{{tag>electrostatic capacitor plate_capacitor capacity exam_ee2_SS2024}}{{include_n>1020}}
 

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<WRAP mexle-task-title>
~~#@ee1_taskctr~~ Capacitor \\
<fs medium>(written test, approx. 12 % of a 120-minute written test, SS2024)</fs> 
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Older capacitive pressure sensors are based on a parallel plate capacitor setup (see left-side image). 

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In the following such a sensor is given with:
  * 	Plate area :  $A=25 ~\rm mm^2$ 
  * 	Distance between both plates: $d=200 ~\rm \mu m$ 
  * 	Air between the plates:  $\varepsilon_{\rm r,air}=1$
  * 	Supply voltage: $3.3 ~\rm V$
  * 	Boundary effects on the end of the layers shall be ignored in the following calculations.
$\varepsilon_{0} =8.854 \cdot 10^{-12} ~\rm  F/m $

1. Calculate the capacity $C$.


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\begin{align*}
C &= \varepsilon_0 \varepsilon_r {{A}\over{d}} \\
  &= 8.854 \cdot 10^{-12} ~\rm As/Vm \cdot 1  \cdot {{25 \cdot 10^{-6} {~\rm m} }\over{200 \cdot 10^{-6} {~\rm m} }}
\end{align*}


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$C = 1.1 ~\rm pF$

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2. Calculate the value of the displacement field between the plates, when a voltage of $U=3.3 ~\rm V$ is applied.


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The displacement field is given by:
\begin{align*}
D &= \varepsilon_0 \varepsilon_r E \\
  &= \varepsilon_0 \varepsilon_r {{U}\over{d}} \\
  &= 8.854 \cdot 10^{-12} ~\rm As/Vm \cdot 1  \cdot {{3.3 {~\rm V} }\over{200 \cdot 10^{-6} {~\rm m} }} \\
\end{align*}

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$D = 146 \cdot 10^{-9} \rm {{C}\over{m^2}}$ 

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3. Calculate the charge difference between both plates for a voltage of $U=3.3 ~\rm V$.


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There are two ways now. Either:
\begin{align*}
Q &= C \cdot U = 1.1 ~\rm pF \cdot 3.3 ~ V = 3.6522... ~pC \\
\end{align*}
Or:
\begin{align*}
Q &= D \cdot A =  146 \cdot 10^{-9} \rm {{C}\over{m^2}} \cdot 25 \cdot 10^{-6} ~m^2 = 3.6522... ~pC \\
\end{align*}


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$Q = 3.65 ~\rm pC $ 

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4. Due to a production problem, the right-side layer is covered with a contaminant, see the right-side image. \\
The contaminant has $\varepsilon_{\rm r,c}>\varepsilon_{\rm r,air}$, while the distance between the plates remains the same. 
Give a generalized formula $C_2=f(A,d,x,\varepsilon_{\rm r,c}, \varepsilon_{\rm r,air})$.


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The resulting capacity $C$ is now a series circuit of $C_{\rm air}$ and $C_{\rm c}$. \\
Therefore:
\begin{align*}
C = {{1}\over{ {{1}\over{C_{\rm air} }} + {{1}\over{ C_{\rm c}}} }}
\end{align*}

With
\begin{align*}
C_{\rm air} &= \varepsilon_0 \varepsilon_{\rm r, air} {{A}\over{d-x}} &&= \varepsilon_0 A {{\varepsilon_{\rm r, air}}\over{d-x}}  \\
C_{\rm c}   &= \varepsilon_0 \varepsilon_{\rm r, c}   {{A}\over{x}}   &&= \varepsilon_0 A {{\varepsilon_{\rm r, c}  }\over{x}}  \\
\end{align*}

This leads to: 
\begin{align*}
C = \varepsilon_0 A {{1}\over{ {{d-x}\over{\varepsilon_{\rm r, air} }} + {{x}\over{ \varepsilon_{\rm r, c}}} }}
\end{align*}


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$C = \varepsilon_0 A {{\varepsilon_{\rm r, air} \cdot  \varepsilon_{\rm r, c} }\over{ {(d-x)\cdot{\varepsilon_{\rm r, c} }} + {x\cdot{ \varepsilon_{\rm r, air}}} }}$ 

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