Съдържание

OPhO 2020, Открит кръг — Задача 46. Sandwiched!

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от OPhO 2020

2020 Online Physics Olympiad (OPhO): Open Contest v1.5 · 25 май 2020 г.

2020 Online Physics Olympiad (OPhO): Open Contest v1.5

OPhO Lead Organizers and Writers: Adithya Balachandran, Eddie Chen, Ashmit Dutta, Ethan Hu, Viraj Jayam, Rishab Parthasarathy, Kushal Thaman, Max Wang, QiLin Xue

OPhO Website Developer: Ben Oommen

Advisors: Dr. Branislav Kisačanin, Kye Shi, Kevin Zhou

Указания

If you wish to request a clarification, please use this form. To see all clarifications, see this document.

Use $g = 9.81\ \mathrm{m/s^2}$ in this contest. See the constants sheet on the following page for other constants.
This test contains 55 short answer questions. Each problem will have three possible attempts.
The weight of each question depends on our scoring system found here. Put simply, the later questions are worth more, and the overall amount of points from a certain question decreases with the number of attempts and days that you take to solve a problem as well as the number of teams who solve it. This means that your score decreases with the number of tries and days you take to solve a given problem.
Any team member is able to submit an attempt. Choosing to split up the work or doing each problem together is up to you. Note that after you have submitted an attempt, your teammates must refresh their page before they are able to see it.
Answers should contain at least three significant figures, unless otherwise specified. All answers within the 1% range will be accepted.
When submitting a response using scientific notation, please use exponential form. In other words, if your answer to a problem is $A \times 10^B$ , please type $AeB$ into the submission portal.
A standard scientific or graphing handheld calculator may be used. Technology and computer algebra systems like Wolfram Alpha or the one in the TI nSpire will not be needed. Attempts to use these tools will be classified as cheating.
You are allowed to use Wikipedia or books in this exam. Asking for help on online forums or your teachers will be considered cheating and may result in a possible ban from future competitions.
Top scorers from this contest will qualify to compete in the Online Physics Olympiad Invitational Contest, which is an olympiad-style exam. More information will be provided to invitational qualifiers after the end of the Open Contest.
In general, answer in SI units (meter, second, kilogram, watt) unless otherwise specified. Please answer all questions in degrees unless otherwise specified.
If the question asks to give your answer as a percent and your answer comes out to be “ $x\%$ ”, please input the value $x$ into the submission form.
Do not put letters in your answer on the submission portal! If your answer is “ $x$ meters” , input only the value $x$ into the submission portal.
Do not communicate information to anyone else apart from your team-members before the exam ends on May 29, 2020 at 11:59 PM UTC.

List of Constants

Proton mass, $m_p = 1.67 \cdot 10^{-27}\ \mathrm{kg}$
Neutron mass, $m_n = 1.67 \cdot 10^{-27}\ \mathrm{kg}$
Electron mass, $m_e = 9.11 \cdot 10^{-31}\ \mathrm{kg}$
Avogadro’s constant, $N_0 = 6.02 \cdot 10^{23}\ \mathrm{mol^{-1}}$
Universal gas constant, $R = 8.31\ \mathrm{J/(mol \cdot K)}$
Boltzmann’s constant, $k_B = 1.38 \cdot 10^{-23}\ \mathrm{J/K}$
Electron charge magnitude, $e = 1.60 \cdot 10^{-19}$
1 electron volt, $1\ \mathrm{eV} = 1.60 \cdot 10^{-19}\ \mathrm{J}$
Speed of light, $c = 3.00 \cdot 10^{8}\ \mathrm{m/s}$
Universal Gravitational constant,
$G = 6.67 \cdot 10^{-11}\ \mathrm{(N \cdot m^2)/kg^2}$
Acceleration due to gravity, $g = 9.81\ \mathrm{m/s^2}$
1 unified atomic mass unit,
$1\ \mathrm{u} = 1.66 \cdot 10^{-27}\ \mathrm{kg} = 931\ \mathrm{MeV/c^2}$
Planck’s constant,
$h = 6.63 \cdot 10^{-34}\ \mathrm{J \cdot s} = 4.41 \cdot 10^{-15}\ \mathrm{eV \cdot s}$
Permittivity of free space,
$\epsilon_0 = 8.85 \cdot 10^{-12}\ \mathrm{C^2/(N \cdot m^2)}$
Coulomb’s law constant,
$k = \frac{1}{4\pi\epsilon_0} = 8.99 \cdot 10^{9}\ \mathrm{(N \cdot m^2)/C^2}$
Permeability of free space, $\mu_0 = 4\pi \cdot 10^{-7}\ \mathrm{T \cdot m/A}$
Magnetic constant,
$\frac{\mu_0}{4\pi} = 1 \cdot 10^{-7}\ \mathrm{(T \cdot m)/A}$
1 atmospheric pressure,
$1\ \mathrm{atm} = 1.01 \cdot 10^{5}\ \mathrm{N/m^2} = 1.01 \cdot 10^{5}\ \mathrm{Pa}$
Wien’s displacement constant, $b = 2.9 \cdot 10^{-3}\ \mathrm{m \cdot K}$
Stefan-Boltzmann constant,
$\sigma = 5.67 \cdot 10^{-8}\ \mathrm{W/m^2/K^4}$

Условие

A point charge $+q$ is placed a distance $a$ away from an infinitely large conducting plate. The force of the electrostatic interaction is $F_0$ . Then, an identical conducting plate is placed a distance $3a$ from the charge, parallel to the first one such that the charge is “sandwiched in.” The new electrostatic force the particle feels is $F'$ . What is $F'/F_0$ ? Round to the nearest hundredths.

Отговори

Решение

Solution: We solve this via the method of image charges. Let us first reflect the charge $+q$ across the closest wall. The force of this interaction will be:

F_0 = \frac{q^2}{4\pi\epsilon_0 a^2}\frac{1}{2^2}

and this will be the only image charge we need to place if there were only one conducting plane. Since there is another conducting plane, another charge will be reflected to a distance $a + 4a$ past the other conducting plane, and thus will be $a + 4a + 3a = 8a$ away from the original charge. All these reflections cause a force that points in the same direction, which we will label as the positive $+$ direction. Therefore:

F_+ = \frac{q^2}{4\pi\epsilon_0 a^2}\left(\frac{1}{2^2} + \frac{1}{8^2} + \frac{1}{10^2} + \frac{1}{16^2}\frac{1}{18^2} + \frac{1}{24^2} + \cdots\right)

Now let us look at what happens if we originally reflect the charge $+q$ across the other wall. Repeating the steps above, we see that through subsequent reflections, each force will point in the negative $-$ direction. Therefore:

F_- = \frac{-q^2}{4\pi\epsilon_0 a^2}\left(\frac{1}{6^2} + \frac{1}{8^2} + \frac{1}{14^2} + \frac{1}{16^2} + \frac{1}{22^2} + \frac{1}{24^2} + \cdots\right)

The net force is a result of the superposition of these two forces, giving us:

\begin{aligned} F' = \frac{q^2}{4\pi\epsilon_0 a^2}\Big(&\frac{1}{2^2} + \frac{1}{8^2} + \frac{1}{10^2} + \frac{1}{16^2}\frac{1}{18^2} + \frac{1}{24^2} + \cdots \\ &-\frac{1}{6^2} - \frac{1}{8^2} - \frac{1}{14^2} - \frac{1}{16^2} - \frac{1}{22^2} - \frac{1}{24^2} - \cdots\Big) \end{aligned}

Even terms can be cancelled out to give:

\begin{aligned} F' &= \frac{q^2}{4\pi\epsilon_0 a^2}\left(\frac{1}{2^2} - \frac{1}{6^2} + \frac{1}{10^2} - \frac{1}{14^2} + \frac{1}{18^2} - \frac{1}{22^2} + \cdots\right) \\ &= \frac{q^2}{4\pi\epsilon_0 a^2}\frac{1}{4}\left(\frac{1}{1^2} - \frac{1}{3^2} + \frac{1}{5^2} - \frac{1}{7^2} + \frac{1}{9^2} - \frac{1}{11^2} + \cdots\right) \end{aligned}

You may recognize the infinite series inside the parentheses to be Catalan’s constant $G \approx 0.916$ . Alternatively, you can use a calculator and evaluate the first seven terms to get a rough answer (but will still be correct since we asked for it to be rounded). Therefore:

F'/F = \frac{1}{1^2} - \frac{1}{3^2} + \frac{1}{5^2} - \frac{1}{7^2} + \frac{1}{9^2} - \frac{1}{11^2} + \cdots = G \approx \boxed{0.916}

Оригинал в Архива: open.pdf · официални решения: open_sol.pdf