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\begin{document}
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\heading{Conductivity}
\auname{I.D. 123456}
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{\bf The problem:}
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\begin{enumerate}
\item For Fig. A find the capacitance between $A,B$.
\item In Fig. B there are $N$ conducting planes of area $A$ in the distance $d$ between each other. The planes are connected as shown on the figure. Find the capacitance.
\item For the infinite triangular resistors lattice of Fig. C where each side of a triangle is of resistance $R$ find the total resistance between any two adjacent nodes.
\end{enumerate}
\begin{figure}[htbp]
\begin{center}
\subfigure[3 Capacitors]{\scalebox{0.4}{\includegraphics{e_42_1_025_p0.jpg}}} \quad
\subfigure[Long capacitor]{\scalebox{0.4}{\includegraphics{e_42_1_025_p1.jpg}}} \quad
\subfigure[Infinite resistors lattice]{\scalebox{0.4}{\includegraphics{e_42_1_025_p2.jpg}}}
\label{fig:my_car}
\end{center}
\end{figure}
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\Dn
{\bf The solution:}
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1. Let us denote the potentials on each plate of the capacitors as on the following figure
\smallskip
\includegraphics[scale=0.5]{e_42_1_025_s_p0.jpg}
Then we clearly see that all the 3 capacitors are in parallel, so that the total capacitance is $3C$.
\Dn\Dn
2. Here like in the previous question we notice that on each node of the upper wire the potential is $V_A$ and on each node of the bottom wire the potential is $V_B$.
Then we have $(N-1)$ pairs of conductors with the potentials difference $V_A-V_B$ between them. The capacitance of each pair is $\frac{\eps_0 A }{d}$, and these capacitors are in parallel. Therefore, the total capacitance is $(N-1)\frac{\eps_0 A }{d}$
\Dn\Dn
3. Consider the infinite 2D
triangular resistive lattice shown in Fig. C. The number of resistors
connected to each node is 6. We use superposition and symmetry
along with two test current sources each of value $I$ to calculate
the effective resistance $R_{eff}$ between two adjacent nodes
by injecting a test current $I$ into any single node on the
resistive lattice from the zero-potential
boundary at infinity and then extracting another identical test
current $I$ from an adjacent node connected to a current sink
kept at zero potential. By using Kirchhoff’s current law and
symmetry, we find that each of the 6 resistors connected to
the original node will receive $I/6$ of the injected current.
Similarly, we find that each of the 6 resistors connected to
the adjacent node will receive $-I/6$ of the extracted current
in the opposite direction. Therefore, by superposition, the
total resulting current flowing in the resistor $R$ connecting the
two adjacent nodes will be $2I/6$, which leads to a voltage
drop across the resistor $R$ of $V=(2I/6)R$.
Thus the effective resistance is $R_{eff}=R/3$.
For more details about similar lattices look at: P. Osterberg, Am. J. Phys. 72 (7), 2004
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