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Условие Отговори Решение

USAPhO 2009, Четвъртфинал — Задача 3

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Всички материали от USAPhO 2009

2009 Quarter-Final Exam · 25 т.

Important Instructions for the Exam Supervisor

Quarterfinal Exam

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Important Instructions for the Exam Supervisor

This examination consists of one part.
The first page that follows is a cover sheet. Examinees may keep the cover sheet during the exam.
Allow 60 minutes to complete the exam. Examinees may read the cover sheet before beginning the exam, but may not look at the examination questions until the 60 minute time period begins.
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 8, 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.
The examinees will need to use a ruler for one of the questions on this exam. They may not share rulers with other examinees.
AAPT must receive the students answer papers no later than Thursday, March 5, 2009. Marking of papers will occur that weekend, and the semifinalists will be selected by March 8, 2009. It will not be possible to mark late papers.

Указания

4 QUESTIONS - 60 MINUTES

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Show all your work. Partial credit will be given.
Start each question on a new sheet of paper. 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

Prob. 1 - P. 1/3
A ruler will be required on this exam.
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 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.
Each of the four questions is worth 25 points. The questions are not necessarily of the same difficulty. Good luck!
In order to maintain exam security, do not communicate any information about the questions (or their answers or solutions) on this contest until after March 10, 2009.

Условие

A parallel plate capacitor is made of two square parallel plates of area $A$ , and separated by a distance $d \ll \sqrt{A}$ . The capacitor is connected to a battery with potential $V$ and allowed to fully charge. The battery is then disconnected. A square metal conducting slab also with area $A$ but thickness $d/2$ is then fully inserted between the plates, so that it is always parallel to the plates. How much work has been done on the metal slab while it is being inserted?

Отговори

Решение

Solution

The capacitance of a thin rectangular capacitor is given by

C = \epsilon_0 A/d

If you forgot this, then derive it. The electric field between the plates is given by

E = \frac{Q}{\epsilon_0 Ad}

so the potential difference between the plates must be

V = Ed = \frac{Q}{\epsilon_0 A}

and the capacitance is

C = Q/V = \epsilon_0 A/d

For the original capacitor we’ll call this $C_0$ .

The energy stored in a capacitor is $U = \frac{1}{2}QV = \frac{1}{2}C{V_0}^2$ , so once charged, the initial energy is

U_0 = \frac{1}{2}\frac{\epsilon_0 A}{d}V_0^2

Note, we are going to need to know the initial charge sooner or later, which is

Q_0 = C_0 V_0 = \frac{\epsilon_0 A V_0}{d}

Inserting the conducting plate can be treated two ways: it results in the creation of two capacitors, in series, each with area $A$ but differing separation. The net charge $Q_0$ won’t change, that’s why we needed it above, but the resulting capacitance of the system will be

1/C_e = 1/C_1 + 1/C_2

For our capacitors, the only thing that changes is $d$ , which we’ll call $x_1$ and $x_2$ . Then

1/C_e = x_1/\epsilon_0 A + x_2/\epsilon_0 A

which gives us

C_e = \frac{\epsilon_0 A}{d/2} = 2C_0

Since the new total separation $x_1 + x_2$ is just $d - d/2$ .

The energy in the system after inserting the plate is

U = \frac{1}{2}Q_0 V = \frac{1}{2}Q_0^2/C_e

since $V$ is not constant!

Finally,

U = \frac{1}{2}Q_0^2/2C_0 = \frac{1}{2}U_0

The work done on the plate is the same as the work done on the system, which is given by

W = U - U_0

W = -\frac{1}{4}\frac{\epsilon_0 A}{d}V_0^2.

This means that the plate is “sucked” into the original capacitor, and we have to fight to hold it out.

Оригинал в Архива: 2009_QtrFinal.pdf · официални решения: 2009_QtrFinalSolutions.pdf