Question

A rod of length $L$ and cross-section area $A$ lies along the x-axis between $x=0$ and $x=L$.The material obeys Ohms law and its resistivity varies along the rod according to $\rho \left(x\right)={\rho }_0{e}^{-x/L}$. The end of the rod at $x=0$ is at a potential $V_0$ and it is zero at $x=L$.

Find the electric potential in the rod as a function of $x$.

Answer 1

At distance x from $x=0$, let us consider shape of area $A$ and length $dx$.

Variation of voltage at $x=x$ & $x=x+dx$ be $V$ & $V+dV$ respectively.

$\therefore$ By Ohm's Law,

$V=\left(\frac{\rho L}{A}\right)I$

$\therefore dV=\frac{\rho \left(x\right)dx}{A}\times I$

Integrating over entire length

$\therefore {\int }_{V_0}^{0}dV={\int }_0^L\frac{{\rho }_0{e}^{-x/L}dx}{A}\times I$

We get $-V_0={\left[\frac{{\rho }_0{e}^{-x/L}}{A(-1/L)}\right]}_0^L$

$\therefore I=\frac{A{V}_0}{{\rho }_{0}L}(e^{-1}-1)$

$\therefore$ For obtaining $V\left(x\right)$

Integrating from $0$ to $x$

${\int }_{V_0}^{V\left(x\right)}dx={\int }_0^x\frac{{\rho }_0{e}^{-\frac{x}{L}}dx}{A}I$

$\therefore V\left(x\right)-V_0=\frac{{\rho }_0}{A}\times \frac{A{V}_0}{{\rho }_{0}L}\frac{(e^{-1}-1)(e^{-\frac{x}{L}}-1)}{-1/L}$

$\therefore V\left(x\right)-V_0=-V_0(e^{-1}-1)(e^{-x/L}-1)$

$\therefore V\left(x\right)=V_0-(V_0{e}^{-1}-V_0)(e^{-x/L}-1)$

$\therefore V\left(x\right)=V_0-[V_0{e}^{(-x/L-1)}-V_0{e}^{-1}-V_0{e}^{-x/L}+V_0]$

$\therefore V\left(x\right)=V_0{e}^{-1}+V_0{e}^{-x/L}-V_0{e}^{-x/L-1}$

$\therefore V\left(x\right)=V_0[e^{-1}+e^{-x/L}-e^{(-x/L-1)}]$