Съдържание

Условие Отговори Решение

USAPhO 2008, Полуфинал — Задача 1. Question A1

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Всички материали от USAPhO 2008

AAPT AIP UNITED STATES PHYSICS TEAM 2008 — Semifinal Exam · 25 т.

Important Instructions for the Exam Supervisor

DO NOT DISTRIBUTE THIS PAGE

This examination consists of three parts.
Part A has four questions and is allowed 90 minutes.
Part B has two questions and is allowed 90 minutes.
Part C has one question and is allowed 20 minutes. The answer for Part C will not be used for team selection, but will be used for special recognition from the Optical Society of America.
The first page that follows is a cover sheet. Examinees may keep the cover sheet for all three parts of the exam.
The three 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 or Part C. 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 look at Part C or go back to Part A. Collect the answers to part B before allowing the examinee to begin Part C. Examinees are allowed a 10 to 15 minutes break between Parts B and C.
Allow 20 minutes to complete Part C. This part is optional; scores on Part C will not be used to select the US Team. Examinees may not go back to Part A or B.
Ideally the test supervisor will divide the question paper into 4 parts: the cover sheet (page 2), Part A (pages 3-7), Part B (pages 8-10), and Part C (page 11). Examinees should be provided the parts individually, although they may keep the cover sheet.
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 March 31, 2008.
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.
Please provide the examinees with graph paper for Part A.

Указания

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 Parts B or C 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 Parts A or C 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 school 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,

School 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.
Part C is an optional part of the test. You will be given 20 additional minutes to complete Part C. Your score on Part C will not affect the selection for the US Team, but can be used for special prizes and recognition to be awarded by the Optical Society of America.
In order to maintain exam security, do not communicate any information about the questions (or their answers/solutions) on this contest until after March 31, 2008.

Possibly Useful Information

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

$g = 9.8\ \mathrm{N/kg}$ $\qquad$ $G = 6.67 \times 10^{-11}\ \mathrm{N \cdot m^2/kg^2}$

$k = 1/4\pi\epsilon_0 = 8.99 \times 10^{9}\ \mathrm{N \cdot m^2/C^2}$ $\qquad$ $k_\mathrm{m} = \mu_0/4\pi = 10^{-7}\ \mathrm{T \cdot m/A}$

$c = 3.00 \times 10^{8}\ \mathrm{m/s}$ $\qquad$ $k_\mathrm{B} = 1.38 \times 10^{-23}\ \mathrm{J/K}$

$N_\mathrm{A} = 6.02 \times 10^{23}\ (\mathrm{mol})^{-1}$ $\qquad$ $R = N_\mathrm{A} k_\mathrm{B} = 8.31\ \mathrm{J/(mol \cdot K)}$

$\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s \cdot m^2 \cdot K^4)}$ $\qquad$ $e = 1.602 \times 10^{-19}\ \mathrm{C}$

$1\mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}$ $\qquad$ $h = 6.63 \times 10^{-34}\ \mathrm{J \cdot s} = 4.14 \times 10^{-15}\ \mathrm{eV \cdot s}$

$m_e = 9.109 \times 10^{-31}\ \mathrm{kg} = 0.511\ \mathrm{MeV/c^2}$ $\qquad$ $(1+x)^n \approx 1 + nx$ for $|x| \ll 1$

$\sin\theta \approx \theta - \frac{1}{6}\theta^3$ for $|\theta| \ll 1$ $\qquad$ $\cos\theta \approx 1 - \frac{1}{2}\theta^2$ for $|\theta| \ll 1$

Условие

Four square metal plates of area $A$ are arranged at an even spacing $d$ as shown in the diagram. (Assume that $A \gg d^2$ .)

Four parallel square plates drawn in perspective, stacked one above another, labelled Plate 1 (top) to Plate 4 (bottom); arrows on the left mark the equal spacings d between neighbouring plates.

Plates 1 and 4 are first connected to a voltage source of magnitude $V_0$ , with plate 1 positive; plates 2 and 3 are then connected together with a wire. The wire is subsequently removed. Finally, the voltage source attached between plates 1 and 4 is replaced with a wire. The steps are summarized in the diagrams below.

Three side views of the four plates labelled Step 1, Step 2 and Step 3: in Step 1 a battery connects plates 1 and 4 and a wire joins plates 2 and 3; in Step 2 only the battery remains; in Step 3 the battery is replaced by a wire joining plates 1 and 4.

Find the resulting potential difference $\Delta V_{12}$ between plates 1 and 2; like wise find $\Delta V_{23}$ and $\Delta V_{34}$ , defined similarly.

Assume, in each case, that a positive potential difference means that the top plate is at a higher potential than the bottom plate.

Отговори

Решение

We treat the plates as three capacitors in series. Each has an identical capacitance $C$ . The figure below then show the three steps.

Three circuit diagrams of capacitors C1, C2, C3 in series: left, a battery across the chain with C2 shorted by a wire; middle, C2 still shorted but no battery; right, the top of C1 joined to the bottom of C3 by a wire.

Since $C_2$ is shorted out originally, then effectively there are only two capacitors in series, so the voltage drop across each is $V_0/2$ , where the a positive potential difference means that the top plate of any given capacitor is positive. The top plate of $C_1$ will then have a positive charge of $q_0 = CV_0/2$ . Note that this means that the bottom plate of the top capacitor will have a negative charge of $-q_0$ . Removing the shorting wire across $C_2$ will not change the charges or potential drops across the other two capacitors. Removing the source $V_0$ will also make no difference.

Shorting the top plate of $C_1$ with the bottom plate of $C_3$ will make a difference. Positive charge will flow out of top plate of $C_1$ into the bottom plate of $C_3$ . Also, negative charge will flow out of the bottom plate of $C_1$ into the top plate of $C_2$ . The result is that $C_1$ will acquire a potential difference of $V_1$ , $C_2$ a potential difference of $V_2$ , and $C_3$ a potential difference of $V_3$ . Let the final charge on the top plate of each capacitor also be labeled as $q_1$ , $q_2$ , and $q_3$ .

The last figure implies that

V_1 + V_2 + V_3 = 0.

By symmetry, we have

V_1 = V_3.

2V_1 = -V_2.

By charge conservation between the bottom plate of $C_1$ and the top plate of $C_2$ we have

-q_0 = -q_1 + q_2.

But $q = CV$ , so

-\frac{1}{2}V_0 = -V_1 + V_2

Combining the above we get

\begin{aligned} -\frac{1}{2}V_0 &= \frac{1}{2}V_2 + V_2, \\ -\frac{1}{3}V_0 &= V_2. \end{aligned}

Finally, solving for $V_1$ , we get $V_1 = V_0/6$ .

Three sketches of four horizontal plates with field arrows: left, downward fields E0 in the top and bottom gaps with the middle plates connected; middle, downward fields in the top and bottom gaps; right, downward fields E1 in the top and bottom gaps and upward field E2 in the middle gap, with the outer plates connected.

In the first picture we require that $2E_0 = V_0/d$ . The charge density on the second plate requires that $\Delta E = E_0$ . In the last picture we have $2E_1 + E_2 = 0$ , since the potential between the top plate and the bottom plate is zero. But we also have, on the second plate, $\Delta E = E_1 - E_2$ . Combining, $E_0 = -\frac{1}{2}E_2 - E_2 = -\frac{3}{2}E_2$ , and therefore $V_2 = -\frac{1}{3}V_0$ , and $V_1 = V_0/6$ .

Оригинал в Архива: olympiad_2008_semi-final.pdf · официални решения: USAPhO-2008-Solutions.pdf