Question

Two points are at a distance $a$ and $b$ apart $(a<b)$ from left end of a uniformly charged rod as shown in the figure. The difference between the potentials at the two given points is proportional to

• A. $ln\left(\frac{b}{a}\right)$
• B. $\frac{b}{a}$
• C. $\sqrt{\frac{b}{a}}$
• D. $$ln\left\{\frac{b(a-L)}{a(b-L)}\right\}$$

Answer 1

The correct option is D $$ln\left\{\frac{b(a-L)}{a(b-L)}\right\}$$

Given that,
Length of the rod $=L$
Let the total charge on the rod is $Q$.
The electric potential ($V$) at a distance $r$ from the end of a uniformly charged rod is given by


$V_p=\frac{kQ}{L}ln\left(\frac{r+L}{r}\right)$
Here, we have


The electric potential at distance $a$ from left end of rod is given by
$V_a=\frac{kQ}{L}ln\left(\frac{a}{a-L}\right)$
The electric potential at distance $b$ from left end of rod is given by
$V_b=\frac{kQ}{L}ln\left(\frac{b}{b-L}\right)$

Then the potential difference is,
$\Delta V=V_b-V_a$

$\Rightarrow \Delta V=\frac{kQ}{L}\{ln\left(\frac{b}{b-L}\right)-ln\left(\frac{a}{a-L}\right)\}$
$\Rightarrow \Delta V=\frac{kQ}{L}\left\{ln\left(\frac{b(a-L)}{a(b-L)}\right)\right\}$
$\Rightarrow V\propto ln\left(\frac{b(a-L)}{a(b-L)}\right)$

Hence, option (d) is correct.

$\begin{array}{c}\text{Why this question ?}\\ \text{Tip:}lnx-lny=ln\frac{x}{y}\end{array}$