Question

A charge +q is fixed at each of the points $x=x_0,x=3x_0,x=5x_0\dots \dots$ infinity, on the x - axis and a charge -q is fixed at each of the points $x=2x_0,x=4x_0,x=6x_0\dots \dots$ infinity. Here $x_0$ is a positive constant. The potential at the origin due to the above system of charges is

• A. \N
• B. $\frac{q}{8\pi {ϵ}_0{x}_{0}ln2}$
• C. $\infty$
• D. $\frac{qln2}{4\pi {ϵ}_0{x}_0}$

Answer 1

The correct option is D $\frac{qln2}{4\pi {ϵ}_0{x}_0}$

$V=\frac{q}{4\pi {ϵ}_0{x}_0}[1+\frac{1}{3}+\frac{1}{5}+\dots ]-\frac{q}{4\pi {ϵ}_0{x}_0}[\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\dots ]$
$=\frac{q}{4\pi {ϵ}_0{x}_0}[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots ]=\frac{q}{4\pi {ϵ}_0{x}_0}lo{g}_{e}2$