Electrostatics > The electric field in the presence of a dielectric > Problem 6.6.22

Statement

$6.6.22.$ 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

$$
\boxed{h = \frac{(\varepsilon - 1) Q^2}{2\varepsilon_0 \varepsilon \rho g S^2}}.
$$