Capillary Action

A tube with a very fine bore (of order about 1 mm) and open at both ends is called a capillary tube. If one end of a capillary tube is dipped into a liquid that wets the tube's surface, partially or completely (such as water in glass), the level of that liquid inside the capillary is observed to rise above the level of the liquid outside; conversely, if the capillary is dipped into a liquid that does not wet its surface (such as mercury in glass), the level inside the capillary falls below the level outside. This phenomenon — the rise or fall of a liquid inside a fine tube dipped into it — is called capillarity. Capillarity is at work in many everyday situations: oil rising up the wick of a lamp, a cloth rag sucking up water, water rising through fine crevices in rock, sap and water rising all the way up to the topmost leaves of a tall tree, and blotting paper absorbing ink.

a) Capillary fall. Consider a capillary tube dipped into a liquid that does not wet its surface — mercury, for example — so its meniscus inside the tube is upper-convex. Mark four points: A, just above the convex meniscus, inside the capillary; B, just below that meniscus, also inside the capillary; C, just above the plane mercury surface outside the capillary, at the same horizontal level as A; and D, just below that outer plane surface, at the same horizontal level as B. Since the points A and C are both at the same level and both exposed to the atmosphere, $p_A = p_C$ (--- 2.25); and between C and D the surface outside the capillary is plane, so $p_C = p_D = p_A$ (--- 2.26). But because the pressure is always greater on the concave side of any curved surface than on the convex side (section 2.4.5), and the meniscus here is convex as seen from inside the capillary, $p_B > p_A$ — hence $p_B > p_D$. Yet B and D sit at exactly the same horizontal level. For the pressure to be equal at the same level (as equilibrium demands), mercury must rush out of the capillary tube until its internal level drops enough to bring $p_B$ back down to match $p_D$ — this is exactly the mechanism behind the observed drop in mercury's level inside a capillary tube.

b) Capillary rise. By the mirror-image version of exactly this same argument — but now for a liquid such as water, whose meniscus inside the capillary is concave rather than convex — the pressure just below the meniscus inside the tube works out lower than the pressure at the same level outside, and liquid is correspondingly drawn up into the capillary from outside until equilibrium is restored — producing the observed rise of water's level inside a capillary tube.

Expression for capillary rise or fall.

Method I (using pressure difference). Let r be the radius of the capillary tube and θ the angle of contact of the liquid with the tube. The radius of curvature R of the meniscus is related to r and θ by $R = r/\cos\theta$. The pressure difference across the curved meniscus, $2T/R$, must equal the hydrostatic pressure $h\rho g$ of the liquid column that has risen (or fallen) by height h:

This gives the expression for capillary rise (or fall) of a liquid: the narrower the tube, the greater the height to which the liquid rises (or falls). If the tube is held in a liquid whose meniscus is convex, the angle of contact θ is obtuse, so $\cos\theta$ is negative, and hence h itself comes out negative — meaning the liquid actually suffers a capillary fall (depression) rather than a rise, exactly matching the mercury-in-glass case worked through above.

Method II (using forces directly). Water rising up a capillary rises against gravity, so the weight of the risen liquid column must be exactly balanced by the vertical component of the surface-tension force acting at the point of contact. The length of liquid actually in contact with the tube wall, around its inner circumference, is $2\pi r$; so the total force due to surface tension is $f_T = T \times 2\pi r$, acting along the tangent to the meniscus at the wall. Its vertical component is:

$f_v = T(2\pi r)\cos\theta \qquad \text{--- (2.29)}$ …