IPhO 2011 · Theory problem 2

An Electrified Soap Bubble

Download PDFIPhO 2011

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 \(\dfrac{\rho_iT_i}{\rho_aT_a}\) in terms of \(\gamma\), \(P_a\) and \(R_0\). [1.7 point]

2.2 Find the numerical value of \(\dfrac{\rho_iT_i}{\rho_aT_a}-1\) using \(\gamma=0.0250\ \mathrm{N\,m^{-1}}\), \(R_0=1.00\ \mathrm{cm}\), and \(P_a=1.013\times10^5\ \mathrm{N\,m^{-2}}\). [0.4 point]

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 \(T_a=300\ \mathrm K\), \(\rho_s=1000\ \mathrm{kg\,m^{-3}}\), \(\rho_a=1.30\ \mathrm{kg\,m^{-3}}\), \(t=100\ \mathrm{nm}\) and \(g=9.80\ \mathrm{m\,s^{-2}}\). [2.0 points]

After the bubble is formed for a while, it will be in thermal equilibrium with the surroundings. 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’s Law is \(F=6\pi\eta R_0u\). [1.6points]

2.5 Calculate the numerical value for \(u\) using \(\eta=1.8\times10^{-5}\ \mathrm{kg\,m^{-1}\,s^{-1}}\). [0.4 point]

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 \(\varepsilon_0\). [2.0points]

2.7 Assume that the total charge is not too large (i.e. \(\dfrac{q^2}{\varepsilon_0R_0^4}\ll P_a\)) and the bubble only experiences a small increase in its radius, find \(\Delta R\) where \(R_1=R_0+\Delta R\).

Given that \((1+x)^n\approx1+nx\) where \(x\ll1\). [0.7 point]

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 \(\varepsilon_0=8.85\times10^{-12}\ \mathrm{farad/m}\). [1.2 points]

Official English text of the problem.

Answers

  • 2.1 \[ \boxed{\frac{\rho_iT_i}{\rho_aT_a}=1+\frac{4\gamma}{P_aR_0}}. \]

  • 2.2 \[ \boxed{\frac{\rho_iT_i}{\rho_aT_a}-1=9.87\times10^{-5}}. \]

  • 2.3 \[ \boxed{ T_{i,\min} =T_a\frac{1+\dfrac{4\gamma}{P_aR_0}} {1-\dfrac{3\rho_st}{\rho_aR_0}} =307\ \mathrm K }. \]

  • 2.4, neglecting the small surface-tension correction: \[ \boxed{u_{\min}\simeq\frac{2\rho_sR_0tg}{3\eta}}. \]

  • 2.5 \[ \boxed{u_{\min}\simeq0.36\ \mathrm{m\,s^{-1}}}. \]

  • 2.6 \[ \boxed{R_1^4-R_0^3R_1-\frac{q^2}{32\pi^2\varepsilon_0P_a}=0}, \qquad R_1\ge R_0. \]

  • 2.7 \[ \boxed{\Delta R\simeq\frac{q^2}{96\pi^2\varepsilon_0P_aR_0^3}}. \]

  • 2.8, using the small-expansion approximation: \[ \boxed{ |q|\simeq \sqrt{\frac{96\pi^2\varepsilon_0P_a\rho_stR_0^3}{\rho_a}} =2.56\times10^{-7}\ \mathrm C }. \] Without linearizing the expansion, the thin-film result is \[ \boxed{ |q|= \sqrt{\frac{96\pi^2\varepsilon_0P_a\rho_stR_0^3}{\rho_a}} \left(1+\frac{3\rho_st}{\rho_aR_0}\right)^{1/6} =2.57\times10^{-7}\ \mathrm C }. \]

Solution

We treat the air as an ideal gas with the same composition inside and outside. In the thin-film approximation, the bubble volume and soap mass are \[ V=\frac43\pi R^3, \qquad m_s\simeq4\pi R^2t\rho_s. \] The film has two soap-air interfaces, but its mass is that of one shell.

2.1

Both interfaces contribute to the surface energy. Neglecting their small difference in radius, \[ E_s=2\gamma(4\pi R^2)=8\pi\gamma R^2. \] For a small outward displacement \(dR\), mechanical equilibrium requires \[ (P_i-P_a)\,dV=dE_s. \] Since \[ dV=4\pi R^2\,dR, \qquad dE_s=16\pi\gamma R\,dR, \] the excess internal pressure is \[ P_i-P_a=\frac{4\gamma}{R_0}. \]

The ideal-gas relation \(P=\rho\mathcal R_{\rm air}T\), with the same specific gas constant on both sides, gives \[ \boxed{ \frac{\rho_iT_i}{\rho_aT_a} =\frac{P_i}{P_a} =1+\frac{4\gamma}{P_aR_0} }. \]

2.2

With \(R_0=1.00\times10^{-2}\ \mathrm m\), \[ P_i-P_a =\frac{4(0.0250)}{1.00\times10^{-2}} =10.0\ \mathrm{Pa}. \] Thus \[ \boxed{ \frac{\rho_iT_i}{\rho_aT_a}-1 =\frac{10.0}{1.013\times10^5} =9.87\times10^{-5} }. \]

2.3

The upward buoyant force is the weight of the displaced ambient air: \[ F_B=\rho_aV_0g, \qquad V_0=\frac43\pi R_0^3. \] At the minimum temperature that permits floating, buoyancy just balances the combined weight of the internal air and soap: \[ \rho_aV_0g=\rho_iV_0g+4\pi R_0^2t\rho_sg. \] Therefore, \[ \rho_i=\rho_a-\frac{3\rho_st}{R_0}. \] Numerically, \[ \frac{3\rho_st}{R_0} =\frac{3(1000)(1.00\times10^{-7})}{1.00\times10^{-2}} =0.0300\ \mathrm{kg\,m^{-3}}, \] so neutral buoyancy requires \[ \rho_i=1.27\ \mathrm{kg\,m^{-3}}. \]

Using part 2.1, \[ T_{i,\min} =\frac{\rho_aT_a}{\rho_i} \left(1+\frac{4\gamma}{P_aR_0}\right) =T_a \frac{1+\dfrac{4\gamma}{P_aR_0}} {1-\dfrac{3\rho_st}{\rho_aR_0}}. \] Consequently, \[ T_{i,\min} =300\frac{1+9.8717\times10^{-5}}{1-0.0300/1.30} =307.117\ \mathrm K, \] or \[ \boxed{T_{i,\min}=307\ \mathrm K}. \] A higher temperature gives a lower internal air density and hence an upward net force.

2.4

At thermal equilibrium, \(T_i=T_a\). Part 2.1 then gives \[ \rho_i-\rho_a=\rho_a\frac{4\gamma}{P_aR_0}. \]

For a stationary bubble, the upward Stokes drag must balance its weight minus buoyancy: \[ 6\pi\eta R_0u_{\min} = 4\pi R_0^2t\rho_sg +\frac43\pi R_0^3(\rho_i-\rho_a)g. \] Hence, retaining surface tension, \[ u_{\min} = \frac{2\rho_sR_0tg}{3\eta} +\frac{8\rho_a\gamma R_0g}{9\eta P_a}. \]

The ratio of the second term to the first is \[ \frac{4\rho_a\gamma}{3P_a\rho_st} =4.28\times10^{-3}. \] Thus surface tension changes the required speed by only about \(0.43\%\). Neglecting this correction gives the expression in the requested parameters: \[ \boxed{u_{\min}\simeq\frac{2\rho_sR_0tg}{3\eta}}. \]

This uses the stipulated Stokes drag law and the instruction to ignore the radius change during cooling.

2.5

Substituting the given values, \[ u_{\min}\simeq \frac{2(1000)(1.00\times10^{-2})(1.00\times10^{-7})(9.80)} {3(1.8\times10^{-5})} =0.362963\ \mathrm{m\,s^{-1}}. \] Therefore, \[ \boxed{u_{\min}\simeq0.36\ \mathrm{m\,s^{-1}}}. \] Including surface tension gives \(0.364516\ \mathrm{m\,s^{-1}}\), which rounds to the same answer at two significant figures.

2.6

We now neglect surface tension as instructed. Before charging, the bubble is at temperature \(T_a\), with internal pressure \(P_a\). The enclosed air mass is conserved, and the final state is also at temperature \(T_a\). The ideal-gas law therefore gives \[ P_iV_1=P_aV_0, \qquad P_i=P_a\frac{R_0^3}{R_1^3}. \]

A uniformly charged spherical shell has electric field \[ E(r)=\frac{q}{4\pi\varepsilon_0r^2} \quad (r>R), \] and zero field inside. Its electrostatic energy is \[ U_e=\frac{\varepsilon_0}{2}\int_R^\infty E(r)^2\,4\pi r^2\,dr =\frac{q^2}{8\pi\varepsilon_0R}. \] At fixed charge, an increase in radius lowers this energy. The corresponding outward electric pressure is \[ p_e=-\frac{dU_e}{dV} =-\frac{dU_e/dR}{4\pi R^2} =\frac{q^2}{32\pi^2\varepsilon_0R^4}. \]

Mechanical equilibrium requires \[ P_i+p_e=P_a. \] At \(R=R_1\), \[ P_a\frac{R_0^3}{R_1^3} +\frac{q^2}{32\pi^2\varepsilon_0R_1^4} =P_a. \] Multiplying by \(R_1^4/P_a\) gives \[ \boxed{ R_1^4-R_0^3R_1-\frac{q^2}{32\pi^2\varepsilon_0P_a}=0 }. \] For nonzero charge, the physical solution has \(R_1>R_0\).

2.7

Let \[ x=\frac{\Delta R}{R_0}\ll1, \qquad R_1=R_0(1+x). \] The equation from part 2.6 becomes \[ R_0^4\big[(1+x)^4-(1+x)\big] =\frac{q^2}{32\pi^2\varepsilon_0P_a}. \] To first order, \[ (1+x)^4-(1+x)\simeq(1+4x)-(1+x)=3x. \] Thus \[ 3R_0^4x=\frac{q^2}{32\pi^2\varepsilon_0P_a}, \] and \[ \boxed{ \Delta R\simeq\frac{q^2}{96\pi^2\varepsilon_0P_aR_0^3} }. \]

2.8

Charging increases the displaced air volume while leaving the total bubble mass unchanged.

Before charging, the internal air density equals \(\rho_a\), because surface tension is now neglected and both its pressure and temperature equal the ambient values. Its mass is therefore \[ m_i=\rho_aV_0. \] The conserved soap mass is \[ m_s=4\pi R_0^2t\rho_s. \] Here \(t\) is the original thickness; the film becomes thinner as it expands.

Neutral buoyancy after charging requires \[ \rho_aV_1g=(\rho_aV_0+m_s)g. \] Hence \[ \rho_a\frac43\pi(R_1^3-R_0^3) =4\pi R_0^2t\rho_s, \] or \[ R_1^3-R_0^3=\frac{3\rho_stR_0^2}{\rho_a}. \]

For a small expansion, \[ R_1^3-R_0^3\simeq3R_0^2\Delta R, \] so the required radius increase is \[ \Delta R\simeq\frac{\rho_st}{\rho_a}. \] Numerically, \[ \Delta R\simeq \frac{1000(1.00\times10^{-7})}{1.30} =7.69\times10^{-5}\ \mathrm m, \qquad \frac{\Delta R}{R_0}\simeq0.00769. \]

Equating this increase to the result of part 2.7 gives \[ \frac{q^2}{96\pi^2\varepsilon_0P_aR_0^3} =\frac{\rho_st}{\rho_a}, \] and therefore \[ \boxed{ |q|\simeq \sqrt{\frac{96\pi^2\varepsilon_0P_a\rho_stR_0^3}{\rho_a}} }. \] Substitution yields \[ |q|\simeq \sqrt{ \frac{ 96\pi^2(8.85\times10^{-12})(1.013\times10^5) (1000)(1.00\times10^{-7})(1.00\times10^{-2})^3 }{1.30} }, \] so \[ \boxed{|q|\simeq2.56\times10^{-7}\ \mathrm C=0.256\ \mu\mathrm C}. \]

The result can also be obtained without linearizing the expansion. Neutral buoyancy gives \[ R_1=R_0\left(1+\frac{3\rho_st}{\rho_aR_0}\right)^{1/3}. \] Combining this with part 2.6, \[ q^2 =32\pi^2\varepsilon_0P_aR_1(R_1^3-R_0^3) =\frac{96\pi^2\varepsilon_0P_a\rho_stR_0^2R_1}{\rho_a}. \] Thus the unexpanded thin-film result is \[ \boxed{ |q|= \sqrt{\frac{96\pi^2\varepsilon_0P_a\rho_stR_0^3}{\rho_a}} \left(1+\frac{3\rho_st}{\rho_aR_0}\right)^{1/6} =2.57\times10^{-7}\ \mathrm C }. \] Either sign of charge produces the same expansion because the electric pressure depends on \(q^2\).

The solution was checked against the official solution.