Question

An infinite no.of electric charges each equal to $5nC$ are placed along X-axis at $x=1cm,x=2cm,x=4cm,x=8cm$ and so on. In this setup, if the consecutive charges have opposite sign, then the electric field in newton/coulomb at $x=0$ is:
$(\frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}N{m}^2/c^2)$

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

Answer 1

Answer: (C)

$36\times {10}^4$

E due to all positive charges will be in $-\hat{x}$ direction and due to all negative charges will be in $+\hat{x}$ direction
E due to positive $=\frac{k(5\times {10}^{-9})}{(1\times {10}^{-2}{)}^2}+\frac{k(5\times {10}^{-9})}{(4\times {10}^{-2}{)}^2}+\frac{k(5\times {10}^{-9})}{(16\times {10}^{-2}{)}^2}+----$

$=k(5\times {10}^{-9})\frac{\frac{1}{(1\times {10}^{-2}{)}^2}}{1-\frac{1}{16}}=\frac{(9\times {10}^9)(5\times {10}^{-9})}{{10}^{-4}}\times \frac{16}{15}=48\times {10}^4(-\hat{x})$

E due to negative $=\frac{k(5\times {10}^{-9})}{(2\times {10}^{-2}{)}^2}+\frac{k(5\times {10}^{-9})}{(8\times {10}^{-2}{)}^2}+----$

$=\frac{k(5\times {10}^{-9})}{{10}^{-4}}\left(\frac{\frac{1}{4}}{1-\frac{1}{16}}\right)=12\times {10}^4\left(\hat{x}\right)$

$\therefore E_{net}=(48-12)\times {10}^4$

$\therefore E_{net}=36\times {10}^4(-\hat{x})N/C$