Съдържание

Условие Отговори Решение

USAPhO 2007, Полуфинал — Задача 3. A3

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Всички материали от USAPhO 2007

AAPT AIP United States Physics Team 2007 — 2007 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. 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 27th.
Good luck!

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

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	$k = 1/4\pi\varepsilon_o = 8.99 \times 10^{9}\ \mathrm{N\cdot m^2/C^2}$
Biot-Savart constant	$k_m = \mu_o/4\pi = 10^{-7}\ \mathrm{T\cdot m/A}$
Speed of light in a vacuum	$c = 3.00 \times 10^{8}\ \mathrm{m/s}$
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	$R = N_A k_B = 8.31\ \mathrm{J/(mol\cdot K)}$
Stefan-Boltzmann constant	$\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$
Elementary charge	$e = 1.602 \times 10^{-19}\ \mathrm{C}$
1 electron volt	$1\ \mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}$
Planck’s constant	$h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s} = 4.14 \times 10^{-15}\ \mathrm{eV\cdot s}$
Electron mass	$m = 9.109 \times 10^{-31}\ \mathrm{kg} = 0.511\ \mathrm{MeV/c^2}$
Binomial expansion	$(1 + x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$
Small angle approximations	$\sin\theta \approx \theta$ ; $\cos\theta \approx 1 - \frac{1}{2}\theta^2$

Условие

A volume $V_f$ of fluid with uniform charge density $\rho$ is sprayed into a room, forming spherical drops. As they float around the room, the drops may break apart into smaller drops or coalesce into larger ones. Suppose that all of the drops have radius $R$ . Ignore inter-drop forces and assume that $V_f \gg R^3$ .

a. Calculate the electrostatic potential energy of a single drop. (Hint: suppose the sphere has radius $r$ . How much work is required to increase the radius by $dr$ ?). [10 т.]

b. What is the total electrostatic energy of the drops? [4 т.]

Your answer to (b) should indicate that the total energy increases with $R$ . In the absence of surface tension, then, the fluid would break apart into infinitesimally small drops. Suppose, however, that the fluid has a surface tension $\gamma$ . (This value is the potential energy per unit surface area, and is positive.)

c. What is the total energy of the drops due to surface tension? [4 т.]

d. What is the equilibrium radius of the drops? [7 т.]

Отговори

Решение

a. If the sphere has radius $r$ , it has charge

q = \frac{4}{3}\pi\rho r^3 \qquad (A3-1)

and thus its surface is at electrostatic potential

V = \frac{q}{4\pi\epsilon_0 r} = \frac{\rho r^2}{3\epsilon_0} \qquad (A3-2)

To increase the radius by $dr$ , an additional charge $dq = 4\pi r^2 dr$ must be brought in from infinity, requiring work

dU = V\,dq\ = \frac{4\pi r^4\rho^2}{3\epsilon_0}dr \qquad (A3-3)

Thus to grow the sphere from $r = 0$ to $r = R$ requires

U = \int_0^R \frac{4\pi r^4\rho^2}{3\epsilon_0}dr = \frac{4\pi R^5\rho^2}{15\epsilon_0} \qquad (A3-4)

b. Each drop has volume $V_d = \frac{4}{3}\pi R^3$ , so the number of drops is

n = \frac{V_f}{V_d} = \frac{V_f}{\frac{4}{3}\pi R^3} \qquad (A3-5)

Since we are ignoring inter-drop forces, the total energy of the drops is simply the sum of the energies of each individual drop:

U_{e,tot} = nU = \frac{V_f}{\frac{4}{3}\pi R^3}\frac{4\pi R^5\rho^2}{15\epsilon_0} = \frac{R^2\rho^2}{5\epsilon_0}V_f \qquad (A3-6)

c. Each drop has surface area $4\pi R^2$ and thus surface tension energy $4\pi R^2\gamma$ . As before, the total energy due to surface tension is just the sum of the energies of the individual drops:

U_{s,tot} = 4\pi R^2\gamma n = 4\pi R^2\gamma\frac{V_f}{\frac{4}{3}\pi R^3} = \frac{3\gamma}{R}V_f \qquad (A3-7)

d. The total potential energy from both sources is

U_{tot} = \left(\frac{R^2\rho^2}{5\epsilon_0} + \frac{3\gamma}{R}\right)V_f \qquad (A3-8)

Equilibrium is reached when the total energy is a minimum; since $U \to \infty$ at both $R \to 0$ and $R \to \infty$ , it must have an interior minimum.

\frac{d}{dR}U_{tot} = \left(\frac{2R\rho^2}{5\epsilon_0} - \frac{3\gamma}{R^2}\right)V_f \qquad (A3-9)

Setting this equal to zero,

\frac{2R\rho^2}{5\epsilon_0} = \frac{3\gamma}{R^2} \qquad (A3-10)

R^3 = \frac{15\gamma\epsilon_0}{2\rho^2} \qquad (A3-11)

R = \left(\frac{15\gamma\epsilon_0}{2\rho^2}\right)^{\frac{1}{3}} \qquad (A3-12)

Оригинал в Архива: olympiad_2007_semi-final.pdf · официални решения: olympiad_2007_semi-final_soln.pdf