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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$

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

Answer : (c) $\;\large\frac{4a}{3}$
Explanation :
$V_{P}=\large\frac{\sigma}{2\in_{0}}\;[sqrt{a^2+H^2}-H]$
$V_{0}=\large\frac{\sigma\;a}{2 \in_{0}} \quad \; (H=0)$
Particle is released from P and it just reaches point O . Therefore , from conservation of energy . As there is no external force
$\bigtriangleup T.E=0$
$q(V_{0}-V_{P})-mgH=0$
$gH=(\large\frac{q}{m})\;\large\frac{\sigma}{2 \in_{0}}\;[a-\sqrt{a^2+H^2}+H]$
$\large\frac{q}{m}=\large\frac{4 \in_{0} g}{\sigma}$
Thus we get $\;gH=2g\;[a+H-\sqrt{a^2+H^2}]$
$\large\frac{H}{2}=a+H-\sqrt{a^2+H^2}$
$\sqrt{a^2+H^2}=a+\large\frac{H}{2} \quad\;or \quad \; a^2+H^2=a^2+\large\frac{H^2}{4}+aH$
$\large\frac{3}{4}H^2=aH \quad or \quad H=\large\frac{4a}{3}$
edited Feb 20, 2014 by yamini.v