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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


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Answer : (b) $\;\large\frac{a}{\sqrt{3}}$
Explanation :
F. B .D of particle is
For the particle to be in equilibrium
$\large\frac{q \sigma}{2 \in_{0}}\;[1-\large\frac{1}{\sqrt{\large\frac{a^2}{H^2}}+1}]$
$1-\large\frac{H}{\sqrt{a^2+H^2}}=\large\frac{m}{q}\;\large\frac{2 \in_{0} g}{\sigma}$
$1-\large\frac{H}{\sqrt{a^2+H^2}}=\large\frac{\sigma}{4 \in_{0} g}\;\large\frac{2 \in_{0} g}{\sigma}$
answered Feb 18, 2014 by yamini.v
edited Feb 20, 2014 by yamini.v

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