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--- electrical_engineering_2:the_electrostatic_field [2024/07/01 11:39]
mexleadmin [Bearbeiten - Panel]
+++ electrical_engineering_2:the_electrostatic_field [2024/07/01 13:08] (aktuell)
mexleadmin [Bearbeiten - Panel]
@@ Zeile 914: / Zeile 914: @@
 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 \\
+U_0 &= U_{\rm g}               + U_{\rm a} \\
-    &= E_\rm g \cdot d_\rm g + E_\rm a \cdot d_\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 \\
+D_{\rm g}                                            &= D_{\rm a} \\
-\varepsilon_0 \varepsilon_{\rm r, g} \cdot E_\rm g &= \varepsilon_0  \cdot E_\rm a \\
+\varepsilon_0 \varepsilon_{\rm r, g} \cdot E_{\rm g} &= \varepsilon_0  \cdot E_{\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 \\
+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) \\
+    &= 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}}
+\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:
+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}}} \\
+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*}
@@ Zeile 941: / Zeile 940: @@
 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 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}\\
@@ Zeile 954: / Zeile 953: @@
 . 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~@#
-$d_{ \rm g} = 5.96~{ \rm mm}$
+$d_{ \rm g} = 2.10~{ \rm mm}$
 #@HiddenEnd_HTML~1532R,Results~@#