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electrical_engineering_and_electronics_1:block11 [2025/10/31 15:34] mexleadminelectrical_engineering_and_electronics_1:block11 [2025/11/08 14:27] (aktuell) mexleadmin
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-====== Block 11 — Influence and displacement field ======+===== Block 11 — Influence and Displacement Field ======
  
 ===== Learning objectives ===== ===== Learning objectives =====
 <callout> <callout>
 After this 90-minute block, you can 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). 
 +  * 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> </callout>
  
 +~~PAGEBREAK~~ ~~CLEARFIX~~
 ===== Preparation at Home ===== ===== Preparation at Home =====
  
Zeile 14: Zeile 21:
  
 For checking your understanding please do the following exercises: For checking your understanding please do the following exercises:
-  * ...+  * 5.4.
 +  * 5.4.4 
 +  * 5.4.5
  
 +~~PAGEBREAK~~ ~~CLEARFIX~~
 ===== 90-minute plan ===== ===== 90-minute plan =====
-  - Warm-up (min):  +  - Warm-up (10 min): 
-    - ....  +    - One-minute recap quiz (Block 10): equipotentials, field lines. 
-  - Core concepts & derivations (min): +    - Demo: conductor in external field → Faraday cage effect (refer to the embedded sim in this block)
-    - ... +  - Core concepts & derivations (45 min): 
-  - Practice (min): ... +    - Induction on conductors: charge displacement, $E_{\rm inside}=0$, field normal to the surface. 
-  - Wrap-up (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 ===== ===== Conceptual overview =====
 <callout icon="fa fa-lightbulb-o" color="blue"> <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> </callout>
  
 +~~PAGEBREAK~~ ~~CLEARFIX~~
 ===== Core content ===== ===== Core content =====
  
 +~~PAGEBREAK~~ ~~CLEARFIX~~
 ==== Electric Field inside of a conductor ==== ==== Electric Field inside of a conductor ====
  
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 </callout> </callout>
  
 +~~PAGEBREAK~~ ~~CLEARFIX~~
 ==== Electrostatic Induction ==== ==== Electrostatic Induction ====
  
Zeile 122: Zeile 152:
   - The internal measurable electric field is compensated   - The internal measurable electric field is compensated
  
-To have an uncompensated field in the following the **electric displacement flux density $\vec{D}$** is introduced.+To have an uncompensated field in the following the **electric displacement flux density $\vec{D}$** is introduced. 
 The electric displacement flux density is only focusing on the __cause__ of the electric fields.  The electric displacement flux density is only focusing on the __cause__ of the electric fields. 
-As we have seen, its effect can differ since the space can also "hinder" the electric field in an effect. +As we have seen, its effect can differ since the space can also "hinder" the electric field in an effect. \\
  
 The electric displacement flux density is only related to the causing charges $Q$. Thie relationship is shown in the following. The electric displacement flux density is only related to the causing charges $Q$. Thie relationship is shown in the following.
Zeile 146: Zeile 176:
 </WRAP> </WRAP>
  
 +~~PAGEBREAK~~ ~~CLEARFIX~~
 +==== Dielectric Constant (Permittivity) ====
 +
 +Dielectric materials reduce the electric field inside them. How much die field is reduced is given by a material dependent constant the **dielectric constant** or **permittivity** $\varepsilon_r$. It is unitless and a ratio related to the unhindered field in vacuum.
 +
 +\begin{align*}
 +{{D}\over{E}} = \varepsilon = \varepsilon_0 \cdot \varepsilon_{ \rm r} \\
 +\boxed{D = \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot E}
 +\end{align*}
 +
 +Some values of the relative permittivity $\varepsilon_{ \rm r}$ for dielectrics are given in <tabref tab01>
 +
 +
 +<WRAP 30em>
 +
 +<tabcaption tab01| relative permittivity>
 +^ material               ^ relative permittivity \\ $\varepsilon_{ \rm r}$ for low frequencies ^
 +| air                    | $\rm 1.0006$  |
 +| paper                  | $\rm 2$       |
 +| PE, PP                 | $\rm 2.3$     |
 +| PS                     | $\rm 2.5$     |
 +| hard paper             | $\rm 5$       |
 +| glass                  | $\rm 6...8$   |
 +| water ($20~°{ \rm C}$) | $\rm 80$      |
 +</tabcaption>
 +</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
 + 600,400 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
 + 600,400 noborder}} \\
 +
 +{{drawio>FieldCylPlates01.svg}}
 +</WRAP></WRAP></WRAP>
 +
 +~~PAGEBREAK~~ ~~CLEARFIX~~
 ==== Dielectric strength of dielectrics ==== ==== Dielectric strength of dielectrics ====
  
Zeile 172: Zeile 266:
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
 ===== Common pitfalls ===== ===== Common pitfalls =====
-  * ...+  * Mixing up **cause** and **effect**: using $\oint \vec{E}\cdot{\rm d}\vec{A}$ to count chargeUse **$\vec{D}$** 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 ===== ===== Exercises =====
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 #@HiddenEnd_HTML~1,Result~@# #@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. 
-  
-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 5.5.4 Spherical capacitor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> 
- 
-Two concentric spherical conducting plates set up a spherical capacitor.  
-The radius of the inner sphere is $r_{ \rm i} = 3~{ \rm mm}$, and the inner radius from the outer sphere is $r_{ \rm o} = 9~{ \rm mm}$.  
- 
-  - What is the capacity of this capacitor, given that air is used as a dielectric? 
-  - What would be the limit value of the capacity when the inner radius of the outer sphere goes to infinity ($r_{ \rm o} \rightarrow \infty$)? 
-<button size="xs" type="link" collapse="Loesung_1_5_4_Tipps">{{icon>eye}} Tips for the solution</button><collapse id="Loesung_1_5_4_Tipps" collapsed="true"> 
-  * What is the displacement flux density of the inner sphere? 
-  * Out of this derive the strength of the electric field $E$ 
-  * What ist the general relationship between $U$ and $\vec{E}$? Derive from this the voltage between the spheres. 
-</collapse> 
- 
-<button size="xs" type="link" collapse="Loesung_1_5_4_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_1_5_4_Endergebnis" collapsed="true"> 
-  - $C = 0.5~{ \rm pF}$  
-  - $C_{\infty} = 0.33~{ \rm pF}$ 
-</collapse> 
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>