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The cross-section area and length of cylindrical conductor are A and $l$ respectively. The specific conductivity varies as $σ (x) = σ_{0} \frac{l}{\sqrt{x}}$ where x is the distance along the axis of the cylinder from one of its ends. Compute the current density if the potential drop along the cylinder is V₀. Also find the electric field at each point in the cylinder in the case described?

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$\frac{3}{2} \frac{σ_{0} V_{0}}{l}, \frac{3}{2} \frac{V_{0} x}{l^{3 / 2}}$

$\frac{σ_{0} V_{0}}{\sqrt{x}}, \frac{V_{0} \sqrt{x}}{2 l^{3 / 2}}$

$\frac{2}{3} \frac{σ_{0} V_{0}}{\sqrt{x}}, \frac{2}{3} \frac{V_{0} \sqrt{x}}{l^{3 / 2}}$

$\frac{3 σ_{0} V_{0}}{2 \sqrt{l}}, \frac{3 V_{0} \sqrt{x}}{2 l^{3 / 2}}$

answer is D.

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Detailed Solution

Resistance of thin elementary disc is

$dR = \frac{ρ d x}{A} = \frac{1}{σ} \frac{d x}{A} = \frac{\sqrt{x}}{σ_{0} l} \frac{d x}{A}$

$R_{total} = \frac{1}{σ_{0} I A} \int_{0}^{l} x^{1 / 2} dx = \frac{1}{σ_{0} I A} \frac{l^{3 / 2}}{3 / 2} = \frac{2}{3} \frac{\sqrt{l}}{A σ_{0}} .$

$J = \frac{I}{A} = \frac{V_{0}}{R} \cdot \frac{1}{A} = \frac{V_{0}}{A} \times (\frac{3 A σ_{0}}{2 \sqrt{l}}) = \frac{3 σ_{0} V_{0}}{2 \sqrt{l}} .$

Again

$J = E σ \Rightarrow E = \frac{J}{σ} = \frac{3 σ_{0} V_{0}}{2 \sqrt{1}} \times \frac{\sqrt{x}}{σ_{0} l} = \frac{3}{2} \frac{V_{0} \sqrt{x}}{l^{3 / 2}} .$

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