Problem 4: Capillary Rise on a Vertical Plate
1. Principle
The liquid rises up the plate due to surface tension $\sigma$. The shape of the free surface (meniscus) is determined by the balance between the pressure difference across the surface and hydrostatic pressure.
At any height $y$, the pressure difference is:
$$ \Delta P = \frac{\sigma}{R} = \rho g y $$Where $R$ is the radius of curvature of the meniscus.
2. Derivation using Integration
Let $\theta$ be the angle the tangent to the liquid surface makes with the horizontal. The curvature $1/R$ can be expressed as $\frac{d\theta}{ds}$, where $s$ is the arc length.
$$ \sigma \frac{d\theta}{ds} = \rho g y $$Using the geometric relation $\sin\theta = \frac{dy}{ds}$, we can substitute $ds = \frac{dy}{\sin\theta}$:
$$ \sigma \sin\theta \frac{d\theta}{dy} = \rho g y $$Separating variables and integrating from the free surface ($y=0, \theta=0$) to the contact point at the plate ($y=h$, $\theta = 90^\circ – \theta_c$):
$$ \sigma \int_0^{\pi/2 – \theta_c} \sin\theta \, d\theta = \rho g \int_0^h y \, dy $$ $$ \sigma \left[ -\cos\theta \right]_0^{\pi/2 – \theta_c} = \rho g \left[ \frac{y^2}{2} \right]_0^h $$ $$ \sigma \left( -\cos(\frac{\pi}{2} – \theta_c) – (-\cos(0)) \right) = \frac{1}{2} \rho g h^2 $$ $$ \sigma \left( -\sin\theta_c + 1 \right) = \frac{1}{2} \rho g h^2 $$3. Final Expression
Rearranging to solve for height $h$:
$$ h^2 = \frac{2\sigma (1 – \sin\theta_c)}{\rho g} $$Note: If the contact angle $\theta_c \approx 0$ (perfect wetting, e.g., clean glass and water), the expression simplifies to $h = \sqrt{\frac{2\sigma}{\rho g}}$.