Explanation

(b) From Coulomb’s law, the force between two point charges $q_{1},$ $q_{2}$ separated by distance $r$ is
$F = \frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}q_{2}}{r^{2}}\ $
The schematic diagram of distribution of charges on $x -$axis is shown.
[image:1_image248.emf]
The total force acting on 1 C charge is given by
$F = \Sigma\frac{1}{4\pi\varepsilon_{0}}.\frac{q_{1}q_{2}}{r^{2}}$
$= \frac{1}{4\pi\varepsilon_{0}} \times \left( \frac{1 \times 1 \times 10^{- 6}}{(1)^{2}} + \frac{1 \times 1 \times 10^{- 6}}{(2)^{2}} + \frac{1 \times 1 \times 10^{- 6}}{(4)^{2}} + \ldots\infty \right)$
=$9 \times 10^{9} \times 10^{- 6}\left( \frac{1}{1 - \frac{1}{4}} \right)$
Using the fact that sum of GP series is
$S_{\infty} = \frac{a}{1 - r}$ , where $a$ is first term of series and $r$ the common ratio.
$\therefore\ F = 9 \times 10^{3} \times \frac{4}{3} = 12000N$