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USAPhO 2016, Полуфинал — Задача 6. Question B2

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Всички материали от USAPhO 2016

USA Physics Olympiad Exam — Part A / Part B · 50 т.

Внимание: Бележка към темата

The native worked key contains printed mathematical inconsistencies at A1(b), B1 and B2(a)(iii)/(c)(ii); the source formulas and adjacent reasoning are retained as printed. The A3 key graphs omit the requested event-letter labels and do not draw the stopped-rod world-line changes; no new answer labels or lines have been invented. Exact originals and detailed source-defect evidence accompany this transcription.

Important Instructions for the Exam Supervisor — DO NOT DISTRIBUTE THIS PAGE

This examination consists of two parts: Part A has four questions and is allowed 90 minutes; Part B has two questions and is allowed 90 minutes.
The first page that follows is a cover sheet. Examinees may keep the cover sheet for both parts of the exam.
The parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining.
Allow 90 minutes to complete Part A. Do not let students look at Part B. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minutes break between parts A and B.
Allow 90 minutes to complete Part B. Do not let students go back to Part A.
Ideally the test supervisor will divide the question paper into 4 parts: the cover sheet (page 2), Part A (pages 3-6), Part B (pages 8-9), and several answer sheets for one of the questions in Part A (pages 11-12). Examinees should be provided parts A and B individually, although they may keep the cover sheet. The answer sheets should be printed single sided!
The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after April 15, 2016.
Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas.

USA Physics Olympiad Exam — Instructions

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN - Work Part A first. You have 90 minutes to complete all four problems. Each question is worth 25 points. Do not look at Part B during this time. - After you have completed Part A you may take a break. - Then work Part B. You have 90 minutes to complete both problems. Each question is worth 50 points. Do not look at Part A during this time. - Show all your work. Partial credit will be given. Do not write on the back of any page. Do not write anything that you wish graded on the question sheets. - Start each question on a new sheet of paper. Put your AAPT ID number, your name, the question number and the page number/total pages for this problem, in the upper right hand corner of each page. For example,

AAPT ID # Doe, Jamie A1 - 1/3

A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas. - Questions with the same point value are not necessarily of the same difficulty. - In order to maintain exam security, do not communicate any information about the questions (or their answers/solutions) on this contest until after April 15, 2016.

Possibly Useful Information. You may use this sheet for both parts of the exam.

Information	Value
$g$	$9.8\ \mathrm{N/kg}$
$k=1/(4\pi\varepsilon_0)$	$8.99\times10^9\ \mathrm{N\cdot m^2/C^2}$
$c$	$3.00\times10^8\ \mathrm{m/s}$
$N_A$	$6.02\times10^{23}\ (\mathrm{mol})^{-1}$
$\sigma$	$5.67\times10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$
$1\ \mathrm{eV}$	$1.602\times10^{-19}\ \mathrm J$
$m_e$	$9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$
$\sin\theta$	$\approx\theta-\frac16\theta^3$ for $\vert \theta\vert \ll1$
$G$	$6.67\times10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
$k_m=\mu_0/(4\pi)$	$10^{-7}\ \mathrm{T\cdot m/A}$
$k_B$	$1.38\times10^{-23}\ \mathrm{J/K}$
$R=N_Ak_B$	$8.31\ \mathrm{J/(mol\cdot K)}$
$e$	$1.602\times10^{-19}\ \mathrm C$
$h$	$6.63\times10^{-34}\ \mathrm{J\cdot s}=4.14\times10^{-15}\ \mathrm{eV\cdot s}$
$(1+x)^n$	$\approx1+nx$ for $\vert x\vert \ll1$
$\cos\theta$	$\approx1-\frac12\theta^2$ for $\vert \theta\vert \ll1$

Условие

a.

A spherical region of space of radius $R$ has a uniform charge density and total charge $+Q$ . An electron of charge $-e$ is free to move inside or outside the sphere, under the influence of the charge density alone. For this first part ignore radiation effects.

i. Consider a circular orbit for the electron where $r<R$ . Determine the period of the orbit $T$ in terms of any or all of $r$ , $R$ , $Q$ , $e$ , and any necessary fundamental constants.

ii. Consider a circular orbit for the electron where $r>R$ . Determine the period of the orbit $T$ in terms of any or all of $r$ , $R$ , $Q$ , $e$ , and any necessary fundamental constants.

iii. Assume the electron starts at rest at $r=2R$ . Determine the speed of the electron when it passes through the center in terms of any or all of $R$ , $Q$ , $e$ , and any necessary fundamental constants.

b.

Accelerating charges radiate. The total power $P$ radiated by charge $q$ with acceleration $a$ is given by

P=C\xi a^n

where $C$ is a dimensionless numerical constant (which is equal to $1/6\pi$ ), $\xi$ is a physical constant that is a function only of the charge $q$ , the speed of light $c$ , and the permittivity of free space $\epsilon_0$ , and $n$ is a dimensionless constant. Determine $\xi$ and $n$ .

c.

Consider the electron in the first part, except now take into account radiation. Assume that the orbit remains circular and the orbital radius $r$ changes by an amount $|\Delta r|\ll r$ .

i. Consider a circular orbit for the electron where $r<R$ . Determine the change in the orbital radius $\Delta r$ during one orbit in terms of any or all of $r$ , $R$ , $Q$ , $e$ , and any necessary fundamental constants. Be very specific about the sign of $\Delta r$ .

ii. Consider a circular orbit for the electron where $r>R$ . Determine the change in the orbital radius $\Delta r$ during one orbit in terms of any or all $r$ , $R$ , $Q$ , $e$ , and any necessary fundamental constants. Be very specific about the sign of $\Delta r$ .

Решение

a.

i. We apply Gauss's law,

\frac{Q_{\mathrm{in}}}{\epsilon_0}=\oint\vec E\cdot d\vec A.

This yields

\frac Q{\epsilon_0}\frac{r^3}{R^3}=4\pi r^2E\quad\Rightarrow\quad E=\frac Q{4\pi\epsilon_0}\frac r{R^3}

Since the motion is circular,

m\frac{4\pi^2r}{T^2}=eE=\frac{eQ}{4\pi\epsilon_0}\frac r{R^3}

and solving for $T$ gives

T=2\pi\sqrt{\frac{4\pi\epsilon_0mR^3}{eQ}}.

It is independent of $r$ since the motion is simple harmonic.

ii. Applying Gauss's law as in the previous part gives

E=\frac Q{4\pi\epsilon_0}\frac1{r^2}

as expected by the shell theorem; one could also just write this down directly. Using the same circular motion equation,

m\frac{4\pi^2r}{T^2}=e\frac{eQ}{4\pi\epsilon_0}\frac1{r^2}

and solving for $T$ gives

T=2\pi\sqrt{\frac{4\pi\epsilon_0mr^3}{eQ}}.

It is proportional to $r^{3/2}$ in accordance with Kepler's third law.

iii. We use the above results to compute the potential difference,

\begin{aligned} \Delta V&=-\int_{2R}^0\vec E\cdot d\vec s,\\ &=\int_{2R}^R\frac Q{4\pi\epsilon_0}\frac1{r^2}+\int_R^0\frac Q{4\pi\epsilon_0}\frac r{R^3},\\ &=\frac Q{4\pi\epsilon_0}\left(\frac{-1}{2R}-\frac{-1}R+\frac{R^2}{2R^3}\right),\\ &=\frac Q{4\pi\epsilon_0R}. \end{aligned}

By energy conservation,

v=\sqrt{\frac2m e\Delta V}=\sqrt{\frac{2eQ}{4\pi\epsilon_0mR}}.

b.

This is a dimensional analysis problem. The most straightforward method is to write out all the dimensions explicitly. Note that $a$ has dimensions of $[\mathrm L]/[\mathrm T]^2$ , $P$ has dimensions of $[\mathrm M][\mathrm L]^2/[\mathrm T]^3$ , $c$ has dimensions of $[\mathrm L]/[\mathrm T]$ , $q$ has dimensions of $[\mathrm C]$ , and $\epsilon_0$ has dimensions of $[\mathrm C]^2[\mathrm T]^2/[\mathrm M][\mathrm L]^3$ . The equation

P=a^\alpha c^\beta\epsilon_0^\gamma q^\delta

has dimensions

[\mathrm M][\mathrm L]^2/[\mathrm T]^3=\left([\mathrm L]/[\mathrm T]^2\right)^\alpha\left([\mathrm L]/[\mathrm T]\right)^\beta\left([\mathrm C]^2[\mathrm T]^2/[\mathrm M][\mathrm L]^3\right)^\gamma([\mathrm C])^\delta

Mass is only balanced if $\gamma=-1$ . As a result, charge is balanced if $\delta=2$ . Proceeding similarly for length and time,

P=\frac1{6\pi}a^2c^{-3}\epsilon_0^{-1}q^2

giving answers of $\xi=q^2/c^3\epsilon_0$ and $n=2$ .

c.

i. The energy radiated away is given by

\Delta E=-PT

where $T$ is determined in the previous sections.

It is possible to compute the actual energy of each orbit, and it is fairly trivial to do for regions $r>R$ , but perhaps there is an easier, more entertaining way. Consider

\Delta E=\Delta K+\Delta U

and for small changes in $r$ ,

\frac{\Delta U}{\Delta r}\approx-F=\frac{eQ}{4\pi\epsilon_0}\frac r{R^3}.

This implies the potential energy increases with increasing $r$ , as expected. Now

\frac{\Delta K}{\Delta r}\approx\frac d{dr}\left(\frac12mv^2\right)=\frac12\frac d{dr}\left|r\frac{mv^2}r\right|

but $mv^2/r=F$ , so

\frac{\Delta K}{\Delta r}\approx\frac12\frac d{dr}|rF|=\frac{eQ}{4\pi\epsilon_0}\frac r{R^3}.

This implies the kinetic energy increases with increasing $r$ , also as expected, as this region acts like a multidimensional simple harmonic oscillator. Combining,

\frac{\Delta E}{\Delta r}\approx2\frac{eQ}{4\pi\epsilon_0}\frac r{R^3}=2ma

Finally,

\Delta r=-\left(\frac1{6\pi}\frac{a^2}{c^3\epsilon_0}e^2\right)\left(2\pi\sqrt{\frac{4\pi\epsilon_0mR^3}{eQ}}\right)\left(\frac1{2ma}\right).

Plugging in the value of $a$ , this can be simplified to

\Delta r=-\frac16\sqrt{\frac{e^5Q}{4\pi\epsilon_0^3R(mc^2)^3}}\frac rR.

Alternatively, we can write the result in terms of dimensionless groups,

\Delta r=-\frac{2\pi}3\left(\frac{e^2}{4\pi\epsilon_0Rmc^2}\right)\sqrt{\frac{eQ}{4\pi\epsilon_0Rmc^2}}r.

ii. Picking up where we left off,

\frac{\Delta U}{\Delta r}\approx-F=\frac{eQ}{4\pi\epsilon_0}\frac1{r^2}.

This implies the potential energy increases with increasing $r$ .

\frac{\Delta K}{\Delta r}\approx\frac12\frac d{dr}|rF|=\frac{\Delta K}{\Delta r}\approx-\frac{eQ}{8\pi\epsilon_0}\frac1{r^2}.

This implies the kinetic energy decreases with increasing $r$ , a somewhat nonintuitive but true statement for circular orbits. Combining,

\frac{\Delta E}{\Delta r}\approx\frac12\frac{eQ}{4\pi\epsilon_0}\frac r{R^3}=\frac{ma}2.

Using the same manipulations as before,

\Delta r=-\left(\frac1{6\pi}\frac{a^2}{c^3\epsilon_0}e^2\right)\left(2\pi\sqrt{\frac{4\pi\epsilon_0mr^3}{eQ}}\right)\left(\frac2{ma}\right).

Plugging in the value of $a$ , this can be simplified to

\Delta r=-\frac23\sqrt{\frac{e^5Q}{4\pi\epsilon_0^3r(mc^2)^3}}.

Alternatively, we can write the result in terms of dimensionless groups,

\Delta r=-\frac{8\pi}3\left(\frac{e^2}{4\pi\epsilon_0Rmc^2}\right)\sqrt{\frac{eQ}{4\pi\epsilon_0Rmc^2}}\frac{R^2}r.

Оригинал в Архива: USAPhO-2016.pdf · официални решения: USAPhO-2016-Solutions.pdf