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--- electrical_engineering_and_electronics_1:block11 [2025/11/01 00:32] – mexleadmin
+++ electrical_engineering_and_electronics_1:block11 [2025/11/08 14:27] (aktuell) – mexleadmin
@@ Zeile 1: / Zeile 1: @@
-====== Block 11 — Influence and Displacement Field ======
+===== Block 11 — Influence and Displacement Field ======
 ===== Learning objectives =====
 <callout>
 After this 90-minute block, you can
-  * ...
+  * explain **electrostatic induction** on conductors and argue why the interior of a conductor is field-free (Faraday cage).
-  * Interpret **dielectric strength** $E_0$ (breakdown) and reason about its impact on design limits.
+  * distinguish the **electric field strength** $\vec{E}$ from the **electric displacement flux density** $\vec{D}$ and state $ \vec{D} = \varepsilon \vec{E} = \varepsilon_0 \varepsilon_{\rm r}\vec{E}$.
+  * apply **Gauss’s law** for the displacement field to simple closed surfaces to relate enclosed charge $Q$ and flux $\oint \vec{D}\cdot {\rm d}\vec{A}$.
+  * determine $E(r)$ for parallel-plate and coaxial geometries starting from $\vec{D}$, then using $\vec{E}=\vec{D}/(\varepsilon_0\varepsilon_{\rm r})$.
+  * reason about **surface charge density** $\varrho_A = \Delta Q/\Delta A$ and the normal field at conductor surfaces.
+  * use typical **relative permittivities** $\varepsilon_{\rm r}$ to estimate field reduction in dielectrics.
+  * interpret **dielectric strength** $E_0$ (breakdown) and reason about its impact on design limits (safe $E$, spacing, material choice).
 </callout>
+~~PAGEBREAK~~ ~~CLEARFIX~~
 ===== Preparation at Home =====
@@ Zeile 15: / Zeile 21: @@
 For checking your understanding please do the following exercises:
-  * ...
+  * 5.4.1
+  * 5.4.4
+  * 5.4.5
+~~PAGEBREAK~~ ~~CLEARFIX~~
 ===== 90-minute plan =====
-  - Warm-up (x min):
+  - Warm-up (10 min):
-    - ....
+    - One-minute recap quiz (Block 10): equipotentials, field lines.
-  - Core concepts & derivations (x min):
+    - Demo: conductor in external field → Faraday cage effect (refer to the embedded sim in this block).
-    - ...
+  - Core concepts & derivations (45 min):
-  - Practice (x min): ...
+    - Induction on conductors: charge displacement, $E_{\rm inside}=0$, field normal to the surface.
-  - Wrap-up (x min): Summary box; common pitfalls checklist.
+    - Polarization of dielectrics; motivation for introducing $\vec{D}$.
+    - Definitions: $\vec{D}$, $\vec{E}$, $\varepsilon_0$, $\varepsilon_{\rm r}$; Gauss’s law with closed surface.
+    - Worked derivations via $\vec{D}$:
+      * Parallel plates: $D=Q/A \;\Rightarrow\; E = D/(\varepsilon_0\varepsilon_{\rm r})$.
+      * Coaxial cylinders: $D(r)=Q/(2\pi l r) \;\Rightarrow\; E(r)=D(r)/(\varepsilon_0\varepsilon_{\rm r})$.
+    - Material data: typical $\varepsilon_{\rm r}$; concept of **dielectric strength** $E_0$ and safe design margins.
+  - Practice (25 min):
+    - Short board tasks using pillbox surfaces to find $\varrho_A$ on a conductor.
+    - Mixed-dielectric capacitor slice: split voltages via constant $D$.
+    - Guided use of the embedded sims to observe field/equipotential behavior and verify normal field at surfaces.
+  - Wrap-up (10 min):
+    - Summary box (key formulas, when to start with $D$ vs. $E$).
+    - Common pitfalls checklist and quick self-test questions.
+    - Preview to Block 12 (capacitors from field viewpoint).
+~~PAGEBREAK~~ ~~CLEARFIX~~
 ===== Conceptual overview =====
 <callout icon="fa fa-lightbulb-o" color="blue">
-  - ...
+  - **Conductors in electrostatics:** free charges move until $E_{\rm inside}=0$; the surface becomes an equipotential and field lines are perpendicular to it. \\ Induced charges live on the surface (surface density $\varrho_A$).
+  - **Dielectrics (polarization):** bound charges shift slightly → the macroscopic effect is a reduced $E$ compared to vacuum. \\ This motivates $\vec{D}$, which “counts causes” (free/enclosed charge) independent of polarization details.
+  - **Displacement field & Gauss’s law:** for any closed surface, the flux of $\vec{D}$ equals the enclosed charge: $Q=\oint \vec{D}\cdot{\rm d}\vec{A}$. \\ Choose the surface to exploit symmetry, get $\vec{D}$ first, then $\vec{E}$ via material law.
+  - **Permittivity:** $\varepsilon=\varepsilon_0\varepsilon_{\rm r}$ links $\vec{D}$ and $\vec{E}$. Larger $\varepsilon_{\rm r}$ → smaller $E$ for the same $D$ (same free charge).
+  - **Design limit:** when $|E|$ exceeds the **dielectric strength** $E_0$, breakdown occurs → current flows. \\ Safe design keeps $|E|\ll E_0$ by material choice and geometry (spacing, shaping to avoid high curvature).
 </callout>
+~~PAGEBREAK~~ ~~CLEARFIX~~
 ===== Core content =====
+~~PAGEBREAK~~ ~~CLEARFIX~~
 ==== Electric Field inside of a conductor ====
@@ Zeile 55: / Zeile 84: @@
 </callout>
+~~PAGEBREAK~~ ~~CLEARFIX~~
 ==== Electrostatic Induction ====
@@ Zeile 147: / Zeile 176: @@
 </WRAP>
+~~PAGEBREAK~~ ~~CLEARFIX~~
 ==== Dielectric Constant (Permittivity) ====
@@ Zeile 153: / Zeile 182: @@
 \begin{align*}
-{{D}\over{E}} = \varepsilon = \varepsilon_{ \rm r} \cdot \varepsilon_0 \\
+{{D}\over{E}} = \varepsilon = \varepsilon_0 \cdot \varepsilon_{ \rm r} \\
-\boxed{D = \varepsilon_{ \rm r} \cdot \varepsilon_0 \cdot E}
+\boxed{D = \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot E}
 \end{align*}
@@ Zeile 174: / Zeile 203: @@
 </WRAP>
+~~PAGEBREAK~~ ~~CLEARFIX~~
+==== Typical Geometries ====
+The "new" $D$-field is a nice tool, which helps to derive the $E$-field more easily. \\
+This shall be shown with the two most common geometries (which are the only one necessary for this course).
+<WRAP group>
+<WRAP half column>
+=== Field of a parallel Plates ===
+\\
+  * Nearly all of the field is between the plate (see <imgref ImgNr294> top), when the distance between the plates is much smaller than the width of the plates. \\  → Idealization: all of the field is inside. There is neither a stray field on the side, nor a field on top / below the structure. \\ \\
+  * All of the $D$-field of the charges is between the plates, and therefore through the area $A$ of the plates (see <imgref ImgNr294> bottom).  \\  → The $D$-Field is given as: $$ D = {{Q}\over{A}}$$
+  * Given $D = \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot E$, the eletric field $E$ is: $$  E = {{Q}\over{ \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot A}} $$\\
+<imgcaption ImgNr294 | field of parallel plates (field line breaks are not correct)>
+</imgcaption> <WRAP>
+{{url>https://www.falstad.com/emstatic/EMStatic.html?rol=$+1+256+128+0+10+348+0.048828125+461%0Ab+0+2+10+30+130+220+134+0%0Ab+0+2+-10+30+110+220+114+0%0A
+,400 noborder}} \\
+{{drawio>FieldParallelPlates01.svg}}
+</WRAP></WRAP>
+<WRAP half column>
+=== Field of a coaxial cylindrical Plates ===
+\\
+  * All of the field is between the plate (see <imgref ImgNr295> top). \\ \\ \\ \\ \\
+  * All of the $D$-field of the charges on the inner plate penetrates through any cylintrical area $A(l,r) = 2 \pi \cdot l \cdot r$.  \\  → The $D$-Field is given as: $$ D = {{Q}\over{A(l,r)}} = {{Q}\over{2 \pi \cdot l \cdot r}}$$
+  * Again given $D = \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot E$, the eletric field $E$ is: $$  E = {{Q}\over{ \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot 2 \pi \cdot l \cdot r}} $$\\
+<imgcaption ImgNr295 | field of coaxial cylindrical plates >
+</imgcaption> <WRAP >
+{{url>https://www.falstad.com/emstatic/EMStatic.html?rol=$+1+256+128+0+10+355+0.048828125+355%0AE+1+2+0+30+30+230+230+40+40+220+220+0%0Ae+1+2+-80+111+111+149+149+0%0A
+,400 noborder}} \\
+{{drawio>FieldCylPlates01.svg}}
+</WRAP></WRAP></WRAP>
+~~PAGEBREAK~~ ~~CLEARFIX~~
 ==== Dielectric strength of dielectrics ====
@@ Zeile 200: / Zeile 266: @@
 ~~PAGEBREAK~~ ~~CLEARFIX~~
 ===== Common pitfalls =====
-  * ...
+  * Mixing up **cause** and **effect**: using $\oint \vec{E}\cdot{\rm d}\vec{A}$ to count charge. Use **$\vec{D}$** for Gauss’s law with charge; convert to $\vec{E}$ only via $\vec{E}=\vec{D}/(\varepsilon_0\varepsilon_{\rm r})$.
+  * Forgetting that the **interior of a conductor is field-free** in electrostatics and that $E$ is **normal** to an ideal conducting surface (no tangential $E$ on the surface).
+  * Assuming induced charges fill the **volume** of a conductor. They reside on the **surface**; use $\varrho_A$, not a volume density.
+  * Ignoring that **$D$ is continuous** in the normal direction across simple dielectric interfaces when no free surface charge is present; consequently, the **electric field changes** with $\varepsilon_{\rm r}$.
+  * Treating $\varepsilon_{\rm r}$ as a constant in all contexts. Real materials can be frequency/temperature dependent; here we use low-frequency values as stated.
+  * Checking breakdown with voltage only. The limit is on **field** $E$; always relate geometry (e.g., plate spacing, curvature) to $E$ and compare to **$E_0$** with units (e.g., $\,{\rm kV/mm}$).
 ===== Exercises =====
@@ Zeile 257: / 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~@#
 </WRAP></WRAP></panel>