# What will the excess pressure of a water drop on a table be?

I'm aware of the formula for excess pressure of a spherical drop given by :$$\frac{2S}{R}$$ Where $$R$$ is the radius of the drop.

But what if the drop was on the floor?

Using the formula for excess pressure for any general axis, one of the radii is infinite so I get excess pressure as $$\frac{S}{R}$$

Is this right?

• Why is one of the radiuses infinite? If it is a spherical drop, both radiuses are the same. Jan 21 at 12:50
• @Ján Lalinský when the drop is on the floor is it not a hemisphere? Jan 21 at 13:10
• Yes but that does not matter. In the Laplace formula that gives pressure difference between two phases, the two radiuses $R_1,R_2$ are radiuses of curvature of two perpendicular arcs lying in the boundary that intersect near the point where the pressure difference is sought. So the curved part of the water droplet has both radiuses finite. The flat part of the droplet has both radiuses infinite so there is no capillary pressure effect on that boundary, the pressure inside the droplet and pressure inside the flat solid are the same. Jan 21 at 14:04