Question

Charges of $q$ coulomb but of alternately opposite signs are placed along the x-axis at $x=1$, $x=2$, $x=4$, ... and so on. The electric potential at the point $x=0$ due to all these charges will be

• A. $\frac{q}{2\pi {ϵ}_0}$
• B. $\frac{q}{3\pi {ϵ}_0}$
• C. $\frac{2q}{3\pi {ϵ}_0}$
• D. $\frac{q}{6\pi {ϵ}_0}$

Answer 1

The correct option is D $\frac{q}{6\pi {ϵ}_0}$

Resultant potential at $x=0$ is given by,
$V=\frac{1}{4\pi {\epsilon }_o}[\frac{(+q)}{1}+\frac{(-q)}{2}+\frac{(+q)}{4}+...]$
This is a series in geometric progression. And we know that the sum of the infinite series is $S=\frac{a}{1-r}$
In the above equation, $a=1$ and $r=\frac{-1}{2}$

Therefore, $V=\frac{q}{4\pi {\epsilon }_o}\frac{1}{1-\left(\frac{-1}{2}\right)}=\frac{q}{6\pi {\epsilon }_o}$