$(a)\;\large\frac{2a}{\sqrt{3}}\qquad(b)\;\large\frac{a}{\sqrt{3}}\qquad(c)\;\large\frac{4a}{\sqrt{3}}\qquad(d)\;\large\frac{a}{3}$

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Answer : (b) $\;\large\frac{a}{\sqrt{3}}$

Explanation :

F. B .D of particle is

$F_{net}=F-mg$

For the particle to be in equilibrium

$F_{net}=0$

$F=mg$

$qE=mg$

$\large\frac{q \sigma}{2 \in_{0}}\;[1-\large\frac{1}{\sqrt{\large\frac{a^2}{H^2}}+1}]$

$1-\large\frac{H}{\sqrt{a^2+H^2}}=\large\frac{m}{q}\;\large\frac{2 \in_{0} g}{\sigma}$

$1-\large\frac{H}{\sqrt{a^2+H^2}}=\large\frac{\sigma}{4 \in_{0} g}\;\large\frac{2 \in_{0} g}{\sigma}$

$\large\frac{1}{2}=\large\frac{H}{\sqrt{a^2+H^2}}$

$a^2+H^2=4H^2$

$H=\large\frac{a}{\sqrt{3}}$

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