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Charge on a soap bubble

  1. Apr 27, 2017 #1
    1. The problem statement, all variables and given/known data

    There is a total of ##Q## charge on soap bubble of radius ##R##. What force is acting on one hemisphere by the other ?

    2. Relevant equations

    Electrostatic pressure is ##P_e = 2\pi \sigma^2## for a spherical shell.

    3. The attempt at a solution

    ##P_e = F/A##

    therefore ##F = AP_e = 2{A\over 2}\pi\sigma^2 = A \pi {Q^2 \over A^2} = {\pi Q^2 \over A} = {\pi Q^2 \over 4 \pi R^2} = {Q^2 \over 4\pi} ##

    Given answer is ##{Q^2 \over \color{red}{8}\pi}##.

    Where did I go wrong ?
  2. jcsd
  3. Apr 27, 2017 #2
    First, your expression for electrostatic pressure is incorrect. It has the wrong dimensions. I suggest considering the force per unit area on the bubble which is, because the E field only has a radial component, $$F_r=E_r\sigma$$
    By Gauss' Law, the R field just outside the surface is,$$E=\frac{Q}{4\pi\epsilon_0 R^2}$$
    and the field inside the surface is zero. If you integrate the z component of force per unit area over one hemisphere you'll get the total force acting on it I.e.$$F=\left (\frac{Q}{4\pi \epsilon_0 R^2}\right )\sigma\left (2\pi R^2\right )\int_0^{\frac{\pi}{2}}cos\left (\theta\right )sin\left (\theta\right )d\theta= \frac{Q^2}{8\pi\epsilon_0 R^2}$$
  4. Apr 27, 2017 #3
    Can you explain how you got this ?

    Think you did,
    Force per unit area is ##F/A = qE/A = \sigma E ##

    Am I correct ?
  5. Apr 27, 2017 #4
    What happened to x, y component ?
  6. Apr 27, 2017 #5
    Yes, the force per unit area on the bubble is the E field multiplied by the change density. The total force between the two hemispheres is in the z direction.
    I made a mistake in the integration by dropping a factor of 1/2. The answer I get is$$\frac{Q^2}{16\pi \epsilon_0 R^2}$$
  7. Apr 27, 2017 #6
    SInce you gave the expression for total then the answer should be ##\displaystyle {Q^2 \over 4R^2}## but this does not match the given answer ?

    I took ##k=1## since the unit used in the book is cgs.
  8. Apr 27, 2017 #7


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    From your integral I get $$\frac{Q^2}{4\pi \epsilon_0 R^2}$$
  9. Apr 27, 2017 #8


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    The force per unit area is ##\frac{1}{2}E_r\sigma##, where ##E_r## is the field at the outer surface of the bubble. The reason for the factor of 1/2 takes some thought.

    For a derivation see http://phys.columbia.edu/~nicolis/Surface_Force.pdf The result is given in equations (13) and (14) of this link. For the bubble, Ebelow of equation (14) is zero.

    A different (somewhat hand-wavy) derivation is given in http://www.physicspages.com/2011/10/31/electrostatic-pressure/
    Last edited: Apr 27, 2017
  10. Apr 28, 2017 #9
    ##F = {1\over 2}E\pi \sigma = {1\over 2}4\pi\sigma^2 = 2\pi\sigma^2##

    I used this and I got wrong answer :(.
  11. Apr 28, 2017 #10
    Oh I need to integrate over the hemisphere with this to get the answer but why can't I just multiple with this with the area ?
  12. Apr 28, 2017 #11


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    The forces on different patches of area point in different directions.

    It is possible to show that the answer can be obtained by multiplying the pressure by the area of a circle of radius equal to the radius of the hemisphere. Consider a solid, uncharged hemisphere acted on all sides by a uniform atmospheric pressure. We know that the hemisphere does not accelerate off in some direction. So, the force of the pressure on the curved surface of the hemisphere must be equal and opposite to the force of the pressure on the flat surface of the hemisphere.
  13. May 3, 2017 #12
    I got ##dF = 2\pi \sigma^2 \cos \theta dA##

    Changing to spherical coordinate,

    ##\displaystyle F = 2\pi\sigma^2R^2 \int^{2\pi}_{0}\int^{\pi/2}_{0}\cos \theta \sin \theta d\theta d\phi = {Q^2 \over 8R^2}##

    But I want to do this without change to spherical coordinates, how can I do that ?

    My issue is I am not able to express ##dA## into something useful.
  14. May 3, 2017 #13


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    Spherical coordinates are the natural coordinates for doing the integration. However, you can avoid having to do any integration by giving a geometrical interpretation to ##dA\cos \theta##. The figure below shows how to interpret this as a projection of a patch of area of the sphere onto a horizontal surface. So, summing all the ##dA\cos \theta## for the hemisphere is the same as summing all the horizontally projected areas. The sum of the projected areas is seen by inspection without having to perform an integration.
  15. May 3, 2017 #14
    Yes Thank you for the answer and the diagram really explains the thing.

    The limits for the integration would be ##0 \to \pi/2## right ?
  16. May 3, 2017 #15


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    Yes, that would be the limits for ##\theta##.
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