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--- electrical_engineering_and_electronics_1:block11 [2025/11/02 16:54] – mexleadmin
+++ electrical_engineering_and_electronics_1:block11 [2025/11/08 14:27] (aktuell) – mexleadmin
@@ Zeile 21: / Zeile 21: @@
 For checking your understanding please do the following exercises:
-  * ...
+  * 5.4.1
+  * 5.4.4
+  * 5.4.5
 ~~PAGEBREAK~~ ~~CLEARFIX~~
@@ Zeile 326: / Zeile 328: @@
 #@HiddenEnd_HTML~1,Result~@#
-</WRAP></WRAP></panel>
-<panel type="info" title="Task 5.5.1 induced Charges"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-A plate capacitor with a distance of $d = 2 ~{ \rm cm}$ between the plates and with air as dielectric ($\varepsilon_{ \rm r}=1$) gets charged up to $U = 5~{ \rm kV}$.
-In between the plates, a thin metal foil with the area $A = 45~{ \rm cm^2}$ is introduced parallel to the plates.
-Calculate the amount of the displaced charges in the thin metal foil.
-<button size="xs" type="link" collapse="Loesung_5_5_1_Tipps">{{icon>eye}} Tips for the solution</button><collapse id="Loesung_5_5_1_Tipps" collapsed="true">
-  * What is the strength of the electric field $E$ in the capacitor?
-  * Calculate the displacement flux density $D$
-  * How can the charge $Q$ be derived from $D$?
-</collapse>
-<button size="xs" type="link" collapse="Loesung_5_5_1_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_5_5_1_Endergebnis" collapsed="true">
-$Q = 10 ~{ \rm nC}$
-</collapse>
-</WRAP></WRAP></panel>
-<panel type="info" title="Task 5.5.2 Manipulating a Capacitor I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-An ideal plate capacitor with a distance of $d_0 = 7 ~{ \rm mm}$ between the plates gets charged up to $U_0 = 190~{ \rm V}$ by an external source.
-The source gets disconnected. After this, the distance between the plates gets enlarged to $d_1 = 7 ~{ \rm cm}$.
-  - What happens to the electric field and the voltage?
-  - How does the situation change (electric field/voltage), when the source is not disconnected?
-<button size="xs" type="link" collapse="Loesung_5_5_2_Tipps">{{icon>eye}} Tips for the solution</button><collapse id="Loesung_5_5_2_Tipps" collapsed="true">
-  * Consider the displacement flux through a surface around a plate
-</collapse>
-<button size="xs" type="link" collapse="Loesung_5_5_2_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_5_5_2_Endergebnis" collapsed="true">
-  - $U_1 = 1.9~{ \rm kV}$, $E_1 = 27~{ \rm kV/m}$
-  - $U_1 = 190~{ \rm V}$, $E_1 = 2.7~{ \rm kV/m}$
-</collapse>
-</WRAP></WRAP></panel>
-<panel type="info" title="Task 5.5.3 Manipulating a Capacitor II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-An ideal plate capacitor with a distance of $d_0 = 6 ~{ \rm mm}$ between the plates and with air as dielectric ($\varepsilon_0=1$) is charged to a voltage of $U_0 = 5~{ \rm kV}$.
-The source remains connected to the capacitor. In the air gap between the plates, a glass plate with $d_{ \rm g} = 4 ~{ \rm mm}$ and $\varepsilon_{ \rm r} = 8$ is introduced parallel to the capacitor plates.
-. Calculate the partial voltages on the glas $U_{ \rm g}$ and on the air gap $U_{ \rm a}$.
-#@HiddenBegin_HTML~1531T,Tipps~@#
-  * Build a formula for the sum of the voltages first
-  * How is the voltage related to the electric field of a capacitor?
-#@HiddenEnd_HTML~1531T,Tipps~@#
-#@HiddenBegin_HTML~1531P,Path~@#
-The sum of the voltages across the glass and the air gap gives the total voltage $U_0$, and each individual voltage is given by the $E$-field in the individual material by $E = {{U}\over{d}}$:
-\begin{align*}
-U_0 &= U_{\rm g}               + U_{\rm a} \\
-    &= E_{\rm g} \cdot d_{\rm g} + E_{\rm a} \cdot d_{\rm a}
-\end{align*}
-The displacement field $D$ must be continuous across the different materials since it is only based on the charge $Q$ on the plates.
-\begin{align*}
-D_{\rm g}                                            &= D_{\rm a} \\
-\varepsilon_0 \varepsilon_{\rm r, g} \cdot E_{\rm g} &= \varepsilon_0  \cdot E_{\rm a}
-\end{align*}
-Therefore, we can put $E_\rm a=  \varepsilon_{\rm r, g} \cdot E_\rm g $ into the formula of the total voltage and rearrange to get $E_\rm g$:
-\begin{align*}
-U_0 &= E_{\rm g} \cdot d_{\rm g} + \varepsilon_{\rm r, g} \cdot E_{\rm g}  \cdot d_{\rm a} \\
-    &= E_{\rm g} \cdot ( d_{\rm g} + \varepsilon_{\rm r, g} \cdot d_{\rm a}) \\
-\rightarrow E_{\rm g}  &= {{U_0}\over{d_{\rm g} + \varepsilon_{\rm r, g} \cdot d_{\rm a}}}
-\end{align*}
-Since we know that the distance of the air gap is $d_{\rm a} = d_0 - d_{\rm a}$ we can calculate:
-\begin{align*}
-E_{\rm g}  &= {{5'000 ~\rm V}\over{0.004 ~{\rm m} + 8 \cdot 0.002 ~{\rm m}}} \\
-         &= 250 ~\rm{{kV}\over{m}}
-\end{align*}
-By this, the individual voltages can be calculated:
-\begin{align*}
-U_{ \rm g} &= E_{\rm g} \cdot d_\rm g &&= 250 ~\rm{{kV}\over{m}} \cdot 0.004~\rm m &= 1 ~{\rm kV}\\
-U_{ \rm a} &= U_0 - U_{ \rm g}      &&= 5  ~{\rm kV} - 1 ~{\rm kV}               &= 4 ~{\rm kV}\\
-\end{align*}
-#@HiddenEnd_HTML~1531P,Path~@#
-#@HiddenBegin_HTML~1531R,Results~@#
-$U_{ \rm a} = 4~{ \rm kV}$, $U_{ \rm g} = 1 ~{ \rm kV}$
-#@HiddenEnd_HTML~1531R,Results~@#
-. What would be the maximum allowed thickness of a glass plate, when the electric field in the air-gap shall not exceed $E_{ \rm max}=12~{ \rm kV/cm}$?
-#@HiddenBegin_HTML~1532P,Path~@#
-Again, we can start with the sum of the voltages across the glass and the air gap, such as the formula we got from the displacement field: $D = \varepsilon_0 \varepsilon_{\rm r, g} \cdot E_{\rm g} = \varepsilon_0  \cdot E_\rm a $. \\
-Now we shall eliminate $E_\rm g$, since $E_\rm a$ is given in the question.
-\begin{align*}
-U_0 &=   E_{\rm g}                               \cdot d_{\rm g} + E_{\rm a} \cdot d_{\rm a} \\
-    &= {{E_\rm a}\over{\varepsilon_{\rm r,g}}}   \cdot d_{\rm g} + E_{\rm a} \cdot d_{\rm a} \\
-\end{align*}
-The distance $d_\rm a$ for the air is given by the overall distance $d_0$ and the distance for glass $d_\rm g$:
-\begin{align*}
-d_{\rm a} = d_0 - d_{\rm g}
-\end{align*}
-This results in:
-\begin{align*}
-U_0                      &= {{E_{\rm a}}\over{\varepsilon_{\rm r,g}}} \cdot d_{\rm g} + E_{\rm a} \cdot (d_0 - d_{\rm g}) \\
-{{U_0}\over{E_{\rm a} }} &= {{1}\over{\varepsilon_{\rm r,g}}} \cdot d_{\rm g} + d_0 - d_{\rm g} \\
-                         &= d_{\rm g} \cdot ({{1}\over{\varepsilon_{\rm r,g}}} - 1) + d_0 \\
-d_{\rm g}                &= { { {{U_0}\over{E_{\rm a} }} - d_0 } \over { {{1}\over{\varepsilon_{\rm r,g}}} - 1 } }     &= { { d_0 - {{U_0}\over{E_{\rm a} }} } \over { 1 - {{1}\over{\varepsilon_{\rm r,g}}} } }
-\end{align*}
-With the given values:
-\begin{align*}
-d_{\rm g}                &= { { 0.006 {~\rm m} - {{5 {~\rm kV} }\over{ 12 {~\rm kV/cm}}} } \over { 1 - {{1}\over{8}} } }    &= {  {{8}\over{7}} } \left( { 0.006 - {{5 }\over{ 1200}} }  \right)  {~\rm m}
-\end{align*}
-#@HiddenEnd_HTML~1532P,Path~@#
-#@HiddenBegin_HTML~1532R,Results~@#
-$d_{ \rm g} = 2.10~{ \rm mm}$
-#@HiddenEnd_HTML~1532R,Results~@#
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