) = CV_0\left(1-\mathrm{e}^{-\frac{d}{2vR_0C}}\mathrm{e}^{-\frac{t}{R_0C}}\right).

The charge attains the largest absolute value at $t = d/(2v)$ . The value of the charge at this time is:

q_0 = CV_0\left(1-\mathrm{e}^{-\frac{d}{vR_0C}}\right).

When $t > \frac{d}{2v}$ , we get:

\frac{q(t)}{C} + R_0\frac{\mathrm{d}q}{\mathrm{d}t} = 0.

From here:

q(t) = q_0\mathrm{e}^{-\frac{t}{R_0C}+\frac{d}{2vR_0C}} = CV_0\left(\mathrm{e}^{\frac{d}{2vR_0C}}-\mathrm{e}^{-\frac{d}{2vR_0C}}\right)\mathrm{e}^{-\frac{t}{RC}}.

Therefore, we obtain:

q(t) = \begin{cases} 0 & \text{for } t < -\frac{d}{2v}; \\ CV_0\left(1-\mathrm{e}^{-\frac{d}{2vR_0C}}\mathrm{e}^{-\frac{t}{R_0C}}\right) & \text{for } -\frac{d}{2v} < t < \frac{d}{2v}; \\ CV_0\left(\mathrm{e}^{\frac{d}{2vR_0C}}-\mathrm{e}^{-\frac{d}{2vR_0C}}\right)\mathrm{e}^{-\frac{t}{R_0C}} & \text{for } t > \frac{d}{2v}. \end{cases}

For a lens to be focusing we require that charge is negative, therefore $V_0 < 0$ . The dependence of charge on time is shown in Figure 6.

Graph of q(t) versus t: zero before -d/2v, then decreasing to the minimum q_0 at t = d/2v (dashed lines), then relaxing exponentially back towards zero. — Figure 6: Charge on the ring as a function of time.

Part E. Focal length of a more realistic lens (2 points)

E.1 (1.7 points)

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

m\ddot{r} = 2eq(t)\beta r,

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

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

As $f/v \gg R_0C$ , 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/(2v)}^{\infty} q(t)\,\mathrm{d}t.

We can use the derived equations for $q(t)$ and find the integrals. The integral $\int_{-d/(2v)}^{d/(2v)} q(t)\,\mathrm{d}t$ is (using the notation $d/v = t_0$ , $R_0C = \tau$ , $CV_0 = Q_0$ ):

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

\int_{t_0/2}^{\infty} Q_0\left(\mathrm{e}^{\frac{t_0}{2\tau}}-\mathrm{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)dt = Q_0 t_0.

Interestingly, it does not depend on $\tau = R_0C$ . Therefore, the acquired vertical velocity of the electron is

v_r = \eta r\frac{CV_0 d}{v} = \frac{2e\beta CV_0 dr}{mv}.

Following the logic similar to part C, we derive the focal length

f = -\frac{E}{eCV_0 d\beta}.

E.2 (0.3 points).

Comparing $f = -E/(eCV_0 d\beta)$ with $f = -E/(eqd\beta)$ from part C we immediately obtain $q_{\mathrm{eff}} = CV_0$ .

Оригинал в Архива: Th2.pdf · официални решения: Th2-Solution.pdf