{{tag>electrostatic capacitor plate_capacitor capacity exam_ee2_SS2022}}{{include_n>1030}}
 
#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Capacitor \\
<fs medium>(written test, approx. 7 % of a 120-minute written test, SS2022)</fs> #@TaskText_HTML@#

Given is the multilayer capacitor shown below, with the following dimensions:
  * Length of layer overlap:  $l=1.5 ~\rm mm$ 
  * Distance between single layers: $d=1.0 ~\rm \mu m$ 
  * Depth of component: $w=0.7 ~\rm mm$
  * Number of layers (as shown in the picture): 3 left-side and 3 right-side layers.

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The material shall have a dielectric permittivity of $\varepsilon_r=3$. \\
The following calculations shall ignore boundary effects on the end of the layers. \\ \\
$\varepsilon_0 = 8.854 \cdot 10^{-12} ~\rm As/Vm$

1. What is the field strength in the dielectric material between the layer, when a voltage of $U=6.3 ~\rm V$ is applied?

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The electric field strength $E$ is given by:
\begin{align*}
E &= {{U}\over{d}} \\
  &= {{6.3 ~\rm V}\over{1 \cdot 10^{-6} ~\rm m}} \\
\end{align*}

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$E = 6.3 {{\rm MV}\over{\rm m}}$
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2. Calculate the capacity $C$.

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The capacity can be derived from the geometry by:
\begin{align*}
C = \varepsilon_0 \varepsilon_r {{A}\over{d}} 
\end{align*}

For the area $A$ we have multiple plates with the area $A_0= l \cdot w$ facing each other.

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How many "multiple plates" $N$ do we have to consider? \\
For this, we have to count facing areas $A_0$. There are $N=5$.

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Therefore, the formula is
\begin{align*}
C &= \varepsilon_0 \varepsilon_r {{N \cdot l \cdot w}\over{d}} \\
  &= 8.854 \cdot 10^{-12} ~\rm As/Vm \cdot 3  \cdot {{5 \cdot 1.5 \cdot 10^{-3} {~\rm m} \cdot 0.7 \cdot 10^{-3} {~\rm m} }\over{1 \cdot 10^{-6} {~\rm m} }}
\end{align*}


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$C = 0.139 ~\rm nF$ 
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3. Due to a production problem the left-side layers are covered with $d_{\rm c}=0.1 ~\rm \mu m$ of air ($\varepsilon_{r, \rm c}=1$), while the thickness of the dielectric material remains the same. \\
What is the new capacity $C_\rm c$?

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The air builds another capacitor in series to the dielectric material. 
Therefore, the capacity can be calculated as 
\begin{align*}
C_{\rm c} &= {{C \cdot C_{\rm Air}}\over{C + C_{\rm Air}}}
\end{align*}

The capacity of air is 
\begin{align*}
C_{\rm Air} &= \varepsilon_0 \varepsilon_{r,\rm Air} {{N \cdot l \cdot w}\over{d_{\rm c}}} \\
            &= 8.854 \cdot 10^{-12} ~\rm As/Vm \cdot 1  \cdot {{5 \cdot 1.5 \cdot 10^{-3} {~\rm m} \cdot 0.7 \cdot 10^{-3} {~\rm m} }\over{0.1 \cdot 10^{-6} {~\rm m} }} \\
            &= 0.465... ~\rm nF
\end{align*}

By this the overall capacity is 
\begin{align*}
C_{\rm c} &= {{0.139... ~\rm nF \cdot 0.465... ~\rm nF}\over{0.139... ~\rm nF + 0.465... ~\rm nF}}
\end{align*}

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$C_{\rm c} = 0.107~\rm nF$ 
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