Chapter 23 Problem 40

Problem

The charge on the rod of figure aboce has a nonuniform linear charge distribution $\lambda =ax$. Determine the potential $V$ for

(a) points along the y axis and

(b) points along the x axis outside the rod

Solution

Choose a differential element of length $dx^{\prime }$ at position $x^{\prime }$. The charge on the element is $dq=\lambda dx^{\prime }=ax^{\prime }dx^{\prime }$

(a)

$r={\sqrt {x^{\prime 2}+y^{2}}}$

$V={\frac {1}{4\pi \epsilon _{0}}}\int {\frac {dq}{r}}$ $={\frac {1}{4\pi \epsilon _{0}}}\int _{-l}^{l}{\frac {ax^{\prime }dx^{\prime }}{\sqrt {x^{\prime 2}+y^{2}}}}$ $=0$

(since the region of integration is symmetric about origin (i.e. -l and +l) , and the integrand is odd (i.e. f(-x)=-f(x)))

(b)

$V={\frac {1}{4\pi \epsilon _{0}}}\int {\frac {dq}{r}}$ $={\frac {1}{4\pi \epsilon _{0}}}\int _{-l}^{l}{\frac {ax^{\prime }dx^{\prime }}{x-x^{\prime }}}$ $={\frac {a}{4\pi \epsilon _{0}}}\int _{-l}^{l}{\frac {x^{\prime }dx^{\prime }}{x-x^{\prime }}}$

$z=x-x^{\prime }$ $;dz=-dx^{\prime }$

$={\frac {a}{4\pi \epsilon _{0}}}\int _{-l}^{l}{\frac {x^{\prime }dx^{\prime }}{x-x^{\prime }}}$ $={\frac {a}{4\pi \epsilon _{0}}}\int _{x+l}^{x-l}{\frac {(x-z)(-dz)}{z}}$ $={\frac {a}{4\pi \epsilon _{0}}}\int _{x+l}^{x-l}\left({\frac {x}{z}}-1\right)dz$ $={\frac {a}{4\pi \epsilon _{0}}}\left(x\ln z-z\right)_{x+l}^{x-l}$

$={\frac {a}{4\pi \epsilon _{0}}}\left(x\ln \left({\frac {x+l}{x-l}}\right)-2l\right)$

when $x>l$