Question

An infinite number of charges each equal to $q,$ are placed along the $x-$axis of $x=1,x=2,x=4,x=8,x=16$ and so on. The electric field at origin $O$ due to this set of charges is $\left(k=\frac{1}{4\pi {\epsilon }_o}\right)$

• A. $kq$
• B. $\frac{4}{3}kq$
• C. $\frac{3}{4}kq$
• D. $\frac{4}{9}kq$

Answer 1

The correct option is B $\frac{4}{3}kq$

Electric field, $\overrightarrow{E}=\frac{Kq}{r^2}$

In this case, due to several charges present at different position, resultant electric field is given by,

$\overrightarrow{E}=\frac{Kq}{{\left(1\right)}^2}+\frac{Kq}{{\left(2\right)}^2}+\frac{Kq}{{\left(4\right)}^2}+\frac{Kq}{{\left(8\right)}^2}+......$

$=Kq[\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{8^2}+......]$

$=Kq\left[\frac{1}{{\left(2^n\right)}^2}\right]$ where $n=0,1,2,......$

and we know that

${\sum }_{n=0}^{\infty }\frac{1}{{\left(2^n\right)}^2}=\frac{4}{3}$

$\therefore \overrightarrow{E}=\frac{4}{3}kqN/c$