Capillarity | PhysicsRank

The same thin tube can make one liquid climb and another retreat. The bend of the surface is a clue: it connects wetting at the wall to a pressure difference and, finally, to a height you can measure.

Start from the surface-tension idea, then follow one pressure path at a time. We will compare drops and bubbles, infer wetting from measurements and test when a familiar height formula has reached its limit.

What you will learn to do

Derive the pressure jump for spherical and cylindrical interfaces.
Distinguish one liquid–gas interface from a two-surface soap film.
Connect contact angle, signed rise and pressure at a meniscus.
Infer a wetting parameter and reject inconsistent measurements.
Separate vertical height from length along a tube and recognize a rim limitation.
Conserve volume before comparing the surface energy of drops.

Useful before this chapterSurface tension, curved interfaces and capillary rise →Pressure, floating and fluid support →Work, stored energy and power →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Capillarity and curved interfaces
Learning goals:
- Derive the pressure jump for spherical and cylindrical interfaces.
- Distinguish one liquid–gas interface from a two-surface soap film.
- Connect contact angle, signed rise and pressure at a meniscus.
- Infer a wetting parameter and reject inconsistent measurements.
- Separate vertical height from length along a tube and recognize a rim limitation.
- Conserve volume before comparing the surface energy of drops.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

BEFORE YOU READ

Bring back one idea

Try a sentence or a sketch before opening the comparison. If this is new, use the explanation to get started.

Why does surface tension times a length have force units, while surface tension times an area has energy units?

Compare your reasoning

Surface tension has units N/m. Multiplying by a contact-line length gives N, while multiplying by added interface area gives N·m = J. Choose force balance for a static shape and energy change for creating area.

Revisit the explanation →

01 / Capillarity and curved interfaces

A curved interface needs a pressure jump

A tiny bubble can require more excess pressure than a much larger one. That sounds backward if pressure is confused with total force. The useful comparison is between a pressure force, which acts over an area, and the tension acting around the boundary of a cut through the interface.

Cut a spherical drop mentally into two hemispheres. The pressure difference between the inside and outside pushes one hemisphere outward across a circular area πR². Surface tension pulls around the circular rim of length 2πR. Equilibrium gives Δp πR² = S 2πR. The radius therefore remains in the denominator after cancellation.

A sphere curves in two perpendicular surface directions. A long cylindrical interface curves around its circumference but is straight along its axis, so only one curvature contributes. The sphere’s factor of two counts curvature directions, not two physical surfaces. A soap film introduces a separate factor of two because it has two interfaces.

\begin{aligned}\Delta p&=S\left(\frac1{R_1}+\frac1{R_2}\right)\\ \Delta p_{\mathrm{sphere}}&=\frac{2S}{R}\end{aligned}

For the convex surfaces considered here, Δp means pressure on the inner side minus pressure on the outer side. The principal curvature radii are positive in this stated case.

Explain this equation

Read it in words

Surface tension multiplied by the sum of the two perpendicular curvatures gives the pressure jump across one interface.

Symbols and units

$S$: Surface tension of this single liquid–gas interface.N/m.
$R_1,\ R_2$: Principal radii of curvature at a point, in perpendicular tangent directions.Metres (m).
$R$: Common curvature radius for a sphere.Metres (m).
$\Delta p$: Inner pressure minus outer pressure for the stated convex orientation.Pascals (Pa).

Signs and operations

Reciprocal radius is curvature. A flat direction has infinite radius and zero reciprocal curvature. For concave surfaces, choose an oriented pressure jump and signed curvature consistently; do not reuse a positive convex jump with reversed sides.

When it applies

Use mechanical equilibrium, a uniform surface tension and a thin interface. The spherical result assumes an approximately spherical interface; gravity can noticeably distort a sufficiently large drop.

Make it concrete

For S = 0.060 N/m and a 1.5 mm spherical radius, Δp = 80 Pa. A cylindrical interface with the same radius has only one nonzero curvature and a 40 Pa jump.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Equation (LaTeX): \begin{aligned}\Delta p&=S\left(\frac1{R_1}+\frac1{R_2}\right)\\ \Delta p_{\mathrm{sphere}}&=\frac{2S}{R}\end{aligned}
Read it as: Surface tension multiplied by the sum of the two perpendicular curvatures gives the pressure jump across one interface.
Symbols:
S: Surface tension of this single liquid–gas interface.; N/m.
R_1,\ R_2: Principal radii of curvature at a point, in perpendicular tangent directions.; Metres (m).
R: Common curvature radius for a sphere.; Metres (m).
\Delta p: Inner pressure minus outer pressure for the stated convex orientation.; Pascals (Pa).
Signs and operations: Reciprocal radius is curvature. A flat direction has infinite radius and zero reciprocal curvature. For concave surfaces, choose an oriented pressure jump and signed curvature consistently; do not reuse a positive convex jump with reversed sides.
Conditions: Use mechanical equilibrium, a uniform surface tension and a thin interface. The spherical result assumes an approximately spherical interface; gravity can noticeably distort a sufficiently large drop.

Help me explain the equation in words, predict how changing a quantity affects the result, and apply it to a small numerical example.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

WORKED EXAMPLE

Same radius, different curvature

A long cylindrical liquid interface and a spherical drop both have radius 1.5 mm and surface tension 0.060 N/m. Compare their pressure jumps, ignoring gravity-induced distortion and end effects.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Capillarity and curved interfaces
Focus: A curved interface needs a pressure jump
Learning goals:
- Derive the pressure jump for spherical and cylindrical interfaces.
- Distinguish one liquid–gas interface from a two-surface soap film.
- Connect contact angle, signed rise and pressure at a meniscus.
- Infer a wetting parameter and reject inconsistent measurements.
- Separate vertical height from length along a tube and recognize a rim limitation.
- Conserve volume before comparing the surface energy of drops.
Equations in this section (LaTeX):
\begin{aligned}\Delta p&=S\left(\frac1{R_1}+\frac1{R_2}\right)\\ \Delta p_{\mathrm{sphere}}&=\frac{2S}{R}\end{aligned}
Meaning: For the convex surfaces considered here, Δp means pressure on the inner side minus pressure on the outer side. The principal curvature radii are positive in this stated case.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

02 / Capillarity and curved interfaces

Count physical interfaces separately from curvature

A liquid drop in gas has one liquid–gas interface. A gas bubble inside a liquid also has one interface: the pressure is higher on the bubble’s gas side. A thin soap bubble has an inner and an outer liquid–gas interface. If the film is thin enough that their radii are effectively equal, their jumps add to approximately 4S/R.

Do not infer the number of interfaces from the everyday word “bubble”. Sketch the materials along a radial path. Gas → liquid gives one crossing; gas → thin liquid film → gas gives two. The pressure just outside must then be chosen at the actual location, which may be deep below a liquid’s free surface.

Which pressure jump belongs to which object?
Object	Interfaces crossed	Ideal pressure jump
Liquid drop in gas	One	Liquid minus surrounding gas: 2S/R
Gas bubble in a liquid	One	Bubble gas minus local liquid: 2S/R
Thin soap bubble in gas	Two	Inner gas minus outer gas: approximately 4S/R

WORKED EXAMPLE

A bubble’s outside pressure is local

A nearly spherical air bubble of radius 1.5 mm is momentarily in mechanical equilibrium with water at depth 0.12 m. Surface tension is 0.045 N/m, density is 1000 kg/m³, g = 10 m/s² and surface air pressure is 100 kPa. Find the bubble’s absolute gas pressure. Neglect motion, gas weight and pressure variation across its small size.

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STOP & THINK

Which ideal spherical object needs the 4S/R rather than 2S/R excess-pressure expression?

A gas bubble in water
A thin soap bubble in air
A liquid drop in air

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Capillarity and curved interfaces
Focus: Count physical interfaces separately from curvature
Learning goals:
- Derive the pressure jump for spherical and cylindrical interfaces.
- Distinguish one liquid–gas interface from a two-surface soap film.
- Connect contact angle, signed rise and pressure at a meniscus.
- Infer a wetting parameter and reject inconsistent measurements.
- Separate vertical height from length along a tube and recognize a rim limitation.
- Conserve volume before comparing the surface energy of drops.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

03 / Capillarity and curved interfaces

Compare pressures before predicting the direction of flow

Connect two soap bubbles with a passage and the gas initially responds to their pressure difference. If surface tension is the same and constant, the smaller bubble has the larger excess pressure. Gas initially flows from the smaller one toward the larger one. Equal atmospheric pressure on their outsides cancels from the comparison.

This is an initial-direction prediction. The later evolution may involve film drainage, changing surface tension, gas compression and loss of the spherical shape. A rubber balloon is a different material model: its membrane tension changes with stretch, so the same radius rule cannot be transferred without checking it.

WORKED EXAMPLE

A smaller soap bubble can push into a larger one

Two thin spherical soap bubbles have radii 1.0 mm and 3.0 mm and equal surface tension 0.030 N/m. Their outsides are at the same pressure. A narrow gas connection is opened. What initial pressure difference drives the gas, and in which direction?

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Capillarity and curved interfaces
Focus: Compare pressures before predicting the direction of flow
Learning goals:
- Derive the pressure jump for spherical and cylindrical interfaces.
- Distinguish one liquid–gas interface from a two-surface soap film.
- Connect contact angle, signed rise and pressure at a meniscus.
- Infer a wetting parameter and reject inconsistent measurements.
- Separate vertical height from length along a tube and recognize a rim limitation.
- Conserve volume before comparing the surface energy of drops.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

04 / Capillarity and curved interfaces

A meniscus connects pressure to height

At a wetting tube wall, the interface climbs higher at the edge than in the middle. A nonwetting interface bends the opposite way. Contact angle is measured through the liquid between the wall and the interface tangent. This angle, rather than surface tension alone, determines the sign of a capillary rise or depression.

For a narrow vertical circular tube, the vertical contact-line force is 2πrS cosθ. In the usual small-meniscus-volume approximation, the extra column has weight ρgπr²h. Equating them gives the signed height. The radius r is the tube’s internal radius, which is generally different from the meniscus curvature radius.

There is a second route. At height h above a large reservoir surface, the liquid pressure is below the reservoir-surface pressure by ρgh. The curved meniscus supplies the same pressure difference. Thus a raised column has liquid pressure below the gas at its top; a depressed column has liquid pressure above the gas. This pressure view also works for an inclined tube, where only vertical height enters hydrostatics.

Schematic vertical sections, not to scale. The contact angle is measured through the liquid. With the same gas pressure above tube and reservoir, a raised concave meniscus has lower liquid pressure; a depressed convex meniscus has higher liquid pressure.

\begin{aligned}h&\simeq\frac{2S\cos\theta}{\rho gr}\\ p_{\mathrm{liquid}}-p_{\mathrm{gas}}&\simeq-\rho gh\end{aligned}

Height is signed relative to the large reservoir’s surface; positive means rise. The approximate formula neglects the small meniscus-volume correction.

Explain this equation

Read it in words

The capillary height is set by the vertical component of surface tension divided by density, gravity and tube radius. The pressure jump follows the same signed height.

Symbols and units

$h$: Vertical height of the meniscus relative to the reservoir surface; negative for a depression.Metres (m).
$S$: Uniform liquid–gas surface tension.N/m.
$\theta$: Contact angle measured through the liquid against the tube wall.Angle; cosine is dimensionless.
$\rho$: Liquid mass density, treated as constant.kg/m³.
$g$: Magnitude of uniform gravitational acceleration.m/s².
$r$: Internal radius of the circular tube, not its diameter or meniscus curvature radius.Metres (m).
$p_{\mathrm{liquid}},\ p_{\mathrm{gas}}$: Pressures immediately on the liquid and gas sides of the meniscus, measured with the same pressure reference.Pascals (Pa).

Signs and operations

An acute contact angle gives positive cosine and a rise; an obtuse angle gives negative cosine and a depression. The minus sign in the pressure relation makes liquid pressure lower at a raised meniscus.

When it applies

Use a narrow, uniform tube communicating with a large reservoir, equal gas pressures, a static interface and a reproducible contact angle. Ignore air buoyancy, the small meniscus-volume correction and effects at a tube rim. It predicts vertical height, not the length along an inclined tube.

Make it concrete

With S = 0.060 N/m, θ = 60°, ρ = 750 kg/m³, g = 10 m/s² and r = 0.40 mm, h ≈ 0.020 m. Liquid pressure at the meniscus is then about 150 Pa below the gas.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Equation (LaTeX): \begin{aligned}h&\simeq\frac{2S\cos\theta}{\rho gr}\\ p_{\mathrm{liquid}}-p_{\mathrm{gas}}&\simeq-\rho gh\end{aligned}
Read it as: The capillary height is set by the vertical component of surface tension divided by density, gravity and tube radius. The pressure jump follows the same signed height.
Symbols:
h: Vertical height of the meniscus relative to the reservoir surface; negative for a depression.; Metres (m).
S: Uniform liquid–gas surface tension.; N/m.
\theta: Contact angle measured through the liquid against the tube wall.; Angle; cosine is dimensionless.
\rho: Liquid mass density, treated as constant.; kg/m³.
g: Magnitude of uniform gravitational acceleration.; m/s².
r: Internal radius of the circular tube, not its diameter or meniscus curvature radius.; Metres (m).
p_{\mathrm{liquid}},\ p_{\mathrm{gas}}: Pressures immediately on the liquid and gas sides of the meniscus, measured with the same pressure reference.; Pascals (Pa).
Signs and operations: An acute contact angle gives positive cosine and a rise; an obtuse angle gives negative cosine and a depression. The minus sign in the pressure relation makes liquid pressure lower at a raised meniscus.
Conditions: Use a narrow, uniform tube communicating with a large reservoir, equal gas pressures, a static interface and a reproducible contact angle. Ignore air buoyancy, the small meniscus-volume correction and effects at a tube rim. It predicts vertical height, not the length along an inclined tube.

Help me explain the equation in words, predict how changing a quantity affects the result, and apply it to a small numerical example.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

WORKED EXAMPLE

Surface tension alone does not determine the rise

A liquid has S = 0.060 N/m and density 750 kg/m³. In a vertical tube of internal radius 0.40 mm, the contact angle is 60°. Take g = 10 m/s² and use the narrow-tube approximation. Find the rise and the liquid-minus-gas pressure at the meniscus.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Capillarity and curved interfaces
Focus: A meniscus connects pressure to height
Learning goals:
- Derive the pressure jump for spherical and cylindrical interfaces.
- Distinguish one liquid–gas interface from a two-surface soap film.
- Connect contact angle, signed rise and pressure at a meniscus.
- Infer a wetting parameter and reject inconsistent measurements.
- Separate vertical height from length along a tube and recognize a rim limitation.
- Conserve volume before comparing the surface energy of drops.
Equations in this section (LaTeX):
\begin{aligned}h&\simeq\frac{2S\cos\theta}{\rho gr}\\ p_{\mathrm{liquid}}-p_{\mathrm{gas}}&\simeq-\rho gh\end{aligned}
Meaning: Height is signed relative to the large reservoir’s surface; positive means rise. The approximate formula neglects the small meniscus-volume correction.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

05 / Capillarity and curved interfaces

Use measurements to test the assumed wetting model

A capillary measurement determines S cosθ unless contact angle is independently known. Treating every clean-looking tube as perfect wetting hides a parameter rather than measuring it. Radius matters just as directly: confusing diameter with radius makes a factor-of-two error in the inferred tension.

For the same liquid, surface and temperature, a plot of h against 1/r should be approximately a straight line through the origin in this model. A nonzero height offset, changing contact angle, contamination or a broad meniscus can spoil that comparison. Repeating a measurement with a second radius is a model check, not merely another way to average the same reading.

WORKED EXAMPLE

Infer the missing contact angle

A liquid of known surface tension 0.060 N/m and density 1000 kg/m³ rises 12 mm in a vertical tube of radius 0.50 mm. With g = 10 m/s² and the narrow-tube model, find the contact angle. What would a reported rise of 40 mm imply under the same assumptions?

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Capillarity and curved interfaces
Focus: Use measurements to test the assumed wetting model
Learning goals:
- Derive the pressure jump for spherical and cylindrical interfaces.
- Distinguish one liquid–gas interface from a two-surface soap film.
- Connect contact angle, signed rise and pressure at a meniscus.
- Infer a wetting parameter and reject inconsistent measurements.
- Separate vertical height from length along a tube and recognize a rim limitation.
- Conserve volume before comparing the surface energy of drops.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

06 / Capillarity and curved interfaces

Keep the vertical height and the apparatus limits separate

Tilting a sufficiently long narrow tube changes the length of the liquid column along it. With the same local wetting geometry, it does not change the hydrostatic height required by the meniscus pressure jump. If α is the angle from the vertical, h = ℓ cosα.

Now suppose the predicted height lies beyond the tube’s open end. The original model assumed a meniscus inside a uniform tube; it cannot by itself describe what happens at an edge. Pinning, the outside surface and a new meniscus curvature can matter. Do not extend the interior-column equation into an automatic perpetual overflow claim.

Likewise, setting g = 0 in a gravity-balanced height formula does not describe the motion or final distribution of liquid in a weightless container. Geometry, wetting energy and finite available liquid still matter; the balance used to derive a finite height is no longer the right one.

WORKED EXAMPLE

The same rise can occupy a longer tube

A narrow tube would produce an 18 mm vertical rise under the stated wetting conditions. It is inclined 60° from the vertical, is long enough and remains connected to the same large reservoir. Find the liquid length along the tube above the reservoir’s level, neglecting the small meniscus correction.

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STOP & THINK

The interior-column model predicts a 30 mm rise, but the open tube ends only 20 mm above the reservoir. What can that calculation establish?

A perpetual overflow at a rate fixed by the height formula
The assumed interior meniscus position is unavailable; a rim/contact-line model is needed
Surface tension must become zero

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Capillarity and curved interfaces
Focus: Keep the vertical height and the apparatus limits separate
Learning goals:
- Derive the pressure jump for spherical and cylindrical interfaces.
- Distinguish one liquid–gas interface from a two-surface soap film.
- Connect contact angle, signed rise and pressure at a meniscus.
- Infer a wetting parameter and reject inconsistent measurements.
- Separate vertical height from length along a tube and recognize a rim limitation.
- Conserve volume before comparing the surface energy of drops.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

07 / Capillarity and curved interfaces

Volume conservation comes before the area comparison

When drops join or split, pressure and energy are two views of the same interface physics. At fixed temperature and composition with constant surface tension, increasing the interface area requires work SΔA. During coalescence the surface energy decreases. That energy can become motion, heat or sound; it is not automatically all the kinetic energy of the final drop.

First conserve liquid volume to find the new radius. Only then compare surface areas. If n equal drops of radius r merge into one, its radius is n to the one-third power times r. Its area is not n times one drop’s area. For a soap bubble, gas compression and two film surfaces require a different bookkeeping problem.

R=n^{1/3}r,\qquad E_{\mathrm{released}}=4\pi S(nr^2-R^2)

This expression concerns coalescing liquid drops with one interface each, conserved liquid volume and negligible gravitational deformation.

WORKED EXAMPLE

Joining drops releases surface energy

Twenty-seven equal liquid drops of radius 0.10 mm coalesce into one spherical drop. Surface tension is constant at 0.050 N/m. Neglect evaporation, gravity-induced distortion and mixing effects. Find the new radius and the decrease in surface energy.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Capillarity and curved interfaces
Focus: Volume conservation comes before the area comparison
Learning goals:
- Derive the pressure jump for spherical and cylindrical interfaces.
- Distinguish one liquid–gas interface from a two-surface soap film.
- Connect contact angle, signed rise and pressure at a meniscus.
- Infer a wetting parameter and reject inconsistent measurements.
- Separate vertical height from length along a tube and recognize a rim limitation.
- Conserve volume before comparing the surface energy of drops.
Equations in this section (LaTeX):
R=n^{1/3}r,\qquad E_{\mathrm{released}}=4\pi S(nr^2-R^2)
Meaning: This expression concerns coalescing liquid drops with one interface each, conserved liquid volume and negligible gravitational deformation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

08 / Capillarity and curved interfaces

Choose a pressure path, a force balance or an energy change

For a pressure question, walk through the materials and add hydrostatic changes and signed interface jumps in order. For a static height question, use the contact-line force or the same meniscus pressure with the vertical hydrostatic head. For joining or splitting, conserve material first and compare interface area.

These routes should agree where their assumptions overlap. A raised column needs a pressure deficit at its top, a soap film needs its two interfaces counted, and a smaller convex radius needs a larger curvature pressure. Check those directions before trusting numerical arithmetic.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Capillarity and curved interfaces
Focus: Choose a pressure path, a force balance or an energy change
Learning goals:
- Derive the pressure jump for spherical and cylindrical interfaces.
- Distinguish one liquid–gas interface from a two-surface soap film.
- Connect contact angle, signed rise and pressure at a meniscus.
- Infer a wetting parameter and reject inconsistent measurements.
- Separate vertical height from length along a tube and recognize a rim limitation.
- Conserve volume before comparing the surface energy of drops.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

BRING THE IDEAS TOGETHER

What should stay with you

Curvature directions and physical interfaces produce different factors of two.
Specify both sides of a pressure jump and use the local surrounding pressure.
Capillary height measures S cosθ and is signed relative to the reservoir.
Geometry and boundary conditions must remain compatible with a formula.
Joining drops lowers surface energy; its later destination needs another model.

RETRIEVE, THEN EXPLAIN

Try it without the examples

Work these out before checking. If an answer surprises you, identify which assumption or step needs another look. A correct choice is most useful when you can explain why the alternatives fail.

SELF CHECK 1

Use the narrow-tube model with S = 0.040 N/m, contact angle 60°, density 800 kg/m³, g = 10 m/s² and internal radius 0.25 mm. What is the signed rise in mm?

Answer in mm.

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SELF CHECK 2

A thin spherical soap bubble of radius 2.0 mm has surface tension 0.025 N/m. What is its gas pressure above the surrounding gas pressure in Pa?

Answer in Pa.

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SELF CHECK 3

A static meniscus is below the reservoir surface while the gas pressures above both are equal. Compared with gas pressure, the liquid pressure at the meniscus is…

Lower
Higher
Necessarily zero

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SELF CHECK 4

A set of small liquid drops joins into one larger drop and releases surface energy. Can the final translational speed be found from that release alone?

Yes, all released energy is translational kinetic energy
No, dissipation and other motion need to be specified
Yes, without knowing the mass

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APPLY THE CHAPTER

Take the ideas into a problem

Try a complete problem next. Commit to a method before revealing a hint, then compare your reasoning with the worked solution.

Count curvature before multiplyingIntermediate · Free sample · Hints and guided reasoning How high does the liquid climb?Intermediate · Free sample · Hints and guided reasoning Separate depth from curvatureIntermediate · Hints and guided reasoning Measure the missing wetting angleIntermediate · Hints and guided reasoning Read the sign at a depressed meniscusIntermediate · Hints and guided reasoning A height is not a tube lengthIntermediate · Hints and guided reasoning Pay for newly created surfaceAdvanced · Hints and guided reasoning Two surfaces support the bubbleIntermediate · Hints and guided reasoning

COME BACK TO THE IDEA

What can you explain later?

Try these from memory at your next study session, then again a few days later. Say your reasoning aloud or sketch it before looking. Adjust the gap to what you can explain.

Next study session

Two gas bubbles have the same radius but lie at different depths. Do they have equal internal gas pressures?

Compare your reasoning

Their curvature excesses above the local liquid pressure are equal if surface tension matches. Their absolute gas pressures differ because the deeper surrounding liquid has greater hydrostatic pressure.

Revisit the explanation →

A few days later: change the situation

A rise measurement is smaller than a perfect-wetting prediction. Does that alone prove that surface tension is lower?

Compare your reasoning

No. The measurement constrains S cosθ, so a nonzero contact angle can reduce the rise at unchanged surface tension. Check radius, the height zero, surface conditions and the narrow-tube approximation before assigning a cause.

Revisit the explanation →

Compare the physical reason, the assumptions and the conclusion. Your wording can differ. If one part is missing, revisit that section and try explaining it again after a break.

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