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electrical_engineering_2:the_electrostatic_field [2023/09/19 23:51] – mexleadminelectrical_engineering_2:the_electrostatic_field [2025/03/20 10:56] (aktuell) – mexleadmin
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-<panel type="info" title="Task 1.1.5 Field lines"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Task 1.1.2 Field lines"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
 Sketch the field line plot for the charge configurations given in <imgref ImgNr04>. \\ Sketch the field line plot for the charge configurations given in <imgref ImgNr04>. \\
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 {{youtube>QWOwK-zyEnE}} {{youtube>QWOwK-zyEnE}}
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 +{{page>task_1.1.3_with_calc&nofooter}}
 +{{page>task_1.1.4&nofooter}}
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 +
  
 =====1.3 Work and Potential ===== =====1.3 Work and Potential =====
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 </callout> </callout>
- 
-==== Tasks ==== 
- 
-{{page>task_1.1.3_with_calc&nofooter}} 
-{{page>task_1.1.4&nofooter}} 
-{{page>task_1.1.5&nofooter}} 
  
 =====1.4 Conductors in the Electrostatic Field ===== =====1.4 Conductors in the Electrostatic Field =====
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 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 $U_0 = 5~{ \rm kV}$.  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 $d_{ \rm g} = 2 ~{ \rm mm}$ and $\varepsilon_{ \rm r} = 8$ is introduced parallel to the capacitor plates.+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 $U_{ \rm a}$. +1. Calculate the partial voltages on the glas $U_{ \rm g}$ and on the air gap $U_{ \rm a}$.
-  - What would be the maximum allowed thickness of a glass plate, when the electric field in the air-gap shall not exceed $E_{ \rm max}=12~{ \rm kV/cm}$?+
  
-<button size="xs" type="link" collapse="Loesung_1_5_3_Tipps">{{icon>eye}} Tips for the solution</button><collapse id="Loesung_1_5_3_Tipps" collapsed="true">+#@HiddenBegin_HTML~1531T,Tipps~@#
   * build a formula for the sum of the voltages first    * build a formula for the sum of the voltages first 
   * How is the voltage related to the electric field of a capacitor?   * How is the voltage related to the electric field of a capacitor?
-</collapse>+#@HiddenEnd_HTML~1531T,Tipps~@#
  
-<button size="xs" type="link" collapse="Loesung_1_5_3_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_1_5_3_Endergebnis" collapsed="true"> +#@HiddenBegin_HTML~1531P,Path~@# 
-  - $U_{ \rm a} = 4~{ \rm kV}$, $U_{ \rm g} = 1 ~{ \rm kV}$  + 
-  - $d_{ \rm g} = 5.96~{ \rm mm}$ +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 $E = {{U}\over{d}}$: 
-</collapse>+\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 re-arrange 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~@#
  
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 <button size="xs" type="link" collapse="Loesung_1_5_4_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_1_5_4_Endergebnis" collapsed="true"> <button size="xs" type="link" collapse="Loesung_1_5_4_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_1_5_4_Endergebnis" collapsed="true">
-  - $C = 0.5~pF$ +  - $C = 0.5~{ \rm pF}$ 
   - $C_{\infty} = 0.33~{ \rm pF}$   - $C_{\infty} = 0.33~{ \rm pF}$
 </collapse> </collapse>
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 \end{align*} \end{align*}
 \begin{align*} \begin{align*}
-\boxed{ U_k = const}+\boxed{ U_k = {\rm const.}}
 \end{align*} \end{align*}
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-<panel type="info" title="Task 1.9.2 Capacitor with glass plate"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 1.9.5 Capacitor with glass plate"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
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