Съдържание

Условие Решение

USAPhO 2001, Полуфинал — A2

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Всички материали от USAPhO 2001

2001 Semi-Final Exam · 25 т.

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.
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. It may not be used in graphing mode. Calculators may not be shared. You may not use any tables, books, or collections of formulas. You may use a ruler or straight edge.
Questions with the same point value are not necessarily of the same difficulty.
Good luck!

Possibly Useful Information

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	$k=1/4\pi\epsilon_0=9.0\times10^9$ 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	$c=3.0\times10^8$ 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	$e=1.6\times10^{-19}$ C
1 electron volt	$1\text{ eV}=1.6\times10^{-19}$ 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	$m=9.1\times10^{-31}\text{ kg}=0.51\text{ MeV}/c^2$
Binomial expansion	$(1+x)^n\simeq1+nx\quad\text{for }\lvert x\rvert\ll1$

Условие

Point charges $q_1$ , $q_2$ , and $Q$ are located at three corners of an $r_1$ by $r_2$ rectangle as shown in the diagram to the right.

Rectangle with q1 top left, Q top right, q2 bottom right and X bottom left; sides r1 and r2.

a. Write a general expression for the electrostatic field at the point marked X in the diagram. Define any quantities you introduce. [10 т.]

b. For what values of $q_1$ and $q_2$ (expressed in terms of $r_1$ , $r_2$ , and $Q$ ) does the field at X vanish? [5 т.]

c. Write a general expression for the electrostatic potential at X and evaluate it for the charges found in Part b. Assume the potential vanishes an infinite distance from the charges. [5 т.]

d. How much total work is done by the electric force as a fourth charge $q_4$ moves from infinity to the point marked X? Assume $q_1$ and $q_2$ are the charges found in Part b. [5 т.]

Решение

a. (10 т.)

We will write the fields $\vec E_1$ , $\vec E_2$ , and $\vec E_3$ due to $q_1$ , $q_2$ , and $Q$ , respectively, in terms of the usual unit vectors $\hat i$ and $\hat j$ . See accompanying diagram.

Charge rectangle with unit vectors i and j, angle theta and field arrows E1,E2,E3.

\vec E_1=-\hat j k\frac{q_1}{r_1^2}

where $k=$ Coulomb's constant

\vec E_2=-\hat i k\frac{q_2}{r_2^2}

\vec E_3=-(\hat i\cos\theta+\hat j\sin\theta)k\frac Q{(r_1^2+r_2^2)}.

Noting that $\cos\theta=\dfrac{r_2}{\sqrt{r_1^2+r_2^2}}$ and $\sin\theta=\dfrac{r_1}{\sqrt{r_1^2+r_2^2}}$ , the total field becomes

\vec E=-k\left(\frac{q_2}{r_2^2}+\frac{Qr_2}{(r_1^2+r_2^2)^{3/2}}\right)\hat i-k\left(\frac{q_1}{r_1^2}+\frac{Qr_1}{(r_1^2+r_2^2)^{3/2}}\right)\hat j.

b. (5 т.)

The field at X will vanish if each component vanishes:

\left(\frac{q_2}{r_2^2}+\frac{Qr_2}{(r_1^2+r_2^2)^{3/2}}\right)=0\quad\text{or}\quad q_2=-\frac{Qr_2^3}{(r_1^2+r_2^2)^{3/2}}

\left(\frac{q_1}{r_1^2}+\frac{Qr_1}{(r_1^2+r_2^2)^{3/2}}\right)=0\quad\text{or}\quad q_1=-\frac{Qr_1^3}{(r_1^2+r_2^2)^{3/2}}

c. (5 т.)

The electrostatic potential at X is

V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}+k\frac Q{(r_1^2+r_2^2)^{1/2}}.

For the charges found in Part b

V=-k\frac{Qr_1^2}{(r_1^2+r_2^2)^{3/2}}-k\frac{Qr_2^2}{(r_1^2+r_2^2)^{3/2}}+k\frac Q{(r_1^2+r_2^2)^{1/2}}=k\frac{Q(-r_1^2-r_2^2+r_1^2+r_2^2)}{(r_1^2+r_2^2)^{3/2}}=0

d. (5 т.)

The total work is done by the electric force as a fourth charge $q_4$ is moved from infinity to the point marked X is

W=-\Delta U=-q_4\Delta V=-q_4(V_X-V_\infty)=-q_4(0-0)=0

Оригинал в Архива: Semifinal 2001.pdf · официални решения: Semifinal 2001 Solutions.pdf