Solution 6: Equilibrium of Soap Films

Theory: Pressure vs. Radius

The stability of the system depends on how the excess pressure $\Delta P$ changes as the air volume $V$ (and height $h$ of the cap) changes.

Excess pressure inside a soap bubble on a tube of radius $r$ is:

$$ \Delta P = \frac{4S}{R} = \frac{8Sh}{r^2 + h^2} $$

Differentiating with respect to $h$:

$$ \frac{d(\Delta P)}{dh} \propto \frac{r^2 – h^2}{(r^2 + h^2)^2} $$

• If $h < r$ (Sub-hemispherical): $\frac{dP}{dh} > 0$. Pressure increases as the bubble grows. This resists expansion, leading to Stable Equilibrium.
• If $h > r$ (Super-hemispherical): $\frac{dP}{dh} < 0$. Pressure decreases as the bubble grows. This encourages further expansion, leading to Unstable Equilibrium.

Case-by-Case Analysis

(a) Flat/Slightly Curved: The curvature is small ($h \ll r$). A small displacement leads to a pressure difference that pushes the air back.
$\rightarrow$ (p) Stable

(b) Concave Inwards: This is geometrically equivalent to the stable region ($h < r$). Pushing the air to one side makes the "receiving" meniscus flatter (higher P) and the "leaving" meniscus deeper (lower P), creating a restoring force.
$\rightarrow$ (p) Stable

(c) Convex (Small Bulge): The shape is less than a hemisphere ($h < r$).
$\rightarrow$ (p) Stable

(d) Convex (Large Bulge): The shape is bulbous ($h > r$). If air moves to one side, that bubble grows, its pressure drops, and it sucks in even more air.
$\rightarrow$ (q) Unstable