Question

A charge $+q$ is fixed at each of the points $x=x_0$ , $x=3x_0$, $x=5x_0$. . . . . $\infty$ on the x - axis and a charge $-q$ is fixed at each of the points $x=2x_0,x=4x_0,x=6x_0$......$\infty$. 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 $\frac{Q}{4\pi {ϵ}_{0}r}$. Then the potential at the origin due to the above system of charges is:

• A. $0$
• B. $\frac{q}{8\pi {ϵ}_0{x}_0\ln 2}$
• C. $\infty$
• D. $\frac{q\ln 2}{4\pi {\epsilon }_0{x}_0}$

Answer 1

The correct option is D $\frac{q\ln 2}{4\pi {\epsilon }_0{x}_0}$

Given potential at a distance r is $=\frac{q}{4\pi {\epsilon }_{0}r.}$

Now total potential at origin due to +q charges are

$=(\frac{q}{4\pi {\epsilon }_0{x}_0}+\frac{q}{4\pi {\epsilon }_0\left(3x_0\right)}+\frac{q}{4\pi {\epsilon }_0\left(5x_0\right)}+-----------)$

$=\frac{q}{4\pi {\epsilon }_0{x}_0}(1+\frac{1}{3}+\frac{1}{5}+-------)$

Total potential due to -q charges at origin is

$=\frac{-q}{4\pi {\epsilon }_0{x}_0}(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+-----)$

Now total $P=\frac{q}{4\pi {\epsilon }_0{x}_0}(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-----)$
we know

$ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+------$

$\therefore ln\left(2\right)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+-----$

$\therefore \text{total potential}=\frac{qln2}{4\pi {\epsilon }_0{x}_0}$