Question

A thin dielectric rod of length $l$ lies along the $\text{x-axis}$ with one end at the origin and the other end at the point $(l,0)$. It is charged uniformly along its length with a total charge $Q$. The potential at a point $(x,0)$ when $x>l$ is:

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

Answer 1

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

Let us consider a small element on rod of infinitesimal length $dr$ at a distance $r$ from the point $\text{P}$ as shown in figure.


Charge on the small element
$dq=\frac{Q}{l}dr$

Potential at $\text{P}$ due to infinitesimal element,

$dV=\frac{1}{4\pi {ϵ}_0}\frac{dq}{r}$

Substituting the value of $dq$ we can write that,

$dV=\frac{1}{4\pi {ϵ}_0}\frac{Q}{l}\frac{dr}{r}$

Total potential $\left(V\right)$ at point $\text{P}$ due rod can be calculated by,
$V=\underset{x}{\overset{x-l}{\int }}\frac{1}{4\pi {ϵ}_0}\frac{Q}{l}\left(\frac{dr}{r}\right)$

$\Rightarrow V=\frac{Q}{4\pi {ϵ}_{0}l}{\text{log}}_{e}r{|}_x^{x-l}$

$\Rightarrow V=\frac{Q}{4\pi {ϵ}_{0}l}\left[{\text{log}}_e\right(x-l)-{\text{log}}_{e}x]$

$\therefore V=\frac{Q}{4\pi {ϵ}_{0}l}{\text{log}}_e\left(\frac{x-l}{x}\right)$

Hence, option (d) is the correct answer.