TS EAMCET 03 May 2019 Shift 2 Solved Paper

An infinite line of uniform charge density is placed along $y$ -axis and a charge $(+1\,\mathrm{C})$ is placed at $x=+3\,\mathrm{m}$, as shown in the figure. For electric field to be zero at $x$, there should be $E_{\;\text{line charge }\;} = E_{\;\text{point charge }\;} \;\; ......(1)$ As we know, for line charge $E_{\;\text{line charge }\;} = \;\frac{2 k \lambda}{x} = \;\frac{2 k}{x}$ Similarly, for point charge, $E_{\;\text{point charge }\;} = \;\frac{k q}{(3-x)^{2}} = \;\frac{k}{(3-x)^{2}} \;\; ...$ From Eqs. (i), (ii) and (iii), we get $\;\frac{2 k}{x} = \;\frac{k}{(3-x)^{2}} \Rightarrow \;\frac{2}{x} = \;\frac{1}{(3-x)^{2}}$ $\Rightarrow \;\; 2 x^{2} - 13 x + 18 = 0$ This is a quadratic equation, which have solution $x_{1}, x_{2} = \;\frac{13 \pm \sqrt{13^{2} - 144}}{4}$ So, $x_{1} = 4.5\,\mathrm{m}$ (not possible, as $0 < x < 3$ ) and $x_{2} = 2\,\mathrm{m}$. Hence, the correct answer is $2\,\mathrm{m}$, which matches option (b).