Съдържание

Условие Отговори Решение

USAPhO 2002, Полуфинал — Задача 6. B2

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Всички материали от USAPhO 2002

2002 Semi-Final Exam

Указания

INSTRUCTIONS

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

Gravitational field at the Earth’s surface	$g = 9.8\ \mathrm{N/kg}$
Newton’s gravitational constant	$G = 6.67 \times 10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
Coulomb’s constant	$k = 1/4\pi\varepsilon_o = 9.0 \times 10^{9}\ \mathrm{N\cdot m^2/C^2}$
Biot-Savart constant	$k_m = \mu_o/4\pi = 10^{-7}\ \mathrm{T\cdot m/A}$
Speed of light in a vacuum	$c = 3.0 \times 10^{8}\ \mathrm{m/s}$
Boltzmann’s constant	$k_B = 1.38 \times 10^{-23}\ \mathrm{J/K}$
Avogadro’s number	$N_A = 6.02 \times 10^{23}\ \mathrm{(mol)^{-1}}$
Ideal gas constant	$R = N_Ak_B = 8.31\ \mathrm{J/(mol\cdot K)}$
Stefan-Boltzmann constant	$\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$
Elementary charge	$e = 1.6 \times 10^{-19}\ \mathrm{C}$
1 electron volt	$1\ \mathrm{eV} = 1.6 \times 10^{-19}\ \mathrm{J}$
Planck’s constant	$h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s} = 4.14 \times 10^{-15}\ \mathrm{eV\cdot s}$
Electron mass	$m = 9.1 \times 10^{-31}\ \mathrm{kg} = 0.51\mathrm{MeV}/c^2$
Binomial expansion	$(1 + x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$

Условие

Cross-section in the x–z plane of a shaded sphere of radius R centered at the origin, with an unshaded spherical hole of radius R/2 touching the top of the sphere; a point charge q sits at the center of the hole. Arrows mark the radii R and R/2; axes x (right) and z (up). — Figure B2-1

The same cross-section as Figure B2-1 but without the point charge: a shaded sphere of radius R with an empty spherical hole of radius R/2 touching the top; arrows mark R and R/2; axes x and z. — Figure B2-2

A total charge $+Q$ is uniformly distributed throughout a sphere of radius $R$ . Charge is removed from a spherical hole of radius $R/2$ centered on the $z = R/2$ point of the z-axis and concentrated into a point charge $q$ which is placed at the center of the hole. See Fig. B1-1 above.

a. Find the charge $q$ in terms of $Q$ . [5 т.]

b. Find the electric field at all points along the $x$ -axis. [10 т.]

c. The charge $q$ is now removed, leaving the charge distribution seen in Fig. B2-2 above. Find the electric field at all points along the $z$ -axis. [10 т.]

d. Find the electrostatic potential at an arbitrary point $x$ , $y$ , $z$ outside the sphere of radius $R$ as shown in Fig. B2-2 above. Assume the potential vanishes at an infinite distance from the charges. [10 т.]

e. Very far from the charge distribution, the electrostatic potential may be written in the following form

V = k\left(\frac{a}{r} + \frac{\vec{b}\cdot\vec{r}}{r^3} + \ldots\right)

where $r$ is the distance from the origin at the center of the large sphere. Using your result from part d) above, determine the constant $a$ and the constant vector $\vec{b}$ . ( $\vec{b}$ is called the electric dipole moment.) [10 т.]

f. Interpret $a$ and $\vec{b}$ in terms of a set of discrete charges and their locations. [5 т.]

Отговори

Решение

(5) a. If the charge $Q$ is uniformly distributed, the charge density is

\rho = \frac{Q}{\frac{4}{3}\pi R^3}.

The charge in a sphere of radius $R/2$ is

q = \rho\tfrac{4}{3}\pi\left(\frac{R}{2}\right)^3 = \tfrac{1}{8}\rho\tfrac{4}{3}\pi R^3 = \tfrac{1}{8}\frac{Q}{\frac{4}{3}\pi R^3}\tfrac{4}{3}\pi R^3 = \tfrac{1}{8}Q.

(10) b. The $x$ -axis is totally outside the cavity. By Gauss’s Law the field outside a spherical symmetric charge distribution of radius $R/2$ centered on $z = R/2$ is the same as that due to point charge at the center with the same total charge. So on the $x$ -axis the field is the same as that due to uniform charge distribution of total charge $Q$ and radius $R$ .

From Gauss’s Law for a spherically symmetric charge distribution

E = k\frac{Q_{enc}}{R^2}

where $k$ is Coulomb’s constant and $Q_{enc}$ is the total charge enclosed by a Gaussian sphere of radius $R$ .

Outside the sphere $\quad Q_{enc} = Q$

For $|x| > R$ $\quad E = k\dfrac{Q}{R^2}$ $\quad$ in a direction out from the origin.

Inside the sphere $\quad Q_{enc} = \rho\tfrac{4}{3}\pi|x|^3 = \dfrac{Q}{\frac{4}{3}\pi R^3}\tfrac{4}{3}\pi|x|^3 = \dfrac{Q|x|^3}{R^3}$

For $|x| < R$ $\quad E = k\dfrac{1}{x^2}\dfrac{Q|x|^3}{R^3} = k\dfrac{Q|x|}{R^3}$ $\quad$ in a direction out from the origin.

(10) c. The total field can be considered the sum of two terms. That due to:

a sphere of radius $R$ centered on the origin with uniformly distributed total charge $+Q$ .

a sphere of radius $R/2$ centered on $z = R/2$ with uniformly distributed total charge $-Q/8$ .

Outside the large sphere both fields are equal to those due to point charges at their centers.

For $|z| > R$ $\quad E = k\dfrac{Q}{z^2} - k\dfrac{Q}{8(z - R/2)^2}$ $\quad$ out from the origin.

Inside the large sphere but outside the cavity, treat the cavity as a point charge at its center and the sphere as an extended distribution

For $-R < z < 0$ $\quad E = k\dfrac{Q|z|}{R^3} - k\dfrac{Q}{8(z - R/2)^2}$ $\quad$ out from the origin.

Inside the cavity, treat both the sphere and cavities as extended charge distributions.

$0 < z < R/2$ $\quad E = k\dfrac{Qz}{R^3} + k\dfrac{Q(R/2 - z)}{8(R/2)^3} = k\dfrac{Qz}{R^3} + k\dfrac{Q(R/2 - z)}{R^3} = k\dfrac{Q}{2R^2}$ $\quad$ out from the origin.

$R/2 < z < R$ $\quad E = k\dfrac{Qz}{R^3} - k\dfrac{Q(z - R/2)}{8(R/2)^3} = k\dfrac{Qz}{R^3} + k\dfrac{Q(R/2 - z)}{R^3} = k\dfrac{Q}{2R^2}$ $\quad$ out from the origin.

(10) d. Outside the sphere, the electrostatic potential is the same as that due to a point charge of $+Q$ at the origin and a point charge of $-Q/8$ at $x = 0$ , $y = 0$ , $z = R/2$ .

Writing the distance from the origin as $\quad r = \sqrt{x^2 + y^2 + z^2}$

and the distance from at $x = 0$ , $y = 0$ , $z = R/2$ . as $\quad r' = \sqrt{x^2 + y^2 + (z - R/2)^2}$ .

The potential is

V = k\frac{Q}{r} - k\frac{q}{r'} = k\frac{Q}{\sqrt{x^2 + y^2 + z^2}} - k\frac{Q}{8\sqrt{x^2 + y^2 + (z - R/2)^2}}

(10) e. In order to get the electrostatic potential in the form shown, expand $r'$ in a binomial series.

r' = \sqrt{x^2 + y^2 + (z - R/2)^2} = \sqrt{x^2 + y^2 + z^2 - zR + R^2/4} = \sqrt{r^2 - zR + R^2/4} = r\sqrt{1 - \frac{zR}{r^2} + \frac{R^2}{4r^2}}

V = k\frac{Q}{r} - k\frac{q}{r'} = k\frac{Q}{r} - k\frac{Q}{8r\sqrt{1 - \frac{zR}{r^2} + \frac{R^2}{4r^2}}} \approx k\frac{Q}{r} - k\frac{Q}{8r}\left(1 + \frac{1}{2}\frac{zR}{r^2} - \frac{1}{2}\frac{R^2}{4r^2} + \ldots\right) \approx k\left(\frac{7Q}{8r} - \frac{QzR}{16r^3} + \ldots\right)

Only terms of order $\dfrac{1}{r}$ and of order $\dfrac{1}{\mathrm{r}^2}$ $\left(\text{such as } \dfrac{z}{r^3}\right)$ have been retained.

Comparing to the expression for $V$ given in 2Be $\quad a = \tfrac{7}{8}Q$

And with $z = \vec{r}\cdot\hat{k}$ $\quad \vec{b}\cdot\vec{r} = -\dfrac{QzR}{16} = -\dfrac{Q\left(\vec{r}\cdot\hat{k}\right)R}{16}.$

Which gives $\quad \vec{b} = -\dfrac{QR}{16}\hat{k}$

(5) f. A point charge of $\frac{7}{8}Q$ at the origin and a dipole consisting of $-\frac{1}{8}Q$ at (0, 0, $R/2$ ) and $+\frac{1}{8}Q$ at the origin.

Автори

The 2002 Semi-Final Examination questions were written by the coaches of the United States Physics Team. The coaches are: Academic Director, Mary Mogge – Professor of Physics at California State Polytechnic University, Pomona, CA; Senior Coach, Leaf Turner – Physicist in the Theoretical Division of Los Alamos National Laboratory, Los Alamos, NM; Warren Turner – Physics Teacher at Massachusetts Academy of Math and Science, Worcester, MA; Robert Shurtz – Physics Teacher at the Hawken School, Gates Mills, OH; Boris Zbarsky –undergraduate student at MIT, Cambridge, MA; Andrew Lin – undergraduate student at Yale University, New Haven, CT. Questions A2, A4, B1, and B2 were suggested by Leaf Turner and Questions A1 and A3 by Mary Mogge.

Оригинал в Архива: Semifinal 2002.pdf · официални решения: Semifinal 2002 Solutions.pdf