Question

An infinite number of identical charges, each $+q\text{coulomb}$, are placed along $\text{x-}$axis at $x=1\text{m},3\text{m},9\text{m},...$ Calculate the electric field at the point $x=0$ due to these charges.

• A. $\frac{1}{4\pi {\epsilon }_0}\frac{9q}{8}{\text{NC}}^{-1}$
• B. $\frac{1}{4\pi {\epsilon }_0}\frac{9q}{10}{\text{NC}}^{-1}$
• C. $\frac{1}{4\pi {\epsilon }_0}\frac{9q}{7}{\text{NC}}^{-1}$
• D. $\frac{1}{4\pi {\epsilon }_0}\frac{9q}{5}{\text{NC}}^{-1}$

Answer 1

The correct option is A $\frac{1}{4\pi {\epsilon }_0}\frac{9q}{8}{\text{NC}}^{-1}$

Electric field at origin will be resultant of all electric fields due to individual charges.

$E_0=E_1+E_2+E_3+...$

Substituting the values of the electric field due to each charge, we get

$E_0=\frac{kq}{1^2}+\frac{kq}{3^2}+\frac{kq}{9^2}+...$

$\Rightarrow E_0=kq[\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{9^2}+.....]$

$\Rightarrow E_0=\frac{q}{4\pi {\epsilon }_0}[1+\frac{1}{9}+\frac{1}{81}+...]...\left(1\right)$

As we know that sum of infinite terms of Geometric progression,

$S_{\infty }=\frac{a}{(1-r)}$

Where $a=1$ is the first term and

$r$ is the ratio of two consecutive terms, $r=\frac{b}{a}=\frac{\frac{1}{9}}{1}=\frac{1}{9}$

Now, substituting the values in $\left(1\right)$,

$E_0=\frac{q}{4\pi {\epsilon }_0}\left[\frac{a}{(1-r)}\right]$

$\Rightarrow E_0=\frac{q}{4\pi {\epsilon }_0}\left[\frac{1}{(1-\frac{1}{9})}\right]$

$\Rightarrow E_0=\frac{9}{8}\left(\frac{q}{4\pi {\epsilon }_0}\right)$

Hence, option (a) is correct answer.