Dielectric Liquid

The plate capacitor $C = C_1 + C_2$ can be divided into two parallel connected capacitors $C_1$ and $C_2$ , each with a plate surface $A_1 = a h$ and $A_2 = a (a - h)$ . Inside the one half of the capacitor is the liquid, while the other half has only air with $\chi \approx 0$ as dielectric. The capacities thus result $C = C_1 + C_2 = \varepsilon_0 (1 + \chi) \frac{a h}{d} + \varepsilon_0 \frac{a (a - h)}{d} = C_0 + \varepsilon_0 \chi \frac{a h}{d}$ where $C_0 = \varepsilon_0 a^2 / d$ denotes the capacity without liquid. The voltage applied from the outside results in an electric energy $W_ \text {el} = \frac {1} {2} C U ^ 2$ stored in the capacitor. Due to the external energy supply, on the dielectric fluid acts a force $F_\text{el} = \frac{d W_\text{el}}{d h} = \varepsilon_0 \chi \frac{a}{2 d} U^2$ which pulls the liquid upwards. The liquid assumes equilibrium when the electric force and the weight $F_\text {g} = m g$ cancel each other, where $m = \rho a d h$ denotes the weight of the liquid inside the capacitor. Therefore, $\begin{aligned} & & F_\text{g} &= F_\text{el} \\ \Rightarrow & & g\rho a d h &= \varepsilon_0 \chi \frac{a}{2 d} U^2 \\ \Rightarrow & & h &= \frac{\varepsilon_0 \chi U^2}{2 g \rho d^2}\\ & & &\approx \frac{9\cdot 10^{-12} \cdot 80 \cdot (10^3)^2}{2 \cdot 10 \cdot 10^3 \cdot (10^{-3})^2}\,\text{m} \\ & & &= 0.036\,\text{m} \end{aligned}$