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| Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
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electrical_engineering_2:the_electrostatic_field [2023/05/12 14:21] mexleadmin |
electrical_engineering_2:the_electrostatic_field [2024/07/01 13:08] (aktuell) mexleadmin [Bearbeiten - Panel] |
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| - | ====== 1. The Electrostatic Field ====== | + | ====== 1 The Electrostatic Field ====== |
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| - | First, we will differentiate some terms: | ||
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| Only electrostatics is discussed in this chapter. For the time being, magnetic fields are thus excluded. | Only electrostatics is discussed in this chapter. For the time being, magnetic fields are thus excluded. | ||
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| 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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| =====1.3 Work and Potential ===== | =====1.3 Work and Potential ===== | ||
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| - | ==== Tasks ==== | ||
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| =====1.4 Conductors in the Electrostatic Field ===== | =====1.4 Conductors in the Electrostatic Field ===== | ||
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| - | --> Answer | + | # |
| $\varrho_2 < \varrho_3 < \varrho_1 < \varrho_4$ | $\varrho_2 < \varrho_3 < \varrho_1 < \varrho_4$ | ||
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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 $E_{ \rm max}=12~{ \rm kV/cm}$? | + | |
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| * 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? | ||
| - | </ | + | # |
| - | <button size=" | + | # |
| - | - $U_{ \rm a} = 4~{ \rm kV}$, $U_{ \rm g} = 1 ~{ \rm kV}$ | + | |
| - | - $d_{ \rm g} = 5.96~{ \rm mm}$ | + | 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 $E = {{U}\over{d}}$: |
| - | </ | + | \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 | ||
| + | \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' | ||
| + | & | ||
| + | \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} & | ||
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| + | \end{align*} | ||
| + | # | ||
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| + | # | ||
| + | $U_{ \rm a} = 4~{ \rm kV}$, $U_{ \rm g} = 1 ~{ \rm kV}$ | ||
| + | # | ||
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| + | 2. What would be the maximum allowed thickness of a glass plate, when the electric field in the air-gap shall not exceed | ||
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| + | # | ||
| + | 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*} | ||
| + | # | ||
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| + | # | ||
| + | $d_{ \rm g} = 2.10~{ \rm mm}$ | ||
| + | # | ||
| </ | </ | ||
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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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