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USAPhO 2019, Полуфинал — Задача 3. Question A3

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Всички материали от USAPhO 2019

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

Внимание: Бележка към темата

Original question scan and matching revised key have been read in full. Printed sign/variable/wording defects in A1, A2, A3 and B1 are retained and documented; key alternatives and all answer sheets are preserved. The scanned B2(c) typo remains source faithful, with revised-key wording difference explicitly recorded.

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.
Divide the exam paper into 4 parts: the instructions (pages 2–3), Part A (pages 4–6), Part B (pages 8–10), and answer sheets for one of the questions in Part A (pages 12–13). The exam should be printed single-sided to facilitate dividing the test and scanning the answer sheets.
Provide students with the instructions for the competition (pages 2–3). Students can keep the pages for both parts of the exam, as they contain a reference list of physical constants.
Provide students with blank sheets of paper as scratch paper. Students are not allowed to bring their own papers.
Then provide students with Part A and the associated answer sheets, and allow 90 minutes to complete Part A. Do not give students Part B during this time, even if they finish with time remaining. At the end of the 90 minutes, collect the solutions to Part A along with the answer sheets and questions.
Students are allowed a 10 to 15 minute break between Parts A and B. Then allow 90 minutes to complete Part B. Do not let students go back to Part A.
At the end of the exam, the supervisor must collect all papers, including the questions, the instructions, and any scratch paper used by the students. Students may not take the exam questions. The examination questions may be returned to the students after April 8th, 2019.
Students are allowed calculators, but they may not use symbolic math, programming, or graphical features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDAs, or cameras may not be used during the exam or while the exam papers are present. Students may not use any tables, books, or collections of formulas.

We acknowledge the following people for their contributions to this year’s exam (in alphabetical order):

JiaJia Dong, Abijith Krishnan, Brian Skinner, and Kevin Zhou.

Instructions for the Student

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

At the beginning of the exam, you shall be provided with the instruction sheets, blank papers (both for your answers and scratch work), and the exam packet.
Work on Part A first. You have 90 minutes to complete three problems. Each question is worth 25 points, but they are not necessarily of the same difficulty. Do not look at Part B during this time.
After you have completed Part A you may take a break. You may consider checking your answers to Part A with the remaining time as you will not be allowed to return to Part A once you start Part B.
Then work on 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 your work and reasoning. Partial credit will be given if you make your reasoning clear. Do not write on the back of any page. Further guidance is given on the next page.
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, PDAs 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.
In order to maintain exam security, do not communicate any information about the questions of this exam, or their solutions until after April 8th, 2019.

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 $\vert \theta\vert \ll1$

Instructions for the Student — Further Guidance

Following is some further guidance on formatting your solutions. Start each question on a new sheet of paper. Put your AAPT ID number, your proctor’s AAPT ID number, the problem number and the page number and total number of pages for this problem, in the upper right hand corner of each page. As an example, the second page of your solution to B3 might look as follows.

student AAPT #29485 proctor AAPT #1038 B3: 2/2

Remember to also write the AAPT ID numbers on the provided answer sheets. Write single- sided to facilitate scanning. You may use either pencil or pen, but in either case, make sure to write sufficiently clearly so your work will be legible after scanning. To preserve anonymity of grading, do not write your name on any sheet.

End of Instructions for the Student

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

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

Divide the exam paper into 4 parts: the instructions (pages 2–3), Part A (pages 4–12), Part B (pages 14–23), and answer sheets for one of the questions in Part A (pages 25–26). The exam should be printed single-sided to facilitate dividing the test and scanning the answer sheets.

Условие

Electric Slide

Two large parallel plates of area $A$ are placed at $x=0$ and $x=d\ll\sqrt A$ in a semiconductor medium. The plate at $x=0$ is grounded, and the plate at $x=d$ is at a fixed potential $-V_0$ , where $V_0>0$ . Particles of positive charge $q$ flow between the two plates. You may neglect any dielectric effects of the medium.

a.

For large $V_0$ , the velocity of the positive charges is determined by a strong drag force, so that

v=\mu E

where $E$ is the local electric field and $\mu$ is the charge mobility.

i. In the steady state, there is a nonzero but time-independent density of charges between the two plates. Let the charge density at position $x$ be $\rho(x)$ . Use charge conservation to find a relationship between $\rho(x)$ , $v(x)$ , and their derivatives.

ii. Let $V(x)$ be the electric potential at $x$ . Derive an expression relating $\rho(x)$ , $V(x)$ , and their derivatives. (Hint: start by using Gauss's law to relate the charge density $\rho(x)$ to the derivative of the electric field $E(x)$ .)

iii. Suppose that in the steady state, conditions have been established so that $V(x)$ is proportional to $x^b$ , where $b$ is an exponent you must find, and the current is nonzero. Derive an expression for the current in terms of $V_0$ and the other given parameters.

b.

For small $V_0$ , the positive charges move by diffusion. The current due to diffusion is given by Fick's Law,

I=-AD\frac{d\rho}{dx}.

Here, $D$ is the diffusion constant, which you can assume to be described by the Einstein relation

D=\frac{\mu k_BT}q,

where $T$ is the temperature of the system.

i. Assume that in the steady state, conditions have been established so that a nonzero, steady current flows, and the electric potential again satisfies $V(x)\propto x^{b'}$ , where $b'$ is another exponent you must find. Derive an expression for the current in terms of $V_0$ and the other given parameters.

ii. At roughly what voltage $V_0$ does the system transition from this regime to the high voltage regime of the previous part?

STOP: Do Not Continue to Part B

If there is still time remaining for Part A, you should review your work for Part A, but do not continue to Part B until instructed by your exam supervisor.

Once you start Part B, you will not be able to return to Part A.

Решение

a.

i. In the steady state, the current is the same everywhere. Consider the region $(x,x+dx)$ . The time it takes for the charge in the second region to leave is $dx/v(x)$ . The amount of charge that leaves is $\rho A\,dx$ . The current is thus given by $\rho Av$ , so $\rho v$ is constant. Alternatively, one can write this as

v\frac{d\rho}{dx}+\rho\frac{dv}{dx}=0.

Both forms were accepted.

ii. Let us find an expression for the electric field at position $x$ . The position $x$ is effectively in between two uniform sheets of charge density. The sheet on the left has charge density $\int_0^x\rho\,dx+\sigma_0$ , where $\sigma_0$ is the charge density on the left plate, and the sheet on the right has charge density $\int_x^d\rho\,dx+\sigma_d$ , where $\sigma_d$ is the charge density on the left plate. Then, the electric field is given by

E=\sigma_0/(2\epsilon_0)+\int_0^x\rho/(2\epsilon_0)\,dx-\int_x^d\rho/(2\epsilon_0)\,dx-\sigma_d/(2\epsilon_0).

Then, by the Fundamental Theorem of Calculus

\frac{dE}{dx}=\frac\rho{\epsilon_0},

\frac{d^2V}{dx^2}=-\frac\rho{\epsilon_0}.

This can also be derived from the differential form of Gauss's Law more easily and is known as Poisson's equation.

iii. We have that

\rho\frac{dv}{dx}+v\frac{d\rho}{dx}=0,

and now that $v=-\mu\frac{dV}{dx}$ , so substituting in Poisson's equation gives us that

\left(\frac{d^2V}{dx^2}\right)^2+\frac{dV}{dx}\left(\frac{d^3V}{dx^3}\right)=0.

Using $V(x)=-V_0(x/d)^b$ gives

b(b-1)b(b-1)=-bb(b-1)(b-2).

The solution with $b=0$ cannot satisfy the boundary conditions, while $b=1$ has zero current. Assuming $b$ is neither of these values, we have $b-1=-(b-2)$ , so $b=3/2$ .

Substituting gives

v=-\frac{3V_0\mu x^{1/2}}{2d^{3/2}}

and

\rho=-\frac{3V_0\epsilon_0}{4d^{3/2}x^{1/2}},

I=\rho Av=\frac{9\epsilon_0\mu AV_0^2}{8d^3},

with the current flowing from left to right.

b.

i. We again have that $V(x)=V_0(x/d)^b$ . Note that from Poisson's equation, $\frac{d\rho}{dx}=-\epsilon_0\frac{d^3V}{dx^3}$ , so we need $b=3$ for this expression to be constant. Therefore,

I=\frac{6\mu k_BTA\epsilon_0V_0}{qd^3}.

ii. We find the crossover voltage by equating our answers in previous parts, to get

\frac{6\mu k_BTA\epsilon_0V_0}{qd^3}=\frac{9\epsilon_0\mu AV_0^2}{8d^3},

V_0=\frac{16k_BT}{3q}.

Оригинал в Архива: 2019_USAPhO.pdf · официални решения: 2019_USAPHO_revised-solutions.pdf