Determine the capacitance $C$ for the plate capacitor drawn on the right with the following data:
$\varepsilon_{0} = 8.854 \cdot 10^{-12} ~{\rm F/m}$
The partial capacitance $C_A$ can be calculated by \begin{align*} C_{\rm A} &= \varepsilon_{0} \varepsilon_{r,A} \cdot \frac{A}{d_A} && | \text{with } A = 3 ~{\rm cm} \cdot 5~{\rm cm} = 6 \cdot 10^{-2} \cdot 8 \cdot 10^{-2} ~{\rm m}^2 = 48 \cdot 10^{-4} ~{\rm m}^2\\ &= 8.854 \cdot 10^{-12} ~{\rm F/m} \cdot \frac{48 \cdot 10^{-4} ~{\rm m}^2}{1.5 \cdot 10^{-3} ~{\rm m}} \\ &= 28.33 \cdot 10^{-12} ~{\rm F} \\ \end{align*}
The partial capacitance $C_B$ can be calculated by \begin{align*} C_{\rm B} &= \varepsilon_{0} \varepsilon_{r,B} \cdot \frac{B}{d_B} \\ &= 100 \cdot 8.854 \cdot 10^{-12} ~{\rm F/m} \cdot \frac{48 \cdot 10^{-4} ~{\rm m}^2}{0,5 \cdot 10^{-3} ~{\rm m}} \\ &= 8.500 \cdot 10^{-9} ~{\rm F} \\ \end{align*}