Съдържание

Условие Отговори Решение

USAPhO 2009, Полуфинал — Задача 6. Question B2

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от USAPhO 2009

United States Physics Team 2009 — Semifinal Exam · 50 т.

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 3 parts: the cover sheet (page 2), Part A (pages 3-4), and Part B (pages 6-7). Examinees should be provided parts A and B 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, 2009.
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 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 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.
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, 2009.

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

$g = 9.8\ \mathrm{N/kg}$	$G = 6.67 \times 10^{-11}\ \mathrm{N \cdot m^2/kg^2}$
$k = 1/4\pi\epsilon_0 = 8.99 \times 10^{9}\ \mathrm{N \cdot m^2/C^2}$	$k_{\mathrm{m}} = \mu_0/4\pi = 10^{-7}\ \mathrm{T \cdot m/A}$
$c = 3.00 \times 10^{8}\ \mathrm{m/s}$	$k_{\mathrm{B}} = 1.38 \times 10^{-23}\ \mathrm{J/K}$
$N_{\mathrm{A}} = 6.02 \times 10^{23}\ \mathrm{(mol)^{-1}}$	$R = N_{\mathrm{A}} k_{\mathrm{B}} = 8.31\ \mathrm{J/(mol \cdot K)}$
$\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s \cdot m^2 \cdot K^4)}$	$e = 1.602 \times 10^{-19}\ \mathrm{C}$
$1\mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}$	$h = 6.63 \times 10^{-34}\ \mathrm{J \cdot s} = 4.14 \times 10^{-15}\ \mathrm{eV \cdot s}$
$m_e = 9.109 \times 10^{-31}\ \mathrm{kg} = 0.511\ \mathrm{MeV/c^2}$	$(1+x)^n \approx 1 + nx$ for $\lvert x \rvert \ll 1$
$\sin\theta \approx \theta - \frac{1}{6}\theta^3$ for $\lvert\theta\rvert \ll 1$	$\cos\theta \approx 1 - \frac{1}{2}\theta^2$ for $\lvert\theta\rvert \ll 1$

Условие

An electric dipole consists of two charges of equal magnitude $q$ and opposite sign, held rigidly apart by a distance $d$ . The dipole moment is defined by $p = qd$ .

Now consider two identical, oppositely oriented electric dipoles, separated by a distance $r$ , as shown in the diagram.

Two vertical dipoles: dipole A on the left with + above and − below, dipole B on the right with − above and + below; each has a small vertical double arrow labelled d beside it, and a horizontal double arrow between them is labelled r.

a. It is convenient when considering the interaction between the dipoles to choose the zero of potential energy such that the potential energy is zero when the dipoles are very far apart from each other. Using this convention, write an exact expression for the potential energy of this arrangement in terms of $q$ , $d$ , $r$ , and fundamental constants.

b. Assume that $d \ll r$ . Give an approximation of your expression for the potential energy to lowest order in $d$ . Rewrite this approximation in terms of only $p$ , $r$ , and fundamental constants.

c. What is the force (magnitude and direction) exerted on one dipole by the other? Continue to make the assumption that $d \ll r$ , and again express your result in terms of only $p$ , $r$ , and fundamental constants.

d. What is the electric field near dipole B produced by dipole A? Continue to make the assumption that $d \ll r$ and express your result in terms of only $p$ , $r$ , and fundamental constants.

Отговори

Решение

a. There are two $+q/-q$ pairs separated by a distance $d$ , each having potential energy

-\frac{q^2}{4\pi\epsilon_0 d}

There are two $+q/-q$ pairs separated by a distance $r$ , each having potential energy

-\frac{q^2}{4\pi\epsilon_0 r}

There are a $+q/+q$ pair and a $-q/-q$ pair separated by a distance $\sqrt{r^2 + d^2}$ , each having potential energy

\frac{q^2}{4\pi\epsilon_0\sqrt{r^2 + d^2}}

Note that the latter two terms go to zero as $r$ becomes large, whereas the first term is not dependent on $r$ . Thus the given zero convention will include only the latter two terms,

U = \frac{q^2}{4\pi\epsilon_0}\left(-\frac{2}{r} + \frac{2}{\sqrt{r^2 + d^2}}\right).

b. Rearranging the result above, we have

U = \frac{2q^2}{4\pi\epsilon_0 r}\left(\frac{1}{\sqrt{1 + \left(\frac{d}{r}\right)^2}} - 1\right).

Using the binomial approximation $(1+x)^n \approx 1 + nx$ ,

U \approx \frac{2q^2}{4\pi\epsilon_0 r}\left(1 - \frac{1}{2}\left(\frac{d}{r}\right)^2 - 1\right) = -\frac{q^2d^2}{4\pi\epsilon_0 r^3} = -\frac{p^2}{4\pi\epsilon_0 r^3}.

c. We can infer by symmetry that the force must point along the line separating the dipoles. Since the potential energy decreases with decreasing distance, the force is attractive. Its magnitude can be determined by taking the derivative of the potential energy:

F = -\frac{dU}{dr} = -3\frac{p^2}{4\pi\epsilon_0 r^4}

with the negative sign confirming that the force is attractive.

One can also use an approach analogous to the previous one, i.e. write down the force exactly and use a binomial approximation as above. One must take care to account for the fact that the force between like-signed charges is not exactly in the same direction as that between opposite-signed charges.

d. The electric field near dipole B produced by dipole A is directed approximately downward. Its magnitude is exactly

E = \frac{q}{4\pi\epsilon_0}\frac{2}{r^2 + (d/2)^2}\sin\left(\frac{d}{2r}\right).

Using $d \ll r$ and the small angle approximation $\sin\theta \approx \theta$ , we have

E \approx \frac{p}{4\pi\epsilon_0 r^3}

where we neglected all higher-order terms in $d/r$ .

Note that at this level of accuracy, the force on dipole B would be exactly zero. In order to extend this calculation to get the answer to part c, we would have to account for the fact that the electric fields at the two charges in dipole B do not point exactly downward. This is a smaller effect than what we’ve found in this part, which is why the answer to part (c) falls off as $1/r^4$ while the answer here falls off as $1/r^3$ .

Оригинал в Архива: 2009_SemiFinal.pdf · официални решения: USAPhO-2009-Solutions.pdf