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electrical_engineering_2:task_1.1.3_with_calc [2024/03/12 23:51] mexleadminelectrical_engineering_2:task_1.1.3_with_calc [Unbekanntes Datum] (aktuell) – gelöscht - Externe Bearbeitung (Unbekanntes Datum) 127.0.0.1
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-<panel type="info" title="Task 1.2.5 Forces on Charges (exam task, ca 8 % of a 60 minute exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> 
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-Given is an arrangement of electric charges located in a vacuum (see picture on the right). \\ 
-The charges have the following values:  \\ 
-$Q_1=7 ~\rm{µC}$ (point charge) \\ 
-$Q_2=5 ~\rm{µC}$ (point charge) \\ 
-$Q_3=0 ~\rm{C}$ (infinitely extended surface charge) 
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-$\varepsilon_0=8.854\cdot 10^{-12}  ~\rm{F/m}$  , $\varepsilon_r=1$ 
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-1. calculate the magnitude of the force of $Q_2$ on $Q_1$, without the force effect of $Q_3$. 
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-<button size="xs" type="link" collapse="Loesung_5_1_3_1_Tipps">{{icon>eye}} Tips for the solution</button><collapse id="Loesung_5_1_3_1_Tipps" collapsed="true"> 
-  * Which equation is to be used for the force effect of charges? 
-  * How can the distance between the two charges be determined? 
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-<button size="xs" type="link" collapse="Loesung_5_1_3_1_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_5_1_3_1_Lösungsweg" collapsed="true"> 
-\begin{align*} 
-F_C &= {{{1} \over {4\pi\cdot\varepsilon}} \cdot {{Q_1 \cdot Q_2} \over {r^2}}} \quad && | \text{with } r=\sqrt{\Delta x^2 + \Delta y^2}  \\ 
-F_C &= {{{1} \over {4\pi\cdot\varepsilon}} \cdot {{Q_1 \cdot Q_2} \over {\Delta x^2 + \Delta y^2}}} \quad && | \text{Insert numerical values, read off distances: } \Delta x = 5~\rm{dm}, \Delta y = 3~\rm{dm}  \\ 
-F_C &= {{{1} \over {4\pi\cdot 8.854\cdot 10^{-12}  ~\rm{F/m}}} \cdot {{7 \cdot 10^{-6} ~\rm{C} \cdot 5 \cdot 10^{-6} ~\rm{C}} \over { (0.5~\rm{m})^2 + (0.2~\rm{m})^2}}}  
-\end{align*} 
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-<button size="xs" type="link" collapse="Loesung_5_1_3_1_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_5_1_3_1_Endergebnis" collapsed="true"> 
-\begin{align*} 
-|\vec{F}_C| = 1.084 ~\rm{N} \rightarrow 1.1 ~\rm{N} 
-\end{align*} 
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-2. is this force attractive or repulsive? 
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-<button size="xs" type="link" collapse="Loesung_5_1_3_2_Tipps">{{icon>eye}} Tips for the solution</button><collapse id="Loesung_5_1_3_2_Tipps" collapsed="true"> 
-  * What force effect do equally or oppositely charged bodies exhibit on each other? 
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-The force is repulsive because both charges have the same sign. \\ \\ \\ 
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-Now let $Q_2=0$ and the surface charge $Q_3$ be designed in such a way that a homogeneous electric field with $E_3=100 ~\rm{kV/m}$ results. \\ What force (magnitude) now results on $Q_1$? 
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-<button size="xs" type="link" collapse="Loesung_5_1_3_3_Tipps">{{icon>eye}} Tips for the solution</button><collapse id="Loesung_5_1_3_3_Tipps" collapsed="true"> 
-  * Which equation is to be applied for the force action in the homogeneous field? 
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-\begin{align*} 
-F_C &= E \cdot Q_1 \quad && | \text{Insert numerical values} \\ 
-F_C &= 100 \cdot 10^3 ~\rm{V/m} \cdot 7 \cdot 10^{-6} ~\rm{C} 
-\end{align*} 
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-<button size="xs" type="link" collapse="Loesung_5_1_3_3_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_5_1_3_3_Endergebnis" collapsed="true"> 
-\begin{align*} 
- |\vec{F}_C| = 0.7 ~\rm{N}  
-\end{align*} \\ 
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