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A solid sphere of radius $R$ carries a charge $(Q+q)$ distributed uniformly over its volume. A very small point like piece of it of mass $m$ gets detached from the bottom of the sphere and falls down vertically under gravity. This piece carries charge $q.$ If it acquires a speed $v$ when it has fallen through a vertical height $y$ (see figure), then :
(assume the remaining portion to be spherical).
$v^{2}=2 y\left[\frac{q Q}{4 \pi \epsilon_{0} R(R+y) m}+g\right]$
$v^{2}=y\left[\frac{q Q}{4 \pi \epsilon_{0} R^{2} y m}+g\right]$
$v^{2}=2 y\left[\frac{q Q R}{4 \pi \epsilon_{0}(R+y)^{3} m}+g\right]$
$v^{2}=y\left[\frac{q Q}{4 \pi \epsilon_{0} R(R+y) m}+g\right]$
Solution
$\frac{ kQq }{ R }+ mgy$
$=\frac{ kQq }{ R + y }+\frac{1}{2} mv ^{2}$
$v ^{2}=2 gy +\frac{2 kQqy }{ mR ( R + y )}$
