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A non-conducting disc of radius a and uniform +ve surface charge density $\; \sigma\;$ is placed on the ground with it 's axis vertical . A particle of mass m and +ve charge q is dropped , along the axis of the disc from a height H with zero initial velocity . Find the value of H if the particle just reaches the disk .$\;\large\frac{q}{m}=4 \in_{0} g/\sigma$ . Find the position of particle at which the particle is in equilibrium

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

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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
$F_{net}=0$
$F=mg$
$qE=mg$
$\large\frac{q \sigma}{2 \in_{0}}\;[1-\large\frac{1}{\sqrt{\large\frac{a^2}{H^2}}+1}]$
$1-\large\frac{H}{\sqrt{a^2+H^2}}=\large\frac{m}{q}\;\large\frac{2 \in_{0} g}{\sigma}$
$1-\large\frac{H}{\sqrt{a^2+H^2}}=\large\frac{\sigma}{4 \in_{0} g}\;\large\frac{2 \in_{0} g}{\sigma}$
$\large\frac{1}{2}=\large\frac{H}{\sqrt{a^2+H^2}}$
$a^2+H^2=4H^2$
$H=\large\frac{a}{\sqrt{3}}$
answered Feb 18, 2014 by yamini.v
edited Feb 20, 2014 by yamini.v
 

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