Question

A thin dielectric rod of length $l$ lies along X-axis. One end of the rod is placed at the origin and the other end of the rod is placed at $(l,0)$. A total change $Q$ is distributed uniformly along the length of the rod. What is the potential at a point $(x,0)$ when $x>l$?

• A. $\frac{Q}{4\pi {ϵ}_0}\cdot {\log }_e\left(\frac{x}{x-l}\right)$
• B. $\frac{Q}{4\pi {ϵ}_{0}l}\cdot {\log }_e\left(\frac{x}{l}\right)$
• C. $\frac{Q}{4\pi {ϵ}_{0}l}\cdot {\log }_e\left(\frac{x-l}{x}\right)$
• D. $\frac{Q}{4\pi {ϵ}_{0}l}$

Answer 1

The correct option is C $\frac{Q}{4\pi {ϵ}_{0}l}\cdot {\log }_e\left(\frac{x-l}{x}\right)$

The potential due to a small element of length $dr$ having charge $dQ$ of the dielectric rod is given by
$dV=\frac{dQ}{4\pi {ϵ}_{0}r}$

$dV=\frac{\lambda dr}{4\pi {ϵ}_{0}r}$ [Since $\lambda =\frac{dQ}{dr}]$
or $V=\frac{\lambda }{4\pi {ϵ}_0}{\int }_x^{x-l}\frac{dr}{r}$
$=\frac{\lambda }{4\pi {ϵ}_0}[{\log }_{e}r{]}_x^{x-1}$
$=\frac{\lambda }{4\pi {ϵ}_0}\cdot \log \left(\frac{x-1}{x}\right)$
$V=\frac{Q}{4\pi {ϵ}_{0}l}\cdot {\log }_e\left(\frac{x-1}{x}\right)$
When $\lambda =\frac{Q}{t}$ (for whole length of rod).