Question

An infinite number of electric charges each equal to $5nC$ (magnitude) are placed along $x$ -axis at $x=1cm$, $x=2cm,x=4cm,x=8m\dots \dots \dots ..$ and so on. In the setup, if the consecutive charges have opposite sign, then the electric field in newton/coulomb at $x=0$ is $\left(\frac{1}{4\pi {\epsilon }_0}=9\times 10^{9}N-m^2/C^2\right)$

• A. $12\times 10^4$
• B. $24\times 10^4$
• C. $36\times 10^4$
• D. $48\times 10^4$

Answer 1

Answer: (C)

$E=\frac{1}{4\pi {\epsilon }_0}\cdot [\frac{5\times 10^{-9}}{{\left(1\times 10^{-2}\right)}^2}-\frac{5\times 10^{-9}}{{\left(2\times 10^{-2}\right)}^2}+\frac{5\times 10^{-9}}{{\left(4\times 10^{-2}\right)}^2}$
$-\frac{\left(5\times 10^{-9}\right)}{{\left(8\times 10^{-2}\right)}^2}+\dots ..]$
$\Rightarrow E=\frac{9\times 10^9\times 5\times 10^{-9}}{10^{-4}}\left[1-\frac{1}{(2{)}^2}+\frac{1}{(4{)}^2}-\frac{1}{(8{)}^2}+\dots \right]$
$\Rightarrow E=45\times 10^4\left[1+\frac{1}{(4{)}^2}+\frac{1}{(16{)}^2}+\dots \right]$
$-45\times 10^4\left[\frac{1}{(2{)}^2}+\frac{1}{(8{)}^2}+\frac{1}{(32{)}^2}+\dots \right]$
$\Rightarrow E=45\times 10^4\left[\frac{1}{1-\frac{1}{16}}\right]$
$-\frac{45\times 10^4}{(2{)}^2}\left[1+\frac{1}{4^2}+\frac{1}{(16{)}^2}+..\right]$
$E=48\times 10^4-12\times 10^4$
$=36\times 10^{4}N/C$