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+{{tag>electrostatic capacitor plate_capacitor capacity exam_ee2_SS2024}}{{include_n>1020}}
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Capacitor \\
+<fs medium>(written test, approx. 12 % of a 120-minute written test, SS2024)</fs> #@TaskText_HTML@#
+Older capacitive pressure sensors are based on a parallel plate capacitor setup (see left-side image).
+{{drawio>ee2:k4wrrhf8v46gct49_question1.svg}}
+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 $
+. Calculate the capacity $C$.
+#@HiddenBegin_HTML~k4wrrhf8v46gct49_11,Path~@#
+\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*}
+#@HiddenEnd_HTML~k4wrrhf8v46gct49_11,Path~@#
+#@HiddenBegin_HTML~k4wrrhf8v46gct49_12,Result~@#
+$C = 1.1 ~\rm pF$
+#@HiddenEnd_HTML~k4wrrhf8v46gct49_12,Result~@#
+. Calculate the value of the displacement field between the plates, when a voltage of $U=3.3 ~\rm V$ is applied.
+#@HiddenBegin_HTML~k4wrrhf8v46gct49_21,Path~@#
+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*}
+#@HiddenEnd_HTML~k4wrrhf8v46gct49_21,Path~@#
+#@HiddenBegin_HTML~k4wrrhf8v46gct49_22,Result~@#
+$D = 146 \cdot 10^{-9} \rm {{C}\over{m^2}}$
+#@HiddenEnd_HTML~k4wrrhf8v46gct49_22,Result~@#
+. Calculate the charge difference between both plates for a voltage of $U=3.3 ~\rm V$.
+#@HiddenBegin_HTML~k4wrrhf8v46gct49_31,Path~@#
+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*}
+#@HiddenEnd_HTML~k4wrrhf8v46gct49_31,Path~@#
+#@HiddenBegin_HTML~k4wrrhf8v46gct49_32,Result~@#
+$Q = 3.65 ~\rm pC $
+#@HiddenEnd_HTML~k4wrrhf8v46gct49_32,Result~@#
+. 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 $C_2=f(A,d,x,\varepsilon_{\rm r,c}, \varepsilon_{\rm r,air})$.
+#@HiddenBegin_HTML~k4wrrhf8v46gct49_41,Path~@#
+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*}
+#@HiddenEnd_HTML~k4wrrhf8v46gct49_41,Path~@#
+#@HiddenBegin_HTML~k4wrrhf8v46gct49_42,Result~@#
+$C = \varepsilon_0 A {{\varepsilon_{\rm r, air} \cdot  \varepsilon_{\rm r, c} }\over{ {(d-x)\cdot{\varepsilon_{\rm r, c} }} + {x\cdot{ \varepsilon_{\rm r, air}}} }}$
+#@HiddenEnd_HTML~k4wrrhf8v46gct49_42,Result~@#
+#@TaskEnd_HTML@#

electrostatic capacitor plate capacitor capacity exam ee2 ss2024