{{tag>electrostatic field_lines exam_ee2_SS2022}}{{include_n>1000}}
 
#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Electrostatics I\\
<fs medium>(written test, approx. 10 % of a 120-minute written test, SS2022)</fs> #@TaskText_HTML@#

Given is the arrangement of electric charges in the picture below.
The values of the point charges are
  * $q_0=-1 ~\rm nC$ 
  * $q_1=-2 ~\rm nC$
  * $q_2=q_3=+5 ~\rm nC$
In the beginning, the area charge is $q_4=0 ~\rm nC$. \\
The permittivity is $\varepsilon_{\rm r} \varepsilon_0 = 8.854 \cdot 10^{-12} ~\rm As/Vm$

{{drawio>ee2:dtoqvpvrbdtcozfk_question1.svg}}

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 $\vec{F}_{01}$. \\
The force $\vec{F}_{01}$ is purely on the $x$-axis and therefore equal to $F_{01,x}$.
\begin{align*}
\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}} \\
             &= 19.97... \cdot 10^{-6} {\rm {{(As)^2 \cdot Vm}\over{As \cdot m^2}}} 
              = 19.97... \cdot 10^{-6} {\rm {{VAs}\over{m}}} 
              = 19.97... \cdot 10^{-6} {\rm {{Ws}\over{m}}} \\
             &= 19.97... {\rm \mu N} \quad \text{(to the right)}
\end{align*}

Similarly, we get for $\vec{F}_{02}$ and $\vec{F}_{03}$
\begin{align*}
\vec{F}_{02} = F_{02,x} &= -28.09... {\rm \mu N} \quad \text{(to the right)} \\
\vec{F}_{03}            &= -22.47... {\rm \mu N} \quad \text{(to the top left)} \\
\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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#@HiddenBegin_HTML~dtoqvpvrbdtcozfk_12,Result~@#
  * $\vec{F}_{01} = \left( {\begin{array}{cccc}  19.97 {~\rm \mu N}  \\ 0                 \\ \end{array} } \right)$
  * $\vec{F}_{02} = \left( {\begin{array}{cccc}  28.09 {~\rm \mu N}  \\ 0                 \\ \end{array} } \right)$
  * $\vec{F}_{03} = \left( {\begin{array}{cccc} -10.05 {~\rm \mu N}  \\ 20.10 {~\rm \mu N} \\ \end{array} } \right)$
 
#@HiddenEnd_HTML~dtoqvpvrbdtcozfk_12,Result~@#

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: \\
\begin{align*}
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} \\
              &= 42.99... {~\rm \mu N} \\
\end{align*}
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#@HiddenBegin_HTML~dtoqvpvrbdtcozfk_22,Result~@#
$F = 43.0 {~\rm \mu N} $
#@HiddenEnd_HTML~dtoqvpvrbdtcozfk_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 $q_0$?

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In the homogenous field the force is calculated by $F = E \cdot q$. \\
Here, this field has to compensate for the force $\vec{F}_{01}$ from $q_1$ on $q_0$:
\begin{align*}
|\vec{F}_{01}|  &= |E_4| \cdot |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 }} \\
                &= 19.97... \cdot 10^{3}{{\rm N}\over{\rm C}} \\
                &= 19.97... \cdot 10^{3}{{\rm VAs/m}\over{\rm As}} \\
                &= 19.97... \cdot 10^{3}{{\rm V}\over{\rm m}} \\
\end{align*}
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#@HiddenBegin_HTML~dtoqvpvrbdtcozfk_32,Result~@#
$E_4 =  19.97 {{\rm kV}\over{\rm m}}$
#@HiddenEnd_HTML~dtoqvpvrbdtcozfk_32,Result~@#

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