6.6.22. One plate of an uncharged capacitor is made of a fine grid and lies on the surface of a liquid of density $\rho$ and dielectric permittivity $\varepsilon$. The area of each plate is $S$. To what height will the liquid level rise in the capacitor if a charge $Q$ is given to it?
Solution
The capacitance of the capacitor (which can be viewed as two capacitors connected in series) is: $$C = \left( \frac{d-h}{\varepsilon_0 S} + \frac{h}{\varepsilon \varepsilon_0 S} \right)^{-1} = \frac{\varepsilon_0 S}{d - h + h/\varepsilon}.$$ The field energy at fixed charge is: $$W_c = \frac{Q^2}{2C} = \frac{Q^2}{2\varepsilon_0 S} \left( d - h + \frac{h}{\varepsilon} \right).$$ The potential energy of the liquid is: $$W_p = \frac{h}{2} m g = \frac{1}{2} S h^2 \rho g.$$ Equilibrium corresponds to a minimum of the total energy; we find this position using the derivative: $$\frac{d}{dh} \bigl[ W_p - W_c \bigr] = 0 \quad\Rightarrow\quad S \rho g h - \frac{Q^2}{2\varepsilon_0 S} \left( 1 - \frac{1}{\varepsilon} \right) = 0.$$ The second derivative: $$\frac{d}{dh} \left[ S \rho g h - \frac{Q^2}{2\varepsilon_0 S} \left( 1 - \frac{1}{\varepsilon} \right) \right] > 0 \quad\Rightarrow\quad h > 0 \text{ – obviously.}$$ Hence the only equilibrium height of rise is: $$h = \frac{(\varepsilon - 1) Q^2}{2\varepsilon_0 \varepsilon \rho g S^2}.$$
Answer
$$\boxed{h = \frac{(\varepsilon - 1) Q^2}{2\varepsilon_0 \varepsilon \rho g S^2}}.$$
