1. Coordinate System and Variables

Let the strip lie in the $xy$-plane with the length along the $x$-axis and width along the $y$-axis. The potential difference $V = 196 \, \text{V}$ is applied between faces AD and BC. This constrains the macroscopic current density $\vec{J}$ to flow purely along the $x$-direction: $$ \vec{J} = J \hat{i} $$

2. Decomposing Current Density

The material has principal resistivity axes: $\hat{n}_1$ (associated with $\rho_{\text{max}}$) and $\hat{n}_2$ (associated with $\rho_{\text{min}}$). We are given that the axis of $\rho_{\text{max}}$ makes an angle $\theta$ with the edge AB ($x$-axis). Resolving $\vec{J}$ along these principal axes: $$ \vec{J} = J \cos\theta \, \hat{n}_1 + J \sin\theta \, \hat{n}_2 $$ Note: In the frame of principal axes, Ohm’s law is scalar for each component.

3. Determining the Electric Field

The electric field components along the principal axes are: $$ \vec{E} = (\rho_{\text{max}} J_{\hat{n}_1}) \hat{n}_1 + (\rho_{\text{min}} J_{\hat{n}_2}) \hat{n}_2 $$ Substituting the current components: $$ \vec{E} = J (\rho_{\text{max}} \cos\theta \, \hat{n}_1 + \rho_{\text{min}} \sin\theta \, \hat{n}_2) $$

4. Calculating the Transverse Field Component ($E_y$)

We need the potential difference between P (on top edge) and Q (on bottom edge). This is determined by the $y$-component of $\vec{E}$. We project the unit vectors $\hat{n}_1$ and $\hat{n}_2$ back onto the $y$-axis:

• The projection of $\hat{n}_1$ on $\hat{j}$ is $\sin\theta$.
• The projection of $\hat{n}_2$ on $\hat{j}$ is $-\cos\theta$ (since $\hat{n}_2$ is perpendicular to $\hat{n}_1$).

Thus: $$ E_y = J [ \rho_{\text{max}} \cos\theta (\sin\theta) + \rho_{\text{min}} \sin\theta (-\cos\theta) ] $$ $$ E_y = J \sin\theta \cos\theta (\rho_{\text{max}} – \rho_{\text{min}}) $$ Similarly, the longitudinal field $E_x$ (which relates to the applied voltage $V$) is: $$ E_x = J [ \rho_{\text{max}} \cos^2\theta + \rho_{\text{min}} \sin^2\theta ] $$

5. Relating $E_y$ to Applied Voltage $V$

The applied voltage is $V = E_x \cdot l$. We can eliminate $J$ by taking the ratio $E_y / E_x$: $$ \frac{E_y}{E_x} = \frac{(\rho_{\text{max}} – \rho_{\text{min}}) \sin\theta \cos\theta}{\rho_{\text{max}} \cos^2\theta + \rho_{\text{min}} \sin^2\theta} $$ The transverse potential difference $V_{\text{PQ}} = E_y \cdot b$. Substituting $E_y = E_x (\text{ratio}) = \frac{V}{l} (\text{ratio})$: $$ V_{\text{PQ}} = V \frac{b}{l} \left[ \frac{(\rho_{\text{max}} – \rho_{\text{min}}) \sin\theta \cos\theta}{\rho_{\text{max}} \cos^2\theta + \rho_{\text{min}} \sin^2\theta} \right] $$

6. Numerical Substitution

Given values:

• $\sin\theta = 3/5 = 0.6 \implies \cos\theta = 0.8$
• $\rho_{\text{min}} = 0.4 \rho_{\text{max}}$
• $l = 9.0$ cm, $b = 1.0$ cm
• $V = 196$ V

Calculate the terms involving resistivity: $$ \text{Numerator term} = (\rho_{\text{max}} – 0.4\rho_{\text{max}}) (0.6)(0.8) = 0.6\rho_{\text{max}} (0.48) = 0.288 \rho_{\text{max}} $$ $$ \text{Denominator term} = \rho_{\text{max}}(0.8)^2 + 0.4\rho_{\text{max}}(0.6)^2 = \rho_{\text{max}}(0.64 + 0.144) = 0.784 \rho_{\text{max}} $$ Now solve for $V_{\text{PQ}}$: $$ V_{\text{PQ}} = 196 \cdot \frac{1}{9} \cdot \frac{0.288}{0.784} $$ Simplify the fraction $\frac{0.288}{0.784} = \frac{288}{784} = \frac{144}{392} = \frac{72}{196}$. $$ V_{\text{PQ}} = 196 \cdot \frac{1}{9} \cdot \frac{72}{196} $$ $$ V_{\text{PQ}} = \frac{72}{9} = 8.0 \, \text{V} $$

Answer

The potential difference between the midpoints P and Q is 8.0 V.