Съдържание

Условие Решение

USAPhO 2006, Полуфинал — A2

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Всички материали от USAPhO 2006

2006 Semi-Final Exam · 25 т.

2006 Semi-Final Exam — Instructions

AAPT / AIP — United States Physics Team 2006.

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Work Part A first. You have 90 minutes to complete all four problems.
After you have completed Part A, you may take a break.
Then work Part B. You have 90 minutes to complete both problems.
Show all your work. Partial credit will be given.
Start each question on a new sheet of paper. Be sure to put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example: 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.
Do not discuss the contents of this exam with anyone until after March 20th.
Good luck!

Possibly Useful Information – (Use for both part A and for part B)

Quantity	Value
Gravitational field at the Earth’s surface	$g=9.8$ N/kg
Newton’s gravitational constant	$G=6.67\times10^{-11}$ N·m $^2$ /kg $^2$
Coulomb’s constant	$k=1/4\pi\epsilon_0=8.99\times10^9$ N·m $^2$ /C $^2$
Biot-Savart constant	$k_m=\mu_0/4\pi=10^{-7}$ T·m/A
Speed of light in a vacuum	$c=3.00\times10^8$ m/s
Boltzmann’s constant	$k_B=1.38\times10^{-23}$ J/K
Avogadro’s number	$N_A=6.02\times10^{23}$ (mol) $^{-1}$
Ideal gas constant	$R=N_Ak_B=8.31$ J/(mol·K)
Stefan-Boltzmann constant	$\sigma=5.67\times10^{-8}$ J/(s·m $^2$ ·K $^4$ )
Elementary charge	$e=1.602\times10^{-19}$ C
1 electron volt	$1\text{ eV}=1.602\times10^{-19}$ J
Planck’s constant	$h=6.63\times10^{-34}\text{ J}\cdot\text{s}=4.14\times10^{-15}\text{ eV}\cdot\text{s}$
Electron mass	$m=9.109\times10^{-31}\text{ kg}=0.511\text{ MeV}/c^2$
Binomial expansion	$(1+x)^n\approx1+nx$ for $\lvert x\rvert\ll1$
Small angle approximations	$\sin\theta\approx\theta$ , $\cos\theta\approx1-\tfrac12\theta^2$

Условие

A parallel plate capacitor is constructed from two vertical square conducting surfaces with sides of length $L$ separated horizontally by a distance $2d$ . Assume that $d\ll L$ , so that fringing effects can be neglected. The plates are charged to $Q$ and $-Q$ , and then electrically isolated. A square conducting slab, also with sides of length $L$ , but with thickness $d$ , is then inserted vertically between the two plates without making contact with either plate. Assume that the slab moves only in the vertical direction. Give your answers to each part only in terms of the variables listed above and any appropriate fundamental constants.

Capacitor with plate separation2d and heightL, inserted slab thicknessd and insertiony, downward gravityg.

a. Find an expression for the capacitance $C$ of the system assuming that the movable slab extends inward between the plates by a distance $y$ . [7 т.]

b. Find an expression for the mass of the slab $m$ given that the equilibrium location (where the force of gravity on the slab is balanced by the attraction into the capacitor) for the slab is $y=L/2$ . [9 т.]

c. The slab is perturbed vertically slightly from its equilibrium position so that it undergoes simple harmonic motion. Find an expression for the frequency of the motion of the slab. [9 т.]

Решение

a. (7 т.)

Treat the system as two capacitors in parallel. One has a spacing of $2d$ and a surface area of $L(L-y)$ , the other a spacing of $2d-d$ and a surface area of $Ly$ . The capacitance is then found from the formula $\epsilon_0A/d$ and the rule for capacitors in parallel to be

C=\frac{\epsilon_0L(L-y)}{2d}+\frac{\epsilon_0Ly}{d}=\frac{\epsilon_0L(L+y)}{2d}.\qquad (A2-1)

b. (9 т.)

The potential energy stored in the capacitor is given by

U=\frac12\frac{Q^2}{C},\qquad (A2-2)

so the electric force of attraction on the slab is

F=-\frac{\partial U}{\partial y}\qquad (A2-3)

=\frac{Q^2}{2C^2}\frac{\partial C}{\partial y}\qquad (A2-4)

=\frac{dQ^2}{\epsilon_0L(L+y)^2}.\qquad (A2-5)

But this is equal to the force from gravity, $mg$ , when $y=L/2$ , so

m=\frac{dQ^2}{\epsilon_0gL(L+L/2)^2}=\frac{4dQ^2}{9\epsilon_0gL^3}.\qquad (A2-6)

c. (9 т.)

We need to expand the expression for the force about the point $y=L/2$ . Letting $y=L/2+x$ , we get

F=\frac{dQ^2}{\epsilon_0L(3L/2+x)^2}\qquad (A2-7)

=\frac{4dQ^2}{9\epsilon_0L^3}\left(1+\frac{2x}{3L}\right)^{-2}\qquad (A2-8)

\approx\frac{4dQ^2}{9\epsilon_0L^3}\left(1-2\frac{2x}{3L}\right)\qquad (A2-9)

=\frac{4dQ^2}{9\epsilon_0L^3}-\frac{16dQ^2}{27\epsilon_0L^4}x.\qquad (A2-10)

We know the first term in the last line balances gravity; it is the second term that is of interest. It is useful to write it in terms of the mass,

\Delta F=-\frac{16dQ^2}{27\epsilon_0L^4}x=-\frac{4mg}{3L}x,\qquad (A2-11)

so that the effective spring constant $k$ is given by

k=\frac{4mg}{3L}.\qquad (A2-12)

The frequency of small oscillations is then

f=\frac1{2\pi}\sqrt{\frac km}=\frac1\pi\sqrt{\frac g{3L}}.\qquad (A2-13)

Оригинал в Архива: Semifinal 2006.pdf · официални решения: Semifinal 2006 Solutions.pdf