Question

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

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

Answer 1

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

Potential at origin due to all +q charges is

$V_{+}=\frac{q}{4\pi {ϵ}_0}[\frac{1}{x_0}+\frac{1}{3x_0}+\frac{1}{5x_0}+.....]=\frac{q}{4\pi {ϵ}_0{x}_0}[1+\frac{1}{3}+\frac{1}{5}+.....]$

Potential at origin due to all -q charges is

$V_{-}=\frac{-q}{4\pi {ϵ}_0}[\frac{1}{2x_0}+\frac{1}{4x_0}+\frac{1}{6x_0}+.....]=\frac{-q}{4\pi {ϵ}_0{x}_0}[\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....]$

Total potential at origin is $V=V_{+}+V_{-}$

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

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

$=\frac{q}{4\pi {ϵ}_0{x}_0}lo{g}_{e}2$ as $\left[lo{g}_e\right(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+....]$