DOCUMENT1
Theoretical Competition: Solution
Question 2
Page 1 of 7
2. SOLUTION
2.1. The bubble is surrounded by air.
Pi , Ti , i
R0
Pa , Ta , a
O
s , t
Cutting the sphere in half and using the projected area to balance the forces
give
Pi R02  Pa R02  2  2 R0 
Pi  Pa 
… (1)
4
R0
The pressure and density are related by the ideal gas law:
PV  nRT or P 
 RT
M
, where M = the molar mass of air.
… (2)
Apply the ideal gas law to the air inside and outside the bubble, we get
M
R
M
 aTa  Pa ,
R
iTi  Pi

iTi
P
4 
 i  1 

aTa
Pa
 R0 Pa 
… (3)
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DOCUMENT1
Theoretical Competition: Solution
Question 2
Page 2 of 7
2.2. Using   0.025 Nm1 , R0 1.0 cm and Pa 1.013 105 Nm 2 , the numerical value
of the ratio is
iTi
4
 1
 1  0.0001
 aTa
R0 Pa
… (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   mass of film+mass of air  g
4


  4 R02  s t   R03 i  g
3


T
4
 4 R02  s tg   R03 a a
3
Ti
… (5)

4 
1 
g
 R0 Pa 
The buoyant force due to air around the bubble is
B
4 3
 R0  a g
3
… (6)
If the bubble floats in still air,
B W
T
4
4
 R03 a g  4 R02  s tg   R03 a a
3
3
Ti
… (7)

4 
1 
g
 R0 Pa 
Rearranging to give
R0  aTa 
4 
1

R0  a  3  st  R0 Pa 
 307.1 K
Ti 
… (8)
The air inside must be about 7.1C warmer.
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Theoretical Competition: Solution
Question 2
Page 3 of 7
2.4. Ignore the radius change  Radius remains R0  1.0 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 R0u
… (9)
If the bubble floats in the updraught,
F W  B
4
4


6 R0u   4 R02  st   R03i  g   R03 a g
3
3


… (10)
When the bubble is in thermal equilibrium Ti  Ta .


4
4  
4
3
6 R0u   4 R02  st   R03 a 1 
 g   R0  a g

3
R
P
3
0 a 


Rearranging to give
 4 
4 2
R0 a g 

R0 Pa 
4 R0  s tg 3

u

6
6
… (11)
2.5. The numerical value is u  0.36 m/s .
The 2nd term is about 3 orders of magnitude lower than the 1st 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 × charge/area)
There are two alternatives to calculate the electric field ON the surface of
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Theoretical Competition: Solution
Question 2
Page 4 of 7
the soap film.
A. From Gauss’s Law
Consider a very thin pill box on the soap surface.
Pi , Ta i
O
R1
E
Pa , Ta , a
q
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.
Eq = total field just outside the pill box =
q
4 0 R
2
1


0
= E + electric field from surface charge 
= E  E
Using Gauss’s Law on the pill box, we have E 

perpendicular to the film
2 0
as a result of symmetry.
Therefore, E  Eq  E 
 

1
q



 0 2 0 2 0 2 0 4 R12
… (12)
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DOCUMENT1
Theoretical Competition: Solution
Question 2
Page 5 of 7
B. From direct integration
R
R

A
oO
chargeq
 q 
q  
2 R sin  .R
2 
 4 R 
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 OA is
 EA 
EA 
1
4 0
q
q
4 R12  2 R12 sin 


 2 R1 sin 
2

4 R12   180
2 0

 0
2
  
cos d   
2 2
sin
q

2

q
4 R12 
2 0
4 R12 
2 0
  
cos   
2 2
… (13)
The repulsive force per unit area of the surface of bubble is
q 4 R12 

 q 
E 

2 
2 0
 4 R1 
2
… (14)
Let Pi  and  i be the new pressure and density when the bubble is electrified.
This electric repulsive force will augment the gaseous pressure Pi .
Pi is related to the original Pi through the gas law.
4
4
Pi  R13  Pi  R03
3
3
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Theoretical Competition: Solution
Question 2
Page 6 of 7
3
3
R 
R 
Pi   0  Pi   0  Pa
 R1 
 R1 
… (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)
q
P
i
3
R 
Pa  0  
 R1 
4 R12 
2
 Pa
2 0
 q 4 R12 
… (16)
2
2 0
 Pa
Rearranging to get
4
 R1   R1 
q2


 0
   
2
4
 R0   R0  32  0 R0 Pa
Note that (17) yields
… (17)
R1
1 when q  0 , as expected.
R0
2.7. Approximate solution for R1 when
q2
32 2 0 R04 Pa
 1
Write R1  R0  R, R  R0
4
R
R  R 
R
Therefore, 1 1  ,  1   1  4
R0
R0  R0 
R0
… (18)
Eq. (17) gives:
R 
q2
96 2 0 R03 Pa
R1  R0 


q2

R
1


0
2
3
2
4
96  0 R0 Pa
 96  0 R0 Pa 
q2
… (19)
… (20)
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DOCUMENT1
Theoretical Competition: Solution
Question 2
Page 7 of 7
2.8. The bubble will float if
B W
… (21)
4
4
 R13 a g  4 R02  stg   R03 i g
3
3

Initially, Ti  Ta  i  a for   0 and R1  R0 1 

R 

R0 
3

4
R 
4
2
3
 R03  1 
  a g  4 R0  s tg   R0  a g
3
R
3
0 

4
  3R   a g  4 R02  stg
3
4
3q 2

 a g  4 R02  stg
2
3 96  0 R0 Pa
q2 
… (22)
96 2 R03  s t 0 Pa
a
q  256  109 C  256 nC
Note that if the surface tension term is retained, we get


2
2
4
 q 96  0 R0 Pa
R1  1 
 2  4  

1  


 3  R0 Pa  




 R0



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