91Ó°ÊÓ

Resistance of a wire is given by

$\mathit{R}\mathbf{=}\mathit{P}\frac{\mathbf{L}}{\mathbf{A}}$

where p is the resistivity and Lis the length of the wire and A is the area of the wire

Current density is given by

$\mathit{J}\mathbf{=}\frac{\mathbf{l}}{\mathbf{A}}$

Resistivity is given by

$\mathit{p}\mathbf{=}\frac{\mathbf{E}}{\mathbf{J}}$

Resistivity is not uniform in the rod, we divide it into many thin slides with Area A and thickness dx

Resistance of this rod is given by

$dR=\frac{p\left(x\right)dx}{A}$

Resistance depends on the distance as $p=p_0{e}^{\frac{-x}{L}}$

Input this into the equation

$dR=\frac{p_{0}e\frac{-x}{L}dx}{A}$

To find total resistance, integrate

$R=∫dR=∫\frac{{\mathrm{ÒÏ}}_0{e}^{\frac{−x}{L}}dx}{A}$

Put limits 0 to L

$R=\frac{{\mathrm{ÒÏ}}_{0}L}{A}\left(1−e^{−1}\right)$

Apply Ohm’s law to get

$I=\frac{V_0}{R}=\frac{V_{0}A}{{\mathrm{ÒÏ}}_{0}L\left(1−e^{−1}\right)}$

We know that electric field is given by

$E=pJ$

Substitute the value for resistivity

$E=P_0{e}^{-\frac{x}{L}}J$

Substitute J

$$\begin{gathered} E={\mathrm{ÒÏ}}_0{e}^{−\frac{x}{L}}\frac{1}{A} \\ E={\mathrm{ÒÏ}}_0{e}^{−\frac{x}{L}}\frac{V_{0}A}{{\mathrm{ÒÏ}}_{0}L\left(1−e^{−1}\right)A} \\ E=\frac{V_0{e}^{−\frac{x}{L}}}{L\left(1−e^{−1}\right)} \end{gathered}$$

Resistivity is not uniform in the rod, we divide it into many thin slides with Area A and thickness dx

Resistance of this rod is given by

$dR=\frac{p\left(x\right)dx}{A}$

Resistance depends on the distance as $p=p_0{e}^{\frac{x}{L}}$

Input this into the equation

$dR=\frac{p_{0}e\frac{-x}{L}dx}{A}$

To find total resistance, integrate

$R=∫dR=∫\frac{{\mathrm{ÒÏ}}_0{e}^{\frac{−x}{L}}dx}{A}$

Put limits 0 to x

$R\left(x\right)=\frac{{\mathrm{ÒÏ}}_{0}L}{A}\left(1−e^{−\frac{x}{L}}\right)$

This is the resistance of a rod of length x

Potential drop is given by

$$\begin{gathered} V\left(x\right)=V_0−IR\left(x\right) \\ V\left(x\right)=V_0−\frac{V_{0}A}{{\mathrm{ÒÏ}}_{0}L\left(1−e^{−1}\right)}\frac{{\mathrm{ÒÏ}}_{0}L}{A}\left(1−e^{−\frac{x}{L}}\right) \\ V\left(x\right)=\frac{V_0\left(e^{\frac{−x}{L}}−e^{−1}\right)}{1−e^{−1}} \end{gathered}$$

Hence the voltage as a function of distance is $V\left(x\right)=\frac{V_0\left(e^{\frac{−x}{L}}−e^{−1}\right)}{1−e^{−1}}$