{{fa>pencil?32}} {{drawio>electrical_engineering_2:schaltung_klws2020_3_1_1.svg}} Determine the capacitance $C$ for the plate capacitor drawn on the right with the following data: * rectangular electrodes with edge length of $6 ~{\rm cm}$ and $8 ~{\rm cm}$. * distance between the plates: $2 ~{\rm mm}$ * dielectric ${\rm A}$: * $\varepsilon_{\rm r,A} = 1 \:\:\rm (air)$ * thickness $d_{\rm A} = 1.5 ~{\rm mm}$ * Dielectric ${\rm B}$: * $\varepsilon_{\rm r,B} = 100 \:\: \rm (ice)$ * thickness $d_{\rm B} = 0.5 ~{\rm mm}$ $\varepsilon_{0} = 8.854 \cdot 10^{-12} ~{\rm F/m}$ {{icon>eye}} Tips for the solution * Which circuit can be used to replace a layered structure with different dielectrics? {{icon>eye}} Solution The total capacitance $C$ can be divided into a partial capacitance $C_{\rm A}$ and a $C_{\rm B}$. These are connected in series. \\ This results in: $C = \frac{C_A \cdot C_B}{C_A + C_B}$ \\ \\ 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*} {{icon>eye}} Result \begin{align*} C = 28.24 \cdot 10^{-12} ~{\rm F} \rightarrow 28~{\rm pF} \end{align*} \\