Q. An infinite number of electric charges each equal to $5\ nano - coulomb$ (magnitude) are placed along $X$-axis at $x = 1\ cm,\ \ x = 2\ cm,\ \ x = 4\ cm\ \ x = 8\ cm\ldots\ldots\ldots$ and so on. In the setup if the consecutive charges have opposite sign, then the electric field in $Newton/Coulomb$ at $X = 0$ is $\left\lbrack \frac{1}{4\pi\varepsilon_{0}} = 9 \times 10^{9}N - m^{2}/c^{2} \right\rbrack$
Explanation
(c) $E = \frac{1}{4\pi\varepsilon_{0}}.\left\lbrack \frac{5 \times 10^{- 9}}{\left( 1 \times 10^{- 2} \right)^{2}} - \frac{5 \times 10^{- 9}}{\left( 2 \times 10^{- 2} \right)^{2}} + \frac{5 \times 10^{- 9}}{\left( 4 \times 10^{- 2} \right)^{2}} \right.\ \left. \ - \frac{\left( 5 \times 10^{- 9} \right)}{\left( 8 \times 10^{- 2} \right)^{2}} + \ldots\cdots \right\rbrack$
$\Rightarrow E = \frac{9 \times 10^{9} \times 5 \times 10^{- 9}}{10^{- 4}}\left\lbrack 1 - \frac{1}{(2)^{2}} + \frac{1}{(4)^{2}} - \frac{1}{(8)^{2}} + \cdots \right\rbrack$
$\Rightarrow E = 45 \times 10^{4}\left\lbrack 1 + \frac{1}{(4)^{2}} + \frac{1}{(16)^{2}} + \cdots \right\rbrack$
$- 45 \times 10^{4}\left\lbrack \frac{1}{(2)^{2}} + \frac{1}{(8)^{2}} + \frac{1}{(32)^{2}} + \cdots \right\rbrack$
$\Rightarrow E = 45 \times 10^{4}\left\lbrack \frac{1}{1 - \frac{1}{16}} \right\rbrack - \frac{45 \times 10^{4}}{(2)^{2}}\left\lbrack 1 + \frac{1}{4^{2}} + \frac{1}{(16)^{2}} + .. \right\rbrack$
$E = 48 \times 10^{4} - 12 \times 10^{4} = 36 \times 10^{4}N/C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.