# Drops, menisci, and lenses

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## Equilibrium of a large drop

Small drops are spherical segments. Large drops are flattened.

de Gennes, 2004, Fig. 2.4

To “spread” the drop requires an force per unit length:$\sigma _{sv}-\left( \sigma _{lv}+\sigma _{sl} \right)$

The hydrostatic pressure integrated over the depth of the drop is a force per unit length pushing to “spread” the drop: $\tilde{P}=\int\limits_{0}^{e}{\rho g\left( e-\tilde{z} \right)d\tilde{z}}=\frac{1}{2}\rho ge^{2}$

At equilibrium the sum of the two is zero: $\sigma _{sv}-\left( \sigma _{lv}+\sigma _{sl} \right)+\frac{1}{2}\rho ge^{2}=0$

Substituting the Young-Dupré equation: $\sigma _{lv}\left( 1-\cos \theta _{e} \right)=\frac{1}{2}\rho ge^{2}$

Re-arranging gives: $\text{ }e=2\kappa ^{-1}\sin \left( \frac{\theta _{e}}{2} \right)$