Question 26.26: In Fig. 26-26a, a 9.00 V battery is connected to a resistive......
In Fig. 26-26a, a 9.00 V battery is connected to a resistive strip that consists of three sections with the same cross-sectional areas but different conductivities. Figure 26-26b gives the electric potential V(x) versus position x along the strip. The horizontal scale is set by x_s = 8.00 mm. Section 3 has conductivity 3.00 × 10^7 (Ω⋅m)^{-1}.What is the conductivity of section (a) 1 and (b) 2?
The absolute values of the slopes (for the straight-line segments shown in the graph of Fig. 26-26(b)) are equal to the respective electric field magnitudes. Thus, applying Eq.26-5 and Eq. 26-13 to the three sections of the resistive strip, we have
\vec{J}=\sigma \vec{E} (26-13)
J=\frac{i}{A} (26-5)
\begin{aligned} & J_1=\frac{i}{A}=\sigma_1 E_1=\sigma_1\left(0.50 \times 10^3 \,V / m \right) \\ & J_2=\frac{i}{A}=\sigma_2 E_2=\sigma_2\left(4.0 \times 10^3 \,V / m \right) \\ & J_3=\frac{i}{A}=\sigma_3 E_3=\sigma_3\left(1.0 \times 10^3 \,V / m. \right) \end{aligned}
We note that the current densities are the same since the values of i and A are the same (see the problem statement) in the three sections, so J_1 = J_2 = J_3 .
(a) Thus we see that \sigma_1=2 \sigma_3=2\left(3.00 \times 10^7(\Omega \cdot m )^{-1}\right)=6.00 \times 10^7(\Omega \cdot m )^{-1} .
(b) Similarly, \sigma_2=\sigma_3 / 4=\left(3.00 \times 10^7(\Omega \cdot m )^{-1}\right) / 4=7.50 \times 10^6(\Omega \cdot m )^{-1} .