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USAPhO 2018, Полуфинал — Задача 4. Question B1

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Всички материали от USAPhO 2018

USA Physics Olympiad Exam — Part A / Part B · 25 т.

Внимание: Бележка към темата

The supervisor and student security dates differ in the original packet. The printed diode key retains its positive negative-voltage exponential and positive discharge derivative despite surrounding text/graphs. The radiation-pressure key retains inconsistent intensity/energy-density terminology and its classical electromagnetic definitions. Exact original sources and detailed defect evidence accompany the transcription.

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

This examination consists of two parts: Part A has three questions and is allowed 90 minutes; Part B also has three 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 3 parts: the cover sheet (page 2), Part A (pages 3-5), Part B (pages 7-9), and several answer sheets for one of the questions in Part A (pages 11-13). 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 21, 2018.
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 three 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 three problems. Each question is worth 25 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 proctor’s AAPT ID number, the question number and the page number/total pages for this problem, in the upper right hand corner of each page. For example,

Student AAPT ID # Proctor AAPT ID # 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 13, 2018.

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

$g$	$9.8\ \mathrm{N/kg}$
$k=1/(4\pi\varepsilon_0)$	$8.99\times10^9\ \mathrm{N\cdot m^2/C^2}$
$c$	$3.00\times10^8\ \mathrm{m/s}$
$N_A$	$6.02\times10^{23}\ (\mathrm{mol})^{-1}$
$\sigma$	$5.67\times10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$
$1\ \mathrm{eV}$	$1.602\times10^{-19}\ \mathrm J$
$m_e$	$9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$
$\sin\theta$	$\approx\theta-\frac16\theta^3$ 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 $\vert \theta\vert \ll1$

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

Ideally the test supervisor will divide the question paper into 3 parts: the cover sheet (page 2), Part A (pages 3-10), Part B (pages 12-18), and several answer sheets for one of the questions in Part A (pages 20-22). Examinees should be provided parts A and B individually, although they may keep the cover sheet. The answer sheets should be printed single sided!

Условие

Part B

The electric potential at the center of a cube with uniform charge density $\rho$ and side length $a$ is

\Phi\approx\frac{0.1894\rho a^2}{\epsilon_0}.

You do not need to derive this.[^1]

For the entirety of this problem, any computed numerical constants should be written to three significant figures.

[^1]: See https://arxiv.org/pdf/chem-ph/9508002.pdf for more details if you are interested.

a. What is the electric potential at a corner of the same cube? Write your answer in terms of $\rho$ , $a$ , $\epsilon_0$ , and any necessary numerical constants.

b. What is the electric potential at the tip of a pyramid with a square base of side length $a$ , height $a/2$ , and uniform charge density $\rho$ ? Write your answer in terms of $\rho$ , $a$ , $\epsilon_0$ , and any necessary numerical constants.

c. What is the electric potential due to a square plate with side length $a$ of uniform charge density $\sigma$ at a height $a/2$ above its center? Write your answer in terms of $\sigma$ , $a$ , $\epsilon_0$ , and any necessary numerical constants.

d. Let $E(z)$ be the electric field at a height $z$ above the center of a square with charge density $\sigma$ and side length $a$ . If the electric potential at the center of the square is approximately $0.281a\sigma/\epsilon_0$ , estimate $E(a/2)$ by assuming that $E(z)$ is linear in $z$ for $0<z<a/2$ . Write your answer in terms of $\sigma$ , $a$ , $\epsilon_0$ , and any necessary numerical constants.

Решение

a.

By dimensional analysis, the answer takes the form

\Phi_c(a,\rho)\approx\frac{C\rho a^2}{\epsilon_0}

for a dimensionless constant $C$ . Note that a cube of side length $a$ consists of 8 cubes of side length $a/2$ , each with a corner at the center of the larger cube. Then

\frac{0.1894\rho a^2}{\epsilon_0}=8\frac{C\rho(a/2)^2}{\epsilon_0}

so $C=0.1894/2=0.0947$ .

b.

A cube of side length $a$ consists of 6 such pyramids. Then we simply compute $0.1894/6$ for

\Phi_p(a,\rho)\approx\frac{0.0316\rho a^2}{\epsilon_0}.

c.

Let the potential due to such a square be $\Phi_s(a,\sigma)$ . Note that adding a square plate of infinitesimal thickness $dz$ and side length $a$ to a square pyramid with base side length $a$ and height $a/2$ yields a square pyramid with base side length $a+2dz$ and height $a/2+dz$ .

The surface charge density of a square plate with thickness $dz$ and volume charge density $\rho$ is $\sigma=\rho\,dz$ . Then by the principle of superposition,

\Phi_s(a,\rho\,dz)=\Phi_p(a+2dz,\rho)-\Phi_p(a,\rho)\approx\frac{0.0316\rho\left((a+2dz)^2-a^2\right)}{\epsilon_0}=\frac{0.126a\rho\,dz}{\epsilon_0}.

so we have

\Phi_s(a,\sigma)\approx\frac{0.126a\sigma}{\epsilon_0}.

d.

The potential difference between height 0 and $a/2$ is

\Delta\Phi=(0.281-0.126)\frac{a\sigma}{\epsilon_0}=\frac{0.155a\sigma}{\epsilon_0}.

On the other hand, we have

\Delta\Phi=\int_0^{a/2}E(z)\,dz\approx\frac a2\frac{E(0)+E(a/2)}2

where we approximated $E(z)$ as linear, and $E(0)=\sigma/2\epsilon_0$ by Gauss's law. Solving for $E(a/2)$ ,

E(a/2)\approx\frac{0.119\sigma}{\epsilon_0}

where the last significant digit is not important. Incidentally, the actual value is exactly $\sigma/6\epsilon_0$ , and this fact has a slick calculation-free proof.

Оригинал в Архива: USAPhO-2018.pdf · официални решения: USAPhO-2018-Solutions.pdf