Съдържание

Условие a–b c–d Решение

USAPhO 2000, Полуфинал — B1

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от USAPhO 2000

2000 Semi-Final Exam · 40 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
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

A hollow spherical shell of radius $b$ and negligible thickness is constructed out of insulating material with total charge $-Q$ . A second similar shell has radius $c$ where $c>2b$ and total charge $+Q$ . Each shell has its charge distributed uniformly over its surface. Assume the electrostatic potential is zero at infinity. As shown at the right, the symbols I, II, and III may be used to denote the regions inside the smaller shell, between the shells, and outside the larger shell, respectively. Express your answers to the following in terms of given quantities and known constants. You may use either spherical or Cartesian coordinates.

a–b

The spherical shells are concentric as shown in Figure a. at the right.

Figure a. Concentric spherical shells, radii b and c, charges -Q and +Q, regions I, II, III and axes.

a. Find the electrostatic field in regions I, II, and III as a function of position. [10 т.]

b. Find the electrostatic potential in regions I, II, and III as a function of position. [10 т.]

c–d

The spherical shells are not concentric as shown in Figure b at the right. The inner shell is shifted in the positive $z$ direction a distance $b$ equal to its radius.

Figure b. Inner charged spherical shell displaced a distance b in the positive z direction.

c. Find the electrostatic field in regions I, II, and III as a function of position. [10 т.]

d. Find the electrostatic potential in regions I, II, and III as a function of position. [10 т.]

Решение

a. (10 т.)

Because the charge distribution is uniform, there is spherical symmetry and we can use Gauss’s Law to calculate the electrostatic field. This yields

\vec E=k\frac{q_{enc}}{R^2}\vec e_R

where $R=\sqrt{x^2+y^2+z^2}$ , $q_{enc}$ is the charge enclosed by a Gaussian surface of radius R, and $\vec e_R$ is a unit vector in the $\boldsymbol{R}$ direction.

Region I: A Gaussian surface of radius $R<b$ encloses no charge, so $E_I=0$

Region II: A Gaussian surface of radius $c>R>b$ encloses $q_{enc}=-Q$

Thus

\vec E_{II}=-k\frac{Q}{R^2}\vec e_R\qquad\text{or}\qquad E_{II}=k\frac{Q}{R^2}\ \text{ radially inward.}\qquad (B1.1)

Region III: A Gaussian surface of radius $R>c$ encloses $q_{enc}=-Q+Q=0$

Therefore $E_{III}=0$

b. (10 т.)

Region III: Since the field is zero, the potential is a constant throughout the region. Its value at any point in III is the same as its value at infinity $V_{III}=0$

Region II:

V_{II}(R)-V_{II}(c)=-\int_c^R\vec E\cdot d\vec l

Using (B1.1), $d\vec l=(-\vec e_R)(-dR)=\vec e_RdR$ , and noting $V_{II}(c)=V_{III}(c)=0$

V_{II}(R)=\int_c^Rk\frac{Q}{r^2}\,dr=\left.-k\frac{Q}{r}\right|_{r=c}^{R}=k\frac{Q}{c}-k\frac{Q}{R}=kQ\left(\frac1c-\frac1R\right)

Region I: Since the field is zero, the potential is constant and is everywhere equal to its value on the boundary $V_I=V_{II}$ at $R=b$ . Thus $V_I=kQ\left(\frac1c-\frac1b\right)$ .

c. (10 т.)

In each region the total field can be obtained by superposition, adding vectorially the fields due to the individual spheres. $\vec E=\vec E_{+Q}+\vec E_{-Q}$ .

Inside each uniformly charged sphere the field due to that sphere is zero.

Region I: This region is inside both spheres so both contributions are zero and $E_I=0$ .

Region II: The field due to the $+Q$ sphere is zero in this region. The field due to the $-Q$ is equivalent to that due to a $-Q$ point charge at $x=0$ , $y=0$ , $z=b$ . Letting $\vec R=x\hat i+y\hat j+(z-b)\hat k$ represent the vector from this point charge to the arbitrary field point, we have

\vec E_{II}=-\frac{kQ\vec R}{R^3}=\frac{kQ(x\hat i+y\hat j+(z-b)\hat k)}{[x^2+y^2+(z-b)^2]^{3/2}}.

Region III: In this region both spheres contribute to the field. Adding to the previous result the contribution equivalent to a $+Q$ point charge at the origin, yields

\vec E_{III}=-\frac{kQ(x\hat i+y\hat j+(z-b)\hat k)}{[x^2+y^2+(z-b)^2]^{3/2}}+\frac{kQ(x\hat i+y\hat j+z\hat k)}{[x^2+y^2+z^2]^{3/2}}.

d. (10 т.)

Region III: Since the field here is the equivalent to that due to two point charges, so is the potential.

V_{III}=kQ\left(\frac1{\sqrt{x^2+y^2+z^2}}-\frac1{\sqrt{x^2+y^2+(z-b)^2}}\right).

Region II: In this region, the contribution due to the inner sphere is equivalent to that due to a point charge. The contribution due to the outer sphere is constant. To find this value, equate across the boundary with Region III.

V_{II}=kQ\left(\frac1c-\frac1{\sqrt{x^2+y^2+(z-b)^2}}\right).

Region I: Since the field is zero, the potential is constant. To find this value, equate across the boundary with Region II.

V_I=kQ\left(\frac1c-\frac1b\right).

Оригинал в Архива: Semifinal 2000.pdf · официални решения: Semifinal 2000 Solutions.pdf