A drop of water of mass 0.20 g is put on a clean, flat glass plate, and a second clean glass plate…

Step 1. The disc (a). The volume is $\dfrac{0.20\ \text{g}}{1\ \text{g/cm}^3} = 0.20\ \text{cm}^3$. Spread to a thickness $d = 0.050\ \text{mm}$, it covers $A = V/d = 40\ \text{cm}^2$ of each plate. So $\pi R^2 = 40\ \text{cm}^2$ and $R = 3.57\ \text{cm}$.

Step 2. The pressure (b). At the rim, the water surface is curved two ways. Seen edge on, it is a half circle of radius $d/2$ that curves into the water (water wets the glass), which lowers the pressure inside by $\dfrac{T}{d/2} = \dfrac{2T}{d} = 2880\ \text{Pa}$. Seen from above, the rim is a circle of radius $R$, which raises it by only $\dfrac{T}{R} = 2.02\ \text{Pa}$. So the water is at $2.88\ \text{kPa}$ below the air's pressure.

Step 3. The force (c). The air pushes on the outside of each plate with the full air pressure, and the water pushes on the inside with less. Over the wetted area the plates are pressed together by

The drop weighs only $0.0020\ \text{N}$, so this is about $5760$ times its weight. A thin film of water makes two plates very hard to pull apart.

Step 4. The work (d). Covering a square metre of glass with water lowers the energy by $T$ (zero contact angle), on each plate. The film covers $V/d$ of each plate, so its surface energy is $E(d) = -\dfrac{2TV}{d}$, apart from a constant. Pushing the plates from $d_1$ to $d_2$, the surface forces do the work

The force $-dE/dd = 2TV/d^2$ is the same as in (c): $\dfrac{2T}{d} \times \dfrac{V}{d}$.