Question

A rod of length $l$ is placed along x-axis. One of its ends is at the origin. The rod has a non-uniform charge density $\lambda =\frac{a}{x}$ , a being a positive constant. The electric potential at the point $P$ (origin) as shown in the figure is

• A. $V=\frac{a}{4\pi {\epsilon }_0}\left(\frac{l}{b(b+l)}\right)$
• B. $V=\frac{a}{4\pi {\epsilon }_0}\left(\frac{b}{l(b+l)}\right)$
• C. $V=\frac{a}{4\pi {\epsilon }_0}\left(\frac{b}{l}\right)$
• D. $V=\frac{a}{4\pi {\epsilon }_0}\left(\frac{l}{b}\right)$

Answer 1

Answer: (A)

$V={\int }_b^{b+l}\frac{kdq}{x}={\int }_b^{b+l}\frac{k\lambda dx}{x}=ka{\int }_b^{b+l}\frac{dx}{x^2}$
$V=ka\left(\frac{1}{b}-\frac{1}{b+l}\right)$
$V=\frac{a}{4\pi {\epsilon }_0}\left(\frac{l}{b(b+l)}\right)$