Dies ist eine alte Version des Dokuments!

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Exercise E1 Capacitor
(written test, approx. 12 % of a 120-minute written test, SS2024)

A $0.5 ~\rm m$ long coax cable is used for signal transmission. The diagram shows the cross-section of the coax cable with the origin in the center of the coax cable. Due to the given load, the following situation appears:

  • Inner conductor: $+3.3 ~\rm mA$, $+10 ~\rm nC$ (current into the plane of the diagram)
  • Outer conductor: $-3.3 ~\rm mA$, $ 0 ~\rm nC$ (current out of the plane of diagram)

ee2:ddjurcpk494go2q1_question1.svg

1. What is the magnitude of the magnetic field strength $H$ at $\rm (-0.1 ~mm | 0)$ and $\rm (0.55 ~mm | 0)$?

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!

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!

Path

\begin{align*} C &= \varepsilon_0 \varepsilon_r {{A}\over{d}} \\ &= 8.854 \cdot 10^{-12} ~\rm As/Vm \cdot 1 \cdot {{25 \cdot 10^{-6} {~\rm m} }\over{200 \cdot 10^{-6} {~\rm m} }} \end{align*}

Result

$C = 1.1 ~\rm pF$

2. Calculate the value of the displacement field between the plates, when a voltage of $U=3.3 ~\rm V$ is applied.

Path

The displacement field is given by: \begin{align*} D &= \varepsilon_0 \varepsilon_r E \\ &= \varepsilon_0 \varepsilon_r {{U}\over{d}} \\ &= 8.854 \cdot 10^{-12} ~\rm As/Vm \cdot 1 \cdot {{3.3 {~\rm V} }\over{200 \cdot 10^{-6} {~\rm m} }} \\ \end{align*}

Result

$D = 146 \cdot 10^{-9} \rm {{C}\over{m^2}}$

3. Calculate the charge difference between both plates for a voltage of $U=3.3 ~\rm V$.

Path

There are two ways now. Either: \begin{align*} Q &= C \cdot U = 1.1 ~\rm pF \cdot 3.3 ~ V = 3.6522... ~pC \\ \end{align*} Or: \begin{align*} 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*}

Result

$Q = 3.65 ~\rm pC $

4. Due to a production problem, the right-side layer is covered with a contaminant, see the right-side image.
The contaminant has $\varepsilon_{\rm r,c}>\varepsilon_{\rm r,air}$, while the distance between the plates remains the same. Give a generalized formula $C_2=f(A,d,x,\varepsilon_{\rm r,c}, \varepsilon_{\rm r,air})$.

Path

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 \begin{align*} C_{\rm air} &= \varepsilon_0 \varepsilon_{\rm r, air} {{A}\over{d-x}} &&= \varepsilon_0 A {{\varepsilon_{\rm r, air}}\over{d-x}} \\ C_{\rm c} &= \varepsilon_0 \varepsilon_{\rm r, c} {{A}\over{x}} &&= \varepsilon_0 A {{\varepsilon_{\rm r, c} }\over{x}} \\ \end{align*}

This leads to: \begin{align*} C = \varepsilon_0 A {{1}\over{ {{d-x}\over{\varepsilon_{\rm r, air} }} + {{x}\over{ \varepsilon_{\rm r, c}}} }} \end{align*}

Result

$C = \varepsilon_0 A {{\varepsilon_{\rm r, air} \cdot \varepsilon_{\rm r, c} }\over{ {(d-x)\cdot{\varepsilon_{\rm r, c} }} + {x\cdot{ \varepsilon_{\rm r, air}}} }}$