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Concern about surface tension in textbook

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I was doing question 63 of the 200 Puzzling Physics Problems (https://dejanphysics.files.wordpress.com/2016/10/gnadig_1.pdf), and I was getting that the change in pressure between inside and outside is:

$\Delta{P} = 2\gamma/R$

Here $R = d/2$, so then

$\Delta{P} = 4\gamma/d$

However, in the answers, they said that the pressure difference is just $\Delta{P} = 2\gamma/d$ (bottom of page 130) . Can someone please explain who is right, and why?

asked Apr 6, 2018 in Physics Problems by andrey333 (130 points)

1 Answer

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Best answer

The air-water surface of the drop has 2 orthogonal radii of curvature : a vertical radius of $r_y=\frac12 d$ and a horizontal radius of $r_x=\frac12 D$. These 2 radii are related to the pressure difference across the surface by the Young-Laplace Equation :
$$\Delta p = \gamma (\frac{1}{r_x}+\frac{1}{r_y})$$

In this case $D \gg d$ so $r_x \gg r_y$. Therefore $\Delta p \approx \frac{\gamma}{r_y}=\frac{2\gamma}{d}$ as stated in the book (S63 page 124).

The formula which you have used applies to a spherical interface which has 2 equal radii $r_x=r_y=r$. Then the pressure difference across the surface is $\frac{2\gamma}{r}=\frac{4\gamma}{d}$. However this is wrong because in this example the surface is not spherical.

See :
https://en.wikipedia.org/wiki/Surface_tension#Surface_curvature_and_pressure

answered Apr 6, 2018 by sammy gerbil (27,948 points)
selected Apr 6, 2018 by andrey333
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