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ee2:task_5u1zbroaz75w39jk_with_calculation [2024/07/15 14:28]
mexleadmin angelegt 		ee2:task_5u1zbroaz75w39jk_with_calculation [2024/07/15 15:18] (aktuell)
mexleadmin angelegt
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 #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Electrostatics I\\ #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Electrostatics I\\
-<fs medium>(written test, approx. 10 % of a 120-minute written test, SS2024)</fs> #@TaskText_HTML@#+<fs medium>(written test, approx. 8 % of a 120-minute written test, SS2024)</fs> #@TaskText_HTML@#
  
 Given is the arrangement of electric charges in the picture below. Given is the arrangement of electric charges in the picture below.
 The values of the point charges are The values of the point charges are
   * $q_0=-1 ~\rm nC$    * $q_0=-1 ~\rm nC$ 
-  * $q_1=-2 ~\rm nC$ +  * $q_1=-5 ~\rm nC$ 
-  * $q_2=q_3=+5 ~\rm nC$+  * $q_2=q_3=+2 ~\rm nC$
 In the beginning, the area charge is $q_4=0 ~\rm nC$. \\ In the beginning, the area charge is $q_4=0 ~\rm nC$. \\
 The permittivity is $\varepsilon_{\rm r} \varepsilon_0 = 8.854 \cdot 10^{-12} ~\rm As/Vm$ The permittivity is $\varepsilon_{\rm r} \varepsilon_0 = 8.854 \cdot 10^{-12} ~\rm As/Vm$
  
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 1. Calculate the single forces $\vec{F}_{01}$, $\vec{F}_{02}$, $\vec{F}_{03}$,  on the charge $q_0$! 1. Calculate the single forces $\vec{F}_{01}$, $\vec{F}_{02}$, $\vec{F}_{03}$,  on the charge $q_0$!
  
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 First, calculate the magnitude of the forces, like $\vec{F}_{01}$. \\ First, calculate the magnitude of the forces, like $\vec{F}_{01}$. \\
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 \begin{align*} \begin{align*}
 \vec{F}_{01} = F_{01,x} &= {{1}\over{4\pi\varepsilon}}\cdot {{q_1\cdot q_0}\over{r^2_{01}}} \\ \vec{F}_{01} = F_{01,x} &= {{1}\over{4\pi\varepsilon}}\cdot {{q_1\cdot q_0}\over{r^2_{01}}} \\
-             &= {{1}\over{4\pi\cdot 8.854 \cdot 10^{-12} ~\rm As/Vm}}\cdot {{1 \cdot 10^{-9} ~\rm C \cdot 2 \cdot 10^{-9} ~\rm C}\over{(3 \cdot 10^{2} ~\rm m)^2}} \\ +             &= {{1}\over{4\pi\cdot 8.854 \cdot 10^{-12} ~\rm As/Vm}}\cdot {{1 \cdot 10^{-9} ~\rm C \cdot 5 \cdot 10^{-9} ~\rm C}\over{(7 \cdot 10^{-3} ~\rm m)^2}} \\ 
-             &= 19.97... \cdot 10^{-6} {\rm {{(As)^2 \cdot Vm}\over{As \cdot m^2}}}  +             &= 917.\;.\!.\!.\! \cdot 10^{-6} {\rm {{(As)^2 \cdot Vm}\over{As \cdot m^2}}}  
-              = 19.97... \cdot 10^{-6} {\rm {{VAs}\over{m}}}  +              = 917.\;.\!.\!.\! \cdot 10^{-6} {\rm {{VAs}\over{m}}}  
-              = 19.97... \cdot 10^{-6} {\rm {{Ws}\over{m}}} \\ +              = 917.\;.\!.\!.\! \cdot 10^{-6} {\rm {{Ws}\over{m}}} \\ 
-             &= 19.97... {\rm \mu N} \quad \text{(to the right)}+             &= 917.\;.\!.\!.\! {\rm \mu N} \quad \text{(to the right)}
 \end{align*} \end{align*}
  
 Similarly, we get for $\vec{F}_{02}$ and $\vec{F}_{03}$ Similarly, we get for $\vec{F}_{02}$ and $\vec{F}_{03}$
 \begin{align*} \begin{align*}
-\vec{F}_{02} = F_{02,x} &= -28.09... {\rm \mu N} \quad \text{(to the right)} \\ +\vec{F}_{02} = F_{02,x} &= -1997.\;.\!.\!.\! {\rm \mu N} \quad \text{(to the right)} \\ 
-\vec{F}_{03}            &= -22.47... {\rm \mu N} \quad \text{(to the top left)} \\+\vec{F}_{03} = F_{03,y} &= -1123.\;.\!.\!.\! {\rm \mu N} \quad \text{(to the top)} \\
 \end{align*} \end{align*}
  
-For $\vec{F}_{03}$, we have to calculate the $x$- and $y$-component. \\ 
-This is possible, by using the angle $\alpha$ between the line through $q_0$ and $q_3$ and the positive $x$-axis (pointing to the right). \\ 
-So, $\alpha$ has to be between $90°$ and $180°$. It can be calculated by: 
-\begin{align*} 
-\alpha = \arctan(\rm {{-4~cm}\over{+2~cm}}) = \pi - 1.1071... = 180° - 63.4...° = 116.6...° 
-\end{align*} 
- 
-Based on this, the $x$- and $y$-component is: 
-\begin{align*} 
-F_{03,x} &= |\vec{F}_{03}| \cdot \cos \alpha = 10.05... {~\rm \mu N} \text{(to the left)} \\ 
-F_{03,y} &= |\vec{F}_{03}| \cdot \sin \alpha = 20.10... {~\rm \mu N} \text{(to the top)} \\ 
-\end{align*} 
  
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-  * $\vec{F}_{01} = \left( {\begin{array}{cccc}  19.97 {~\rm \mu N}  \\ 0                 \\ \end{array} } \right)$ +  * $\vec{F}_{01} = \left( {\begin{array}{cccc}   917 {~\rm \mu N}  \\ 0                 \\ \end{array} } \right)$ 
-  * $\vec{F}_{02} = \left( {\begin{array}{cccc}  28.09 {~\rm \mu N}  \\ 0                 \\ \end{array} } \right)$ +  * $\vec{F}_{02} = \left( {\begin{array}{cccc}  1997 {~\rm \mu N}  \\ 0                 \\ \end{array} } \right)$ 
-  * $\vec{F}_{03} = \left( {\begin{array}{cccc} -10.05 {~\rm \mu N}  \\ 20.10 {~\rm \mu N} \\ \end{array} } \right)$+  * $\vec{F}_{03} = \left( {\begin{array}{cccc}     0 {~\rm \mu N}  \\ -1123 {~\rm \mu N} \\ \end{array} } \right)$
    
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 2. What is the magnitude of the resulting force?  2. What is the magnitude of the resulting force? 
  
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 With all the $x$- and $y$-components, we can calculate the resulting magnitude of the force with the Pythagorean Theorem: \\ With all the $x$- and $y$-components, we can calculate the resulting magnitude of the force with the Pythagorean Theorem: \\
 \begin{align*} \begin{align*}
 F = |\vec{F}| &= \sqrt{\left( \sum_i F_{i,x} \right)^2 + \left( \sum_i F_{i,y} \right)^2} \\ F = |\vec{F}| &= \sqrt{\left( \sum_i F_{i,x} \right)^2 + \left( \sum_i F_{i,y} \right)^2} \\
-              &= \sqrt{\left( 19.97 {~\rm \mu N} + 28.09 {~\rm \mu N} -10.05 {~\rm \mu N} \right)^2 + \left( 20.10 {~\rm \mu N} \right)^2} \\ +              &= \sqrt{\left( +917 {~\rm \mu N} - 1997 {~\rm \mu N} \right)^2 + \left( 1123 {~\rm \mu N} \right)^2} \\ 
-              &= 42.99... {~\rm \mu N} \\+              &= 1560.\;.\!.\!.\!  {~\rm \mu N} \\
 \end{align*} \end{align*}
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-#@HiddenBegin_HTML~dtoqvpvrbdtcozfk_22,Result~@# +#@HiddenBegin_HTML~5u1zbroaz75w39jk_22,Result~@# 
-$F = 43.0 {~\rm \mu N} $ +$F = 1560 {~\rm \mu N} $ 
-#@HiddenEnd_HTML~dtoqvpvrbdtcozfk_22,Result~@#+#@HiddenEnd_HTML~5u1zbroaz75w39jk_22,Result~@#
  
 3. Now the charges $q_2=q_3$ are set to 0. The area charge $q_4$ generates a homogenous field of $E_4$. Which value needs $E_4$ to have to get a resulting force of $0 ~\rm N$ on $q_0$? 3. Now the charges $q_2=q_3$ are set to 0. The area charge $q_4$ generates a homogenous field of $E_4$. Which value needs $E_4$ to have to get a resulting force of $0 ~\rm N$ on $q_0$?
  
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 In the homogenous field the force is calculated by $F = E \cdot q$. \\ In the homogenous field the force is calculated by $F = E \cdot q$. \\
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 |\vec{F}_{01}|  &= |E_4| \cdot |q_0| \\ |\vec{F}_{01}|  &= |E_4| \cdot |q_0| \\
 \rightarrow E_4 &= {{|\vec{F}_{01}|}\over{|q_0|}} \\ \rightarrow E_4 &= {{|\vec{F}_{01}|}\over{|q_0|}} \\
-                &= {{19.97... \cdot 10^{-6} {~\rm N}}\over{1 \cdot 10^{-9} ~\rm C }} \\ +                &= {{917.\;.\!.\!.\!  \cdot 10^{-6} {~\rm N}}\over{1 \cdot 10^{-9} ~\rm C }} \\ 
-                &= 19.97... \cdot 10^{3}{{\rm N}\over{\rm C}} \\ +                &=   917.\;.\!.\!.\!  \cdot 10^{3}{{\rm N}\over{\rm C}} \\ 
-                &= 19.97... \cdot 10^{3}{{\rm VAs/m}\over{\rm As}} \\ +                &=   917.\;.\!.\!.\!  \cdot 10^{3}{{\rm VAs/m}\over{\rm As}} \\ 
-                &= 19.97... \cdot 10^{3}{{\rm V}\over{\rm m}} \\+                &=   917.\;.\!.\!.\!  \cdot 10^{3}{{\rm V}\over{\rm m}} \\
 \end{align*} \end{align*}
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-#@HiddenBegin_HTML~dtoqvpvrbdtcozfk_32,Result~@# +#@HiddenBegin_HTML~5u1zbroaz75w39jk_32,Result~@# 
-$E_4 =  19.97 {{\rm kV}\over{\rm m}}$ +$E_4 =  917 {{\rm kV}\over{\rm m}}$ 
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