A charge of 2C is placed on the x-axis at 1m from the origin along –ve x-axis. Infinite number of charges each of magnitude 2C are placed on x-axis at 1m, 2m, 4m, ..... from origin along +ve x-axis. The first charge is positive and alternate charges are of opposite in nature. The electric field intensity at the origin

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$\frac{1}{{πϵ}_{o}} along + ve x - axis$

$\frac{1}{{πϵ}_{0}} along - ve x - axis$

$\frac{1}{10 {πϵ}_{0}} along - ve x - axis$

$\frac{1}{10 {πϵ}_{0}} along + veX - axis$

answer is A.

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Detailed Solution

$E at x = 0 is E_{(- 1)} + E_{total}$
$\begin{array}{l} = \frac{1}{4 {πε}_{0}} [\frac{2}{(1)^{2}} - \frac{2}{(1)^{2}} + \frac{2}{2^{2}} - \frac{2}{4^{2}} + \frac{2}{8^{2}}] \\ = \frac{1}{4 {πε}_{0}} [\frac{2}{4} - \frac{2}{16} + \frac{2}{64} - \frac{2}{256} + \dots .] \\ = \frac{2}{4 {πε}_{0}} [(\frac{1}{4} + \frac{1}{64}) - (\frac{1}{16} + \frac{1}{256})] \\ = \frac{2}{4 {πε}_{0}} [\frac{\frac{1}{4}}{1 - \frac{1}{16}} - \frac{\frac{1}{16}}{1 - 1 / 6}) \\ = \frac{2}{4 {πε}_{0}} (\frac{16}{4 \times 15} - \frac{16}{16 \times 15}) \\ = \frac{2}{4 {πε}_{0}} \frac{3}{15} = \frac{6}{60 {πε}_{0}} = \frac{1}{10 {πε}_{0}} \end{array}$
Along +ve x–axis