Applying Young-Laplace equation on meniscus formed due to rise of liquid on a single plate
Let's say we have a single plate with liquid on both sides rising up due to surface tension. The meniscus formed has a radius of curvature $R$.
I'm trying to find the excess pressure, i.e, the pressure difference between the exterior and interior of the meniscus.
Since there is only one curved surface (the meniscus which extends across the length of the plate) whose radius of curvature is $R$, from the Young-Laplace equation, the excess pressure should be
$$\triangle P = \frac{\gamma}{R}$$
where $\gamma$ is the surface tension.
However, I know that the excess pressure should be
$$\triangle P = \frac{2 \gamma}{R}$$
The only way we get the factor of $2$ is if there are two curved surfaces with radius $R$. I'm new to using the Young-Laplace equation, and I just don't understand how there are two curved surfaces with radius $R$. I'm somewhat sure that the two meniscus' on either sides of the plate aren't the two curved surfaces since the plate is dividing them and they have no contact.
Here's an image which shows what kind of a setup I'm referring to (credits: Vincent Émyde)
0 answers
3 comments
Does a meniscus really have a constant radius of curvature? — Olin Lathrop about 1 month ago
@OlinLathrop yeah it does not, what I meant was the mean radius of curvature, which I believe should work fairly well. — TripleFault about 1 month ago
The premise that “the meniscus” is a bounded object is wrong. The solution nowhere vanishes (although decays exponentially) and the only sensible answer about “the mean radius of curvature” is ∞. — Incnis Mrsi 15 days ago