Question

A charge is fixed at each of the points $x=x_0,x=3x_0,x=5x_0\dots \dots$ infinite, on the x-axis and a charge -q is fixed at each of the points $x=2x_0,x=4x_0,x=6x_0,\dots \dots$ infinite. Here $x_0$ is a positive constant. Take the electric potential at a point due to a charge Q at a distance r from it to be $\frac{Q}{\left(4\pi {ϵ}_{0}r\right)}$. Then, the potential at the origin due to the above system of charges is

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

Answer 1

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

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