In determining the shape of a meniscus, we have to minimize the energy per unit length along the direction perpendicular to the cross-section of the meniscus: $$\frac{E}{L}=\int^L_0 dx [\gamma \sqrt{1+(\partial_x h)^2}+\frac{1}{2}\Delta\rho g h^2]$$ where $h(x)$ is the height of the meniscus at $x$, $\gamma$ is the surface tension factor and $\Delta \rho$ is the difference between the densities of the fluid and the vapor above it.
I understand where the first term comes from -- it's the contribution of energy due to surface tension. But I don't understand how the second term is obtained. It looks like a contribution of gravitational potential energy, but I don't know how this precise form is arrived at. In particular, I don't understand why $h$ is squared.
Thank you.
Answer
Indeed, the second term is the potential gravitational energy. Gravitational energy for a continuous body could be calculated as $$ E_g = \int_\text{body} \rho g\, z \,d^3x, $$ where we assume that gravitational acceleration $g$ has only $z$-component. The 'body' you have in your problem would stretch to length $L$ along the $y$ axis and lays between the surfaces $z=0$ and $z= h(x)$. So the volumetric integral reduces to: $$ E_g = \int_0^L dy \int_{x_1}^{x_2}\left(\int_0^{h(x)} \rho \,g\, z \,dz\,\right)dx=L \int_{x_1}^{x_2} \frac12 \rho\, g\,h^2 dx. $$ In order to receive your expression we need to substitute $\rho$ with $\Delta \rho$ to account for the fact that fluid displaces gas, which also has potential energy.
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