Unterschiede
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Beide Seiten der vorigen Revision Vorhergehende Überarbeitung | |||
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] |
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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 $E = {{U}\over{d}}$: | 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*} | \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*} | \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. | 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*} | \begin{align*} | ||
- | D_\rm g &= D_\rm a \\ | + | D_{\rm g} & |
- | \varepsilon_0 \varepsilon_{\rm r, g} \cdot E_\rm g &= \varepsilon_0 | + | \varepsilon_0 \varepsilon_{\rm r, g} \cdot E_{\rm g} &= \varepsilon_0 |
- | \varepsilon_0 \varepsilon_{\rm r, g} \cdot E_\rm g &= \varepsilon_0 | + | |
\end{align*} | \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 $E_\rm g$: | 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*} | \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} & |
\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 $d_{\rm a} = d_0 - d_{\rm a}$ we can calculate: |
\begin{align*} | \begin{align*} | ||
- | E_\rm g &= {{5' | + | E_{\rm g} & |
& | & | ||
\end{align*} | \end{align*} | ||
Zeile 941: | Zeile 940: | ||
By this, the individual voltages can be calculated: | By this, the individual voltages can be calculated: | ||
\begin{align*} | \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} & | U_{ \rm a} &= U_0 - U_{ \rm g} &&= 5 ~{\rm kV} - 1 ~{\rm kV} & | ||
Zeile 954: | Zeile 953: | ||
2. What would be the maximum allowed thickness of a glass plate, when the electric field in the air-gap shall not exceed $E_{ \rm max}=12~{ \rm kV/cm}$? | 2. What would be the maximum allowed thickness of a glass plate, when the electric field in the air-gap shall not exceed $E_{ \rm max}=12~{ \rm kV/cm}$? | ||
+ | |||
+ | # | ||
+ | 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 | ||
+ | Now we shall eliminate $E_\rm g$, since $E_\rm a$ is given in the question. | ||
+ | \begin{align*} | ||
+ | U_0 & | ||
+ | &= {{E_\rm a}\over{\varepsilon_{\rm r, | ||
+ | \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} &= { { {{U_0}\over{E_{\rm a} }} - d_0 } \over { {{1}\over{\varepsilon_{\rm r,g}}} - 1 } } & | ||
+ | \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) | ||
+ | \end{align*} | ||
+ | # | ||
+ | |||
# | # | ||
- | $d_{ \rm g} = 5.96~{ \rm mm}$ | + | $d_{ \rm g} = 2.10~{ \rm mm}$ |
# | # | ||