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electrical_engineering_and_electronics_1:block12 [2025/11/08 14:25] – mexleadminelectrical_engineering_and_electronics_1:block12 [2026/01/10 12:53] (aktuell) – mexleadmin
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 ====== Block 12 - Capacitors and Capacitance ====== ====== Block 12 - Capacitors and Capacitance ======
  
-===== Learning objectives =====+===== 12.0 Intro ===== 
 + 
 +==== 12.0.1 Learning Objectives ====
 <callout> <callout>
 After this 90-minute block, you can  After this 90-minute block, you can 
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 </callout> </callout>
  
-===== Preparation at Home =====+==== 12.0.2 Preparation at Home ====
  
 Well, again  Well, again 
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 For checking your understanding please do the following exercises: For checking your understanding please do the following exercises:
-  * ...+  * 5.5.1 
 +  * 5.5.3 
 +  * 5.5.4
  
-===== 90-minute plan =====+==== 12.0.3 90-minute plan ====
   - Warm-up (8 min):   - Warm-up (8 min):
     - Quick recall quiz: $C=\dfrac{Q}{U}$, $\vec{D}=\varepsilon\vec{E}$, units of $E$ (${\rm V/m}$), $D$ (${\rm C/m^2}$).      - Quick recall quiz: $C=\dfrac{Q}{U}$, $\vec{D}=\varepsilon\vec{E}$, units of $E$ (${\rm V/m}$), $D$ (${\rm C/m^2}$). 
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   - Wrap-up (2 min): Summary box + pitfalls checklist; connect to next block (capacitor circuits).   - Wrap-up (2 min): Summary box + pitfalls checklist; connect to next block (capacitor circuits).
  
-===== Conceptual overview =====+==== 12.0.4 Conceptual overview ====
 <callout icon="fa fa-lightbulb-o" color="blue"> <callout icon="fa fa-lightbulb-o" color="blue">
   - A **capacitor** is two conductors separated by a dielectric. It stores **charge** and **energy** in the electric field; no conduction current flows through the ideal dielectric. :contentReference[oaicite:15]{index=15}   - A **capacitor** is two conductors separated by a dielectric. It stores **charge** and **energy** in the electric field; no conduction current flows through the ideal dielectric. :contentReference[oaicite:15]{index=15}
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 </callout> </callout>
  
-===== Core content =====+===== 12.1 Core content =====
  
- +==== 12.1.1 Capacitor ====
-==== Capacitor ====+
  
   * A capacitor can "store" charges. The total charge of a two-plate capacitor is in general 0.    * A capacitor can "store" charges. The total charge of a two-plate capacitor is in general 0. 
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 $\rightarrow$ Thus, for any arrangement of two conductors separated by an insulating material, a capacitance can be specified. $\rightarrow$ Thus, for any arrangement of two conductors separated by an insulating material, a capacitance can be specified.
  
-==== Capacitance $C$ ====+==== 12.1.2 Capacitance $C$ ====
  
 The capacitance $C$ can be derived as follows: The capacitance $C$ can be derived as follows:
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 </WRAP></WRAP> </WRAP></WRAP>
  
-==== Symbols ====+==== 12.1.3 Symbols ====
  
   * The symbol of a general capacitor is given be two parallel lines nearby each other. \\   * The symbol of a general capacitor is given be two parallel lines nearby each other. \\
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 {{drawio>electrical_engineering_and_electronics_1:CapSymbols01.svg}} {{drawio>electrical_engineering_and_electronics_1:CapSymbols01.svg}}
  
-==== Designs and types of capacitors ====+==== 12.1.4 Designs and types of capacitors ====
  
 To calculate the capacitance of different designs, the definition equations of $\vec{D}$ and $\vec{E}$ are used. This can be viewed in detail, e.g., in [[https://www.youtube.com/watch?v=kAXg1xMkR_4&ab_channel=PatrickKaploo|this video]]. \\ Based on the geometry, different equations result (see also <imgref ImgNr17>). To calculate the capacitance of different designs, the definition equations of $\vec{D}$ and $\vec{E}$ are used. This can be viewed in detail, e.g., in [[https://www.youtube.com/watch?v=kAXg1xMkR_4&ab_channel=PatrickKaploo|this video]]. \\ Based on the geometry, different equations result (see also <imgref ImgNr17>).
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 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-===== Common pitfalls =====+===== 12.2 Common pitfalls =====
  
   * **Mixing up geometry symbols.** Use $d$ (or $l$) strictly as **plate spacing** and $A$ as **active area** in $C=\varepsilon_0\varepsilon_{\rm r}\dfrac{A}{d}$. Check units to catch mistakes.    * **Mixing up geometry symbols.** Use $d$ (or $l$) strictly as **plate spacing** and $A$ as **active area** in $C=\varepsilon_0\varepsilon_{\rm r}\dfrac{A}{d}$. Check units to catch mistakes. 
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   * **Real-part issues.** Ignoring polarity of electrolytics and tolerance spreads ($\pm 10~\%$ and more) causes design errors; pick suitable component types.    * **Real-part issues.** Ignoring polarity of electrolytics and tolerance spreads ($\pm 10~\%$ and more) causes design errors; pick suitable component types. 
  
-===== Exercises ===== +===== 12.3 Exercises =====
- +
- +
- +
-<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. +
-  +
-1. 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~@# +
- +
- +
-2. 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~@# +
- +
-</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%> <panel type="info" title="Task 5.5.1 induced Charges"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
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 \end{align*} \end{align*}
  
-Since we know that the distance of the air gap is $d_{\rm a} = d_0 - d_{\rm a}$ we can calculate:+Since we know that the distance of the air gap is $d_{\rm a} = d_0 - d_{\rm g}$ we can calculate:
 \begin{align*} \begin{align*}
 E_{\rm g}  &= {{5'000 ~\rm V}\over{0.004 ~{\rm m} + 8 \cdot 0.002 ~{\rm m}}} \\ E_{\rm g}  &= {{5'000 ~\rm V}\over{0.004 ~{\rm m} + 8 \cdot 0.002 ~{\rm m}}} \\