An infinite number of electric charges each equal to 5 nano-coulomb (magnitude) are placed along $X$-axis at $x=1 \mathrm{~cm}, x=2 \mathrm{~cm}$, $x=4 \mathrm{~cm}, x=8 \mathrm{~cm} . . .$. and so on. In this set up if the consecutive charges have opposite sign, then the electric field in newton/coulomb at $x=0$ is : $\left(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~N}-\mathrm{m}^2 / \mathrm{C}^2\right)$

• A

  $12 \times 10^4$
• B

  $24 \times 10^4$
• Correct answer
  C

  $36 \times 10^4$
• D

  $48 \times 10^4$

Solution

Electric field intensity due to a point charge. $E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r^2}$
The consecutive charges are of opposite signs. $\therefore$ Net electric field at $x=0$, is $E=\frac{1}{4 \pi \varepsilon_0}\left[\frac{Q}{r_1^2}-\frac{Q}{r_2^2}+\frac{Q}{r_3^2}-\frac{Q}{r_4^2} \ldots . . \infty\right]$ $=\frac{Q}{4 \pi \varepsilon_0}\left[\frac{1}{r_1^2}-\frac{1}{r_2^2}+\frac{1}{r_3^2}-\frac{1}{r_4^2}+\ldots . \infty\right\rceil$ $=9 \times 10^9 \times 5 \times 10^{-9}$ $\left[\frac{1}{\left(1 \times 10^{-2}\right)^2}-\frac{1}{\left(2 \times 10^{-2}\right)^2}+\frac{1}{\left(4 \times 10^{-2}\right)^2}+\ldots \infty\right\rfloor$ $=\frac{45}{10^{-4}}\left[\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{8^2}+\ldots . \infty\right]$ $=45 \times 10^4\left[\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{8^2} \ldots \infty\right]$ $\left\lceil\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{8^2} \ldots \infty\right\rfloor=\frac{\frac{1}{1^2}}{1-\left(\frac{-1}{2}\right)^2}$ $\left[\right.$ Sum of infinite GP, $\left.\mathrm{S}_{\infty}=\frac{a}{1-r}\right]$ $=\frac{1}{1+\frac{1}{4}}=\frac{4}{5}$ $\therefore \quad E=45 \times 10^4 \times \frac{4}{5}=36 \times 10^4 \mathrm{~N} / \mathrm{C}$

Hence, the correct answer is option: 'C'.