Question

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

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

Answer 1

Answer: (C)

$E=\frac{1}{4 \pi \varepsilon_{0}} \cdot\left[\frac{5 \times 10^{-9}}{\left(1 \times 10^{-2}\right)^{2}}-\frac{5 \times 10^{-9}}{\left(2 \times 10^{-2}\right)^{2}}+\frac{5 \times 10^{-9}}{\left(4 \times 10^{-2}\right)^{2}}\right.$
$ \left.-\frac{\left(5 \times 10^{-9}\right)}{\left(8 \times 10^{-2}\right)^{2}}+\ldots . .\right] $
$\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}}+\ldots\right] $
$ \Rightarrow E=45 \times 10^{4}\left[1+\frac{1}{(4)^{2}}+\frac{1}{(16)^{2}}+\ldots\right]$
$-45 \times 10^{4}\left[\frac{1}{(2)^{2}}+\frac{1}{(8)^{2}}+\frac{1}{(32)^{2}}+\ldots\right] $
$\Rightarrow E=45 \times 10^{4}\left[\frac{1}{1-\frac{1}{16}}\right] $
$-\frac{45 \times 10^{4}}{(2)^{2}}\left[1+\frac{1}{4^{2}}+\frac{1}{(16)^{2}}+. .\right]$
$E=48 \times 10^{4}-12 \times 10^{4}$
$=36 \times 10^{4} N / C$