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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. Find the energy of the particle at $\;H=\large\frac{a}{\sqrt{3}}\;$ assuming the gravitational energy at $\;H=0\;$ to be zero

$(a)\;mga\qquad(b)\;\sqrt{3} mga\qquad(c)\;\large\frac{mga}{\sqrt{3}}\qquad(d)\;None\;of\;these$

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Answer : (b) $\;\sqrt{3} mga$
Explanation :
Electrostatic energy at H=0 is:
$E_{e}=\large\frac{\sigma q a }{2 \in_{0}}$
$E_{g}=0$
At $\;H=\large\frac{a}{\sqrt{3}}\;, \quad\; E_{e}=\large\frac{\sigma q}{2 \in_{0}}\;[\sqrt{a^2+H^2}-H]$
$E_{e}=\large\frac{\sigma q}{2 \in_{0}}\;[\sqrt{a^2+\large\frac{a^2}{3}}-\large\frac{a}{\sqrt{3}}]$
$E_{e}=\large\frac{\sigma q}{2 \in_{0}} \;[\large\frac{2a}{\sqrt{3}}-\large\frac{a}{\sqrt{3}}]$
$E_{e}=\large\frac{\sigma q}{2 \in_{0}}\;\large\frac{a}{\sqrt{3}}$
$E_{g}=\large\frac{mga}{\sqrt{3}}$
$T.E=E_{e}+E_{g}$
$T.E=\large\frac{mga}{\sqrt{3}}+\large\frac{\sigma q a}{2 \in_{0} \sqrt{3}}$
$T.E=\large\frac{mga}{\sqrt{3}}+\large\frac{2mga}{\sqrt{3}}$
$T.E=\sqrt{3} mg\;.$
answered Feb 19, 2014 by yamini.v
 

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