Question

A charge +q is fixed at each of the points $x=x_0,x=3x_0,x=5x_0.......$ up to infinity on x axis and a charge (-q) is fixed on each of points of $x=2x_0,x=4x_0,x=6x_0.......$ up to infinity, here $x_0$ is a positive constant. Take the potential at a point due to a charge Q at a distance r from it $\frac{Q}{4\pi {\epsilon }_{0}r}$, then the potential at the origin due to above system of charges will be :

• A. $0$
• B. $\infty$
• C. $\frac{qlo{g}_{e}2}{4\pi {\epsilon }_0{x}_0}$
• D. $\frac{q}{8\pi {\epsilon }_0{x}_{0}lo{g}_{e}2}$

Answer 1

The correct option is C $\frac{qlo{g}_{e}2}{4\pi {\epsilon }_0{x}_0}$

The potential at origin of the system due to all charges is

$V=\frac{q}{4\pi {ϵ}_0{x}_0}[1+\frac{1}{3}+\frac{1}{5}+...]-\frac{q}{4\pi {ϵ}_0{x}_0}[\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...]$

$=\frac{q}{4\pi {ϵ}_0{x}_0}[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...]$

$=\frac{q}{4\pi {ϵ}_0{x}_0}lo{g}_e(1+1)$ using $lo{g}_e(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...]$

$=\frac{q}{4\pi {ϵ}_0{x}_0}lo{g}_{e}2$