A drop of water of mass $18\times {10}^{-6}kg$ falls away from the bottom of a charged conducting sphere ( $2.5\times {10}^{-6}$ C ) of  radius 20cm carrying with it charge of  ${10}^{-9}$ coulomb . What is the speed of the drop when it has fallen 30cm?

${\epsilon }_0=\left(\frac{1}{4\pi \times 9\times {10}^9}\right)g=9.8m/s^2$

From law of conservation of energy 

mechanical energy at A= Mechanical energy at B

Take point B as reference point (zero gravitational potential energy),

$$\begin{gathered} \therefore mg\left(0.3\right)+\left(\frac{1}{4\pi {\epsilon }_0}\right)\frac{q_1{q}_2}{\left(0.2\right)}=0+\frac{1}{2}m{v}_{ }^2+\frac{q_1{q}_2}{4\pi {\epsilon }_0}\left(\frac{1}{0.2+0.3}\right) \\ Whrem=18\times {10}^{-6}kg \\ q_1=2.5\times {10}^{-6}c \\ q_2={10}^{-9}c \\ bysolvingitv=3.65m/s \end{gathered}$$