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electrical_engineering_2:task_1.1.3_with_calc [2022/03/10 12:28] – ↷ Seitename wurde von electrical_engineering_2:task_5.1.3_with_calc auf electrical_engineering_2:task_1.1.3_with_calc geändert tfischerelectrical_engineering_2:task_1.1.3_with_calc [2024/03/12 23:51] (aktuell) – mexleadmin
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 Given is an arrangement of electric charges located in a vacuum (see picture on the right). \\ Given is an arrangement of electric charges located in a vacuum (see picture on the right). \\
 The charges have the following values:  \\ The charges have the following values:  \\
-$Q_1=7 μC$ (point charge) \\ +$Q_1=7 ~\rm{µC}$ (point charge) \\ 
-$Q_2=5 μC$ (point charge) \\ +$Q_2=5 ~\rm{µC}$ (point charge) \\ 
-$Q_3=0 C$ (infinitely extended surface charge)+$Q_3=0 ~\rm{C}$ (infinitely extended surface charge)
  
-$\varepsilon_0=8.854\cdot 10^{-12}  F/m$  , $\varepsilon_r=1$+$\varepsilon_0=8.854\cdot 10^{-12}  ~\rm{F/m}$  , $\varepsilon_r=1$
  
 1. calculate the magnitude of the force of $Q_2$ on $Q_1$, without the force effect of $Q_3$. 1. calculate the magnitude of the force of $Q_2$ on $Q_1$, without the force effect of $Q_3$.
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 \begin{align*} \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 {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 = 5dm, \Delta y = 3dm  \\ +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}  F/m}} \cdot {{7 \cdot 10^{-6} C \cdot 5 \cdot 10^{-6} C} \over { (0.5m)^2 + (0.2m)^2}}} +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*} \end{align*}
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 \begin{align*} \begin{align*}
-|\vec{F}_C| = 1.084 N \rightarrow 1.1 N+|\vec{F}_C| = 1.084 ~\rm{N} \rightarrow 1.1 ~\rm{N}
 \end{align*} \end{align*}
  \\  \\
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 </collapse> </collapse>
  
-Now let $Q_2=0$ and the surface charge $Q_3$ be designed in such a way that a homogeneous electric field with $E_3=100 kV/m$ results. \\ What force (magnitude) now results on $Q_1$?+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 $Q_1$?
  
 <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"> <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">
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 \begin{align*} \begin{align*}
 F_C &= E \cdot Q_1 \quad && | \text{Insert numerical values} \\ F_C &= E \cdot Q_1 \quad && | \text{Insert numerical values} \\
-F_C &= 100 \cdot 10^3 V/m \cdot 7 \cdot 10^{-6} C+F_C &= 100 \cdot 10^3 ~\rm{V/m} \cdot 7 \cdot 10^{-6} ~\rm{C}
 \end{align*} \end{align*}
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 \begin{align*} \begin{align*}
- |\vec{F}_C| = 0.7 N + |\vec{F}_C| = 0.7 ~\rm{N} 
 \end{align*} \\ \end{align*} \\
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