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IPhO 2011, Теория — Задача 2. An Electrified Soap Bubble

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Theoretical Competition: 12 July 2011 — Question 2 · 12 юли 2011 г.

Условие

A spherical soap bubble with internal air density $\rho_i$ , temperature $T_i$ and radius $R_0$ is surrounded by air with density $\rho_a$ , atmospheric pressure $P_a$ and temperature $T_a$ . The soap film has surface tension $\gamma$ , density $\rho_s$ and thickness $t$ . The mass and the surface tension of the soap do not change with the temperature. Assume that $R_0 \gg t$ .

The increase in energy, $dE$ , that is needed to increase the surface area of a soap-air interface by $dA$ , is given by $dE = \gamma dA$ where $\gamma$ is the surface tension of the film.

2.1 Find the ratio $\dfrac{\rho_i T_i}{\rho_a T_a}$ in terms of $\gamma$ , $P_a$ and $R_0$ . [1,7 т.]

2.2 Find the numerical value of $\dfrac{\rho_i T_i}{\rho_a T_a} - 1$ using $\gamma = 0.0250\ \mathrm{Nm^{-1}}$ , $R_0 = 1.00\ \mathrm{cm}$ , and $P_a = 1.013 \times 10^{5}\ \mathrm{Nm^{-2}}$ . [0,4 т.]

2.3 The bubble is initially formed with warmer air inside. Find the minimum numerical value of $T_i$ such that the bubble can float in still air. Use $T_a = 300\ \mathrm{K}$ , $\rho_s = 1000\ \mathrm{kgm^{-3}}$ , $\rho_a = 1.30\ \mathrm{kgm^{-3}}$ , $t = 100\ \mathrm{nm}$ and $g = 9.80\ \mathrm{ms^{-2}}$ . [2 т.]

After the bubble is formed for a while, it will be in thermal equilibrium with the surrounding. This bubble in still air will naturally fall towards the ground.

2.4 Find the minimum velocity $u$ of an updraught (air flowing upwards) that will keep the bubble from falling at thermal equilibrium. Give your answer in terms of $\rho_s$ , $R_0$ , $g$ , $t$ and the air’s coefficient of viscosity $\eta$ . You may assume that the velocity is small such that Stokes’s law applies, and ignore the change in the radius when the temperature lowers to the equilibrium. The drag force from Stokes’ Law is $F = 6\pi\eta R_0 u$ . [1,6 т.]

2.5 Calculate the numerical value for $u$ using $\eta = 1.8 \times 10^{-5}\ \mathrm{kgm^{-1}\,s^{-1}}$ . [0,4 т.]

The above calculations suggest that the terms involving the surface tension $\gamma$ add very little to the accuracy of the result. In all of the questions below, you can neglect the surface tension terms.

2.6 If this spherical bubble is now electrified uniformly with a total charge $q$ , find an equation describing the new radius $R_1$ in terms of $R_0, P_a, q$ and the permittivity of free space $\varepsilon_0$ . [2 т.]

2.7 Assume that the total charge is not too large (i.e. $\dfrac{q^2}{\varepsilon_0 R_0^4} \ll P_a$ ) and the bubble only experiences a small increase in its radius, find $\Delta R$ where $R_1 = R_0 + \Delta R$ . Given that $(1+x)^n \approx 1 + nx$ where $x \ll 1$ . [0,7 т.]

2.8 What must be the magnitude of this charge $q$ in terms of $t, \rho_a, \rho_s, \varepsilon_0, R_0, P_a$ in order that the bubble will float motionlessly in still air? Calculate also the numerical value of $q$ . The permittivity of free space $\varepsilon_0 = 8.85 \times 10^{-12}\ \mathrm{farad/m}$ . [1,2 т.]

Отговори

Решение

2. SOLUTION

2.1. The bubble is surrounded by air.

A shaded sphere (the bubble) with centre O and radius R_0 drawn from O to the surface; inside the labels P_i, T_i, ρ_i; outside the labels P_a, T_a, ρ_a; an arrow labelled ρ_s, t points to the soap film.

Cutting the sphere in half and using the projected area to balance the forces give

P_i\pi R_0^2 = P_a\pi R_0^2 + 2(2\pi R_0\gamma)

P_i = P_a + \frac{4\gamma}{R_0} \qquad \ldots (1)

The pressure and density are related by the ideal gas law:

PV = nRT \quad \text{or} \quad P = \frac{\rho RT}{M}, \text{ where } M = \text{the molar mass of air.} \qquad \ldots (2)

Apply the ideal gas law to the air inside and outside the bubble, we get

\rho_i T_i = P_i\frac{M}{R}

\rho_a T_a = P_a\frac{M}{R},

\frac{\rho_i T_i}{\rho_a T_a} = \frac{P_i}{P_a} = \left[1 + \frac{4\gamma}{R_0 P_a}\right] \qquad \ldots (3)

2.2. Using $\gamma = 0.025\ \mathrm{Nm^{-1}}$ , $R_0 = 1.0\ \mathrm{cm}$ and $P_a = 1.013 \times 10^{5}\ \mathrm{Nm^{-2}}$ , the numerical value of the ratio is

\frac{\rho_i T_i}{\rho_a T_a} = 1 + \frac{4\gamma}{R_0 P_a} = 1 + 0.0001 \qquad \ldots (4)

(The effect of the surface tension is very small.)

2.3. Let $W$ = total weight of the bubble, $F$ = buoyant force due to air around the bubble

W = (\text{mass of film+mass of air})g

= \left(4\pi R_0^2\rho_s t + \frac{4}{3}\pi R_0^3\rho_i\right)g \qquad \ldots (5)

= 4\pi R_0^2\rho_s t g + \frac{4}{3}\pi R_0^3\frac{\rho_a T_a}{T_i}\left[1 + \frac{4\gamma}{R_0 P_a}\right]g

The buoyant force due to air around the bubble is

B = \frac{4}{3}\pi R_0^3\rho_a g \qquad \ldots (6)

If the bubble floats in still air,

B \ge W

\frac{4}{3}\pi R_0^3\rho_a g \ge 4\pi R_0^2\rho_s t g + \frac{4}{3}\pi R_0^3\frac{\rho_a T_a}{T_i}\left[1 + \frac{4\gamma}{R_0 P_a}\right]g \qquad \ldots (7)

Rearranging to give

T_i \ge \frac{R_0\rho_a T_a}{R_0\rho_a - 3\rho_s t}\left[1 + \frac{4\gamma}{R_0 P_a}\right] \qquad \ldots (8)

\ge 307.1\ \mathrm{K}

The air inside must be about $7.1^{\circ}\mathrm{C}$ warmer.

2.4. Ignore the radius change $\rightarrow$ Radius remains $R_0 = 1.0\ \mathrm{cm}$

(The radius actually decreases by 0.8% when the temperature decreases from 307.1 K to 300 K. The film itself also becomes slightly thicker.)

The drag force from Stokes’ Law is $F = 6\pi\eta R_0 u \qquad \ldots (9)$

If the bubble floats in the updraught,

F \ge W - B

6\pi\eta R_0 u \ge \left(4\pi R_0^2\rho_s t + \frac{4}{3}\pi R_0^3\rho_i\right)g - \frac{4}{3}\pi R_0^3\rho_a g \qquad \ldots (10)

When the bubble is in thermal equilibrium $T_i = T_a$ .

6\pi\eta R_0 u \ge \left(4\pi R_0^2\rho_s t + \frac{4}{3}\pi R_0^3\rho_a\left[1 + \frac{4\gamma}{R_0 P_a}\right]\right)g - \frac{4}{3}\pi R_0^3\rho_a g

Rearranging to give

u \ge \frac{4R_0\rho_s t g}{6\eta} + \frac{\frac{4}{3}R_0^2\rho_a g\left(\frac{4\gamma}{R_0 P_a}\right)}{6\eta} \qquad \ldots (11)

2.5. The numerical value is $u \ge 0.36\ \mathrm{m/s}$ .

The 2 $^{\mathrm{nd}}$ term is about 3 orders of magnitude lower than the 1 $^{\mathrm{st}}$ term.

From now on, ignore the surface tension terms.

2.6. When the bubble is electrified, the electrical repulsion will cause the bubble to expand in size and thereby raise the buoyant force.

The force/area is (e-field on the surface $\times$ charge/area)

There are two alternatives to calculate the electric field ON the surface of the soap film.

A. From Gauss’s Law

Consider a very thin pill box on the soap surface.

Left: the charged bubble of radius R_1 with centre O, inside labels P_i′, T_a, ρ_i′, outside P_a, T_a, ρ_a, charge q, and a small pill box on its top surface; a curved arrow leads to an enlarged view on the right showing the pill box straddling the film with the field E pointing upward.

$E$ = electric field on the film surface that results from all other parts of the soap film, excluding the surface inside the pill box itself.

$E_q$ = total field just outside the pill box $= \dfrac{q}{4\pi\varepsilon_0 R_1^2} = \dfrac{\sigma}{\varepsilon_0}$

$= E$ + electric field from surface charge $\sigma$

$= E + E_\sigma$

Using Gauss’s Law on the pill box, we have $E_\sigma = \dfrac{\sigma}{2\varepsilon_0}$ perpendicular to the film as a result of symmetry.

\text{Therefore,} \quad E = E_q - E_\sigma = \frac{\sigma}{\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{2\varepsilon_0} = \frac{1}{2\varepsilon_0}\frac{q}{4\pi R_1^2} \qquad \ldots (12)

B. From direct integration

A sphere with centre O and point A on its surface; a thin hatched ring at angle θ from OA, of width Rδθ, with the radius R drawn from O to the ring; an arrow labels the ring as charge δq.

\delta q = \left(\frac{q}{4\pi R^2}\right)2\pi R\sin\theta . R\delta\theta

To find the magnitude of the electrical repulsion we must first find the electric field intensity $E$ at a point on (not outside) the surface itself.

Field at A in the direction $\overrightarrow{\mathrm{OA}}$ is

\delta E_A = \frac{1}{4\pi\varepsilon_0}\frac{\left(q/4\pi R_1^2\right)2\pi R_1^2\sin\theta\,\delta\theta}{\left(2R_1\sin\frac{\theta}{2}\right)^2}\sin\frac{\theta}{2} = \frac{\left(q/4\pi R_1^2\right)}{2\varepsilon_0}\cos\frac{\theta}{2}\,\delta\left(\frac{\theta}{2}\right)

E_A = \frac{\left(q/4\pi R_1^2\right)}{2\varepsilon_0}\int_{\theta=0}^{\theta=180^{\circ}}\cos\frac{\theta}{2}\,d\left(\frac{\theta}{2}\right) = \frac{\left(q/4\pi R_1^2\right)}{2\varepsilon_0} \qquad \ldots (13)

The repulsive force per unit area of the surface of bubble is

\left(\frac{q}{4\pi R_1^2}\right)E = \frac{\left(q/4\pi R_1^2\right)^2}{2\varepsilon_0} \qquad \ldots (14)

Let $P_i'$ and $\rho_i'$ be the new pressure and density when the bubble is electrified.

This electric repulsive force will augment the gaseous pressure $P_i'$ .

$P_i'$ is related to the original $P_i$ through the gas law.

P_i'\frac{4}{3}\pi R_1^3 = P_i\frac{4}{3}\pi R_0^3

P_i' = \left(\frac{R_0}{R_1}\right)^3 P_i = \left(\frac{R_0}{R_1}\right)^3 P_a \qquad \ldots (15)

In the last equation, the surface tension term has been ignored.

From balancing the forces on the half-sphere projected area, we have (again ignoring the surface tension term)

P_i' + \frac{\left(q/4\pi R_1^2\right)^2}{2\varepsilon_0} = P_a

P_a\left(\frac{R_0}{R_1}\right)^3 + \frac{\left(q/4\pi R_1^2\right)^2}{2\varepsilon_0} = P_a \qquad \ldots (16)

Rearranging to get

\left(\frac{R_1}{R_0}\right)^4 - \left(\frac{R_1}{R_0}\right) - \frac{q^2}{32\pi^2\varepsilon_0 R_0^4 P_a} = 0 \qquad \ldots (17)

Note that (17) yields $\dfrac{R_1}{R_0} = 1$ when $q = 0$ , as expected.

2.7. Approximate solution for $R_1$ when $\dfrac{q^2}{32\pi^2\varepsilon_0 R_0^4 P_a} \ll 1$

Write $R_1 = R_0 + \Delta R,\ \Delta R \ll R_0$

\text{Therefore,} \quad \frac{R_1}{R_0} = 1 + \frac{\Delta R}{R_0}, \quad \left(\frac{R_1}{R_0}\right)^4 \approx 1 + 4\frac{\Delta R}{R_0} \qquad \ldots (18)

Eq. (17) gives:

\Delta R \approx \frac{q^2}{96\pi^2\varepsilon_0 R_0^3 P_a} \qquad \ldots (19)

R_1 \approx R_0 + \frac{q^2}{96\pi^2\varepsilon_0 R_0^3 P_a} \approx R_0\left(1 + \frac{q^2}{96\pi^2\varepsilon_0 R_0^4 P_a}\right) \qquad \ldots (20)

2.8. The bubble will float if

B \ge W

\frac{4}{3}\pi R_1^3\rho_a g \ge 4\pi R_0^2\rho_s t g + \frac{4}{3}\pi R_0^3\rho_i g \qquad \ldots (21)

Initially, $T_i = T_a \Rightarrow \rho_i = \rho_a$ for $\gamma \to 0$ and $R_1 = R_0\left(1 + \dfrac{\Delta R}{R_0}\right)$

\frac{4}{3}\pi R_0^3\left(1 + \frac{\Delta R}{R_0}\right)^3\rho_a g \ge 4\pi R_0^2\rho_s t g + \frac{4}{3}\pi R_0^3\rho_a g

\frac{4}{3}\pi(3\Delta R)\rho_a g \ge 4\pi R_0^2\rho_s t g \qquad \ldots (22)

\frac{4}{3}\pi\frac{3q^2}{96\pi^2\varepsilon_0 R_0 P_a}\rho_a g \ge 4\pi R_0^2\rho_s t g

q^2 \ge \frac{96\pi^2 R_0^3\rho_s t\varepsilon_0 P_a}{\rho_a}

q \approx 256 \times 10^{-9}\ \mathrm{C} \approx 256\ \mathrm{nC}

Note that if the surface tension term is retained, we get

R_1 \approx \left(1 + \frac{q^2/96\pi^2\varepsilon_0 R_0^4 P_a}{\left[1 + \frac{2}{3}\left(\frac{4\gamma}{R_0 P_a}\right)\right]}\right)R_0

Оригинал в Архива: 2.pdf · официални решения: 2_sol.pdf