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

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 $\;gH=2g\;[a+H-\sqrt{a^2+H^2}]$

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

$\sqrt{a^2+H^2}=a+\large\frac{H}{2} \quad\;or \quad \; a^2+H^2=a^2+\large\frac{H^2}{4}+aH$

$\large\frac{3}{4}H^2=aH \quad or \quad H=\large\frac{4a}{3}$

Ask Question

Tag:MathPhyChemBioOther

Take Test

...