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ee2:task_ddjurcpk494go2q1_with_calculation [2024/07/15 16:57]
mexleadmin angelegt 		ee2:task_ddjurcpk494go2q1_with_calculation [2024/07/15 21:37] (aktuell)
mexleadmin
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 {{tag>electric_field magnetic_field exam_ee2_SS2024}}{{include_n>1040}} {{tag>electric_field magnetic_field exam_ee2_SS2024}}{{include_n>1040}}
    
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Capacitor \\+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Fields of an coax Cable\\
 <fs medium>(written test, approx. 12 % of a 120-minute written test, SS2024)</fs> #@TaskText_HTML@# <fs medium>(written test, approx. 12 % of a 120-minute written test, SS2024)</fs> #@TaskText_HTML@#
  
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 {{drawio>ee2:ddjurcpk494go2q1_question1.svg}} {{drawio>ee2:ddjurcpk494go2q1_question1.svg}}
  
-1. What is the magnitude of the magnetic field strength $H$ at $\rm (-0.1 ~mm |  0)$ and $\rm (0.55 ~mm |  0)$?+1. What is the magnitude of the magnetic field strength $H$ at $\rm (0.1 ~mm |  0)$ and $\rm (0.55 ~mm |  0)$?
  
-2. Plot the graph of the magnitude of $H(x)$ from $\rm (-0.6 ~mm |  0)$ to $\rm (0.6 ~mm |  0)$ in one diagram. Use proper dimensions and labels for the diagram! +#@HiddenBegin_HTML~ddjurcpk494go2q1_11,Path~@#
- +
-3. What is the magnitude of the electric displacement field $D$ at $\rm (-0.1 ~mm |  0)$ and $\rm (0.55 ~mm |  0)$? +
- +
-4. Plot the graph of the magnitude of $D(x)$ from $\rm (-0.6 ~mm |  0)$ to $\rm (0.6 ~mm |  0)$ in one diagram. Use proper dimensions and labels for the diagram!+
  
 +The magnitude of the magnetic field strength $H$ can be calculated by: $H = {{I}\over{2 \pi \cdot r}} $ \\
 +So, we get for $H_{\rm i}$ at $\rm (0.1 ~mm |  0)$, and $H_{\rm o}$ at $\rm (0.55 ~mm |  0)$:
  
-#@HiddenBegin_HTML~k4wrrhf8v46gct49_11,Path~@# 
 \begin{align*} \begin{align*}
-C &= \varepsilon_0 \varepsilon_r {{A}\over{d}} \\ +H_{\rm i} &= {{I}\over{2 \pi \cdot r_{\rm i}}} \\ 
-  &= 8.854 \cdot 10^{-12} ~\rm As/Vm \cdot 1  \cdot {{25 \cdot 10^{-6} {~\rm m} }\over{200 \cdot 10^{-6} {~\rm m} }}+          &= {{+3.3 A}\over{2 \pi \cdot { 0.1 \cdot 10^{-3}~\rm m}}} \\ 
 +H_{\rm o} &= {{I}\over{2 \pi \cdot r_{\rm o}}} \\ 
 +          &= {{+3.3 A}\over{2 \pi \cdot { 0.55 \cdot 10^{-3}~\rm m}}} \\
 \end{align*} \end{align*}
  
-#@HiddenEnd_HTML~k4wrrhf8v46gct49_11,Path~@#+Hint: For the direction, one has to consider the right-hand rule.  
 +By this, we see that the $H$-field on the right side points downwards. \\ 
 +Therefore, the sign of the $H$-field is negative. \\ 
 +But here, only the magnitude was questioned! 
 +#@HiddenEnd_HTML~ddjurcpk494go2q1_11,Path~@#
  
-#@HiddenBegin_HTML~k4wrrhf8v46gct49_12,Result~@# +#@HiddenBegin_HTML~ddjurcpk494go2q1_12,Result~@# 
-$C = 1.1 ~\rm pF$ +  * for $(0.1 ~/rm mm | 0)$ : $H_{\rm i} = 5.25... ~\rm A/m$ 
-#@HiddenEnd_HTML~k4wrrhf8v46gct49_12,Result~@#+  * for $(0.55 ~/rm mm| 0)$ : $H_{\rm o} = 0.955... ~\rm A/m$ 
 +#@HiddenEnd_HTML~ddjurcpk494go2q1_12,Result~@#
  
-2. Calculate the value of the displacement field between the plates, when a voltage of $U=3.3 ~\rm V$ is applied.+2. Plot the graph of the magnitude of $H(x)$ with $x \in \rm [-0.6~mm, +0.6~mm]$ from $\rm (-0.6 ~mm |  0)$ to $\rm (0.6 ~mm |  0)$ in one diagram. Use proper dimensions and labels for the diagram!
  
-#@HiddenBegin_HTML~k4wrrhf8v46gct49_21,Path~@#+#@HiddenBegin_HTML~ddjurcpk494go2q1_21,Path~@# 
 +  * In general, the $H$-field is proportional to ${{1}\over{r}}$ for the situation between both conductors. 
 +  * For $H(x)$ with $x$ within the inner conductor, one has to consider how much current is conducted within a circle with the radius $x$. \\ This is proportional to the area within this radius. Therefore, Ther formula $H = {{I}\over{2 \pi \cdot r}} $ gets $H(x) = {{I}\over{2 \pi \cdot x}} \cdot {{\pi x^2}\over{\pi (0.1~\rm mm)^2}}$. This leads to a formula proportional to $x$. 
 +  * For $x$ within the outer conductor one also gets a linear proportionality with a similar approach. 
 +#@HiddenEnd_HTML~ddjurcpk494go2q1_21,Path~@#
  
-The displacement field is given by:+#@HiddenBegin_HTML~ddjurcpk494go2q1_22,Result~@# 
 +{{drawio>ee2:ddjurcpk494go2q1_answer1.svg}} 
 +#@HiddenEnd_HTML~ddjurcpk494go2q1_22,Result~@# 
 + 
 +3. What is the magnitude of the electric displacement field $D$ at $\rm (-0.1 ~mm |  0)$ and $\rm (0.55 ~mm |  0)$? 
 + 
 +#@HiddenBegin_HTML~ddjurcpk494go2q1_31,Path~@# 
 +The magnitude of the electric displacement field $D$ can be calculated by: $\int D {\rm d}A = Q$. \\ 
 +Here, for any position radial to the center, the surrounding area is the surface of a cylindrical shape (here for simplicity without the round endings). \\ 
 +This leads to:
 \begin{align*} \begin{align*}
-D &= \varepsilon_0 \varepsilon_r E \\ +D(x) &= {{Q}\over{A}} \\ 
-  &= \varepsilon_0 \varepsilon_r {{U}\over{d}} \\ +     &= {{Q}\over{l \cdot 2\pi \cdot x}} \\
-  &= 8.854 \cdot 10^{-12} ~\rm As/Vm \cdot 1  \cdot {{3.3 {~\rm V} }\over{200 \cdot 10^{-6} {~\rm m} }} \\+
 \end{align*} \end{align*}
-#@HiddenEnd_HTML~k4wrrhf8v46gct49_21,Path~@# 
  
-#@HiddenBegin_HTML~k4wrrhf8v46gct49_22,Result~@# 
-$D = 146 \cdot 10^{-9} \rm {{C}\over{m^2}}$  
-#@HiddenEnd_HTML~k4wrrhf8v46gct49_22,Result~@# 
  
-3. Calculate the charge difference between both plates for a voltage of $U=3.3 ~\rm V$.+So, we get for $D_{\rm i}$ at $\rm (0.1 ~mm |  0)$, and $D_{\rm o}$ at $\rm (0.55 ~mm |  0)$:
  
-#@HiddenBegin_HTML~k4wrrhf8v46gct49_31,Path~@# 
-There are two ways now. Either: 
 \begin{align*} \begin{align*}
-Q &= C \cdot U = 1.1 ~\rm pF \cdot 3.3 ~ V = 3.6522... ~pC \\ +D_{\rm i} &= {{Q                 }\over{2 \pi \cdot r_{\rm i} \cdot l}} \\ 
-\end{align*} +          &= {{10 \cdot 10^{-9} C}\over{2 \pi \cdot { 0.1 \cdot 10^{-3}~\rm m} \cdot 0.5 ~\rm m}} \\ 
-Or: +D_{\rm o} &= {{Q                 }\over{2 \pi \cdot r_{\rm o} \cdot l}} \\ 
-\begin{align*} +          &= {{10 \cdot 10^{-9} C}\over{2 \pi \cdot { 0.55 \cdot 10^{-3}~\rm m}\cdot 0.5 ~\rm m}} \\
-Q &= D \cdot A =  146 \cdot 10^{-9} \rm {{C}\over{m^2}} \cdot 25 \cdot 10^{-6} ~m^2 = 3.6522... ~pC \\+
 \end{align*} \end{align*}
  
-#@HiddenEnd_HTML~k4wrrhf8v46gct49_31,Path~@#+Hint: For the direction, one has to consider the sign of the enclosed charge.  
 +By this, we see that the $D$-field is positive. \\ 
 +But here, again only the magnitude was questioned!
  
-#@HiddenBegin_HTML~k4wrrhf8v46gct49_32,Result~@# +#@HiddenEnd_HTML~ddjurcpk494go2q1_31,Path~@#
-$Q = 3.65 ~\rm pC $  +
-#@HiddenEnd_HTML~k4wrrhf8v46gct49_32,Result~@#+
  
-4. Due to a production problem, the right-side layer is covered with a contaminant, see the right-side image. \\ +#@HiddenBegin_HTML~ddjurcpk494go2q1_32,Result~@# 
-The contaminant has $\varepsilon_{\rm r,c}>\varepsilon_{\rm r,air}$, while the distance between the plates remains the same.  +  * for $(0.1 ~\rm mm | 0)$ : $D_{\rm i} = 31.8... ~\rm uC/m^2$ 
-Give a generalized formula $C_2=f(A,d,x,\varepsilon_{\rm r,c}, \varepsilon_{\rm r,air})$.+  * for $(0.55 ~\rm mm| 0)$ : $D_{\rm o} = 5.78... ~\rm uC/m^2$ 
 +#@HiddenEnd_HTML~ddjurcpk494go2q1_32,Result~@#
  
-#@HiddenBegin_HTML~k4wrrhf8v46gct49_41,Path~@# +4. Plot the graph of the magnitude of $D(x)$ with $x \in \rm [-0.6~mm, +0.6~mm]$ from $\rm (-0.6 ~mm |  0)$ to $\rm (0.6 ~mm |  0)$ in one diagram. Use proper dimensions and labels for the diagram!
-The resulting capacity $C$ is now a series circuit of $C_{\rm air}$ and $C_{\rm c}$. \\ +
-Therefore: +
-\begin{align*} +
-C = {{1}\over{ {{1}\over{C_{\rm air} }} + {{1}\over{ C_{\rm c}}} }} +
-\end{align*}+
  
-With +#@HiddenBegin_HTML~ddjurcpk494go2q1_41,Path~@# 
-\begin{align*} +  * In general, the $D$-field is proportional to ${{1}\over{r}}$ for the situation between both conductors. 
-C_{\rm air} &= \varepsilon_0 \varepsilon_{\rm r, air} {{A}\over{d-x}} &&= \varepsilon_0 A {{\varepsilon_{\rm r, air}}\over{d-x}}  \\ +  * Since the charges are on the surface of the conductor, there is no $D$-field within the conductor. 
-C_{\rm c}   &= \varepsilon_0 \varepsilon_{\rm r, c}   {{A}\over{x}}   &&= \varepsilon_0 A {{\varepsilon_{\rm r, c}  }\over{x}}  \\ +#@HiddenEnd_HTML~ddjurcpk494go2q1_41,Path~@#
-\end{align*}+
  
-This leads to:  +#@HiddenBegin_HTML~ddjurcpk494go2q1_42,Result~@# 
-\begin{align*} +{{drawio>ee2:ddjurcpk494go2q1_answer2.svg}} 
-C = \varepsilon_0 A {{1}\over{ {{d-x}\over{\varepsilon_{\rm r, air} }} + {{x}\over{ \varepsilon_{\rm r, c}}} }} +#@HiddenEnd_HTML~ddjurcpk494go2q1_42,Result~@#
-\end{align*}+
  
-#@HiddenEnd_HTML~k4wrrhf8v46gct49_41,Path~@# 
  
-#@HiddenBegin_HTML~k4wrrhf8v46gct49_42,Result~@# 
-$C = \varepsilon_0 A {{\varepsilon_{\rm r, air} \cdot  \varepsilon_{\rm r, c} }\over{ {(d-x)\cdot{\varepsilon_{\rm r, c} }} + {x\cdot{ \varepsilon_{\rm r, air}}} }}$  
-#@HiddenEnd_HTML~k4wrrhf8v46gct49_42,Result~@# 
  
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#