A drop of water of mass $$18$$×$$10$$−$$6kg$$ falls away from the bottom of a charged conducting sphere $$($$ $$2.5$$×$$10$$−$$6$$ C $$)$$ of radius $$20cm$$ carrying with it charge of $$10$$−$$9$$ coulomb . What is the speed of the drop when it has fallen $$30cm$$?ε$$0=14$$π×$$9$$×$$109$$ $$g=9.8m/s2$$

Question

A drop of water of mass $$18$$×$$10$$−$$6kg$$    falls away from the bottom of a charged conducting sphere $$($$ $$2.5$$×$$10$$−$$6$$ C $$)$$ of  radius $$20cm$$ carrying with it charge of  $$10$$−$$9$$ coulomb . What is the speed of the drop when it has fallen $$30cm$$?ε$$0=14$$π×$$9$$×$$109$$                  $$g=9.8m/s2$$

Infinity Learn · Accepted Answer

From law of conservation of energy

mechanical energy at A= Mechanical energy at B

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

$∴ m g (0.3) + (\frac{1}{4 π ε_{0}}) \frac{q_{1} q_{2}}{(0.2)} = 0 + \frac{1}{2} m v_{}^{2} + \frac{q_{1} q_{2}}{4 π ε_{0}} (\frac{1}{0.2 + 0.3}) W h r e m = 18 \times 10^{- 6} k g q_{1} = 2.5 \times 10^{- 6} c q_{2} = 10^{- 9} c b y s o l v i n g i t v = 3.65 m / s$

Electrostatic potential energy

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Electrostatic potential energy

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A drop of water of mass 18×10−6kg falls away from the bottom of a charged conducting sphere ( 2.5×10−6 C ) of radius 20cm carrying with it charge of 10−9 coulomb . What is the speed of the drop when it has fallen 30cm?ε0=14π×9×109 g=9.8m/s2

From law of conservation of energy mechanical energy at A= Mechanical energy at BTake point B as reference point (zero gravitational potential energy),∴mg0.3+14πε0q1q20.2=0+12mv 2+ q1q24πε010.2+0.3 Whre m=18×10−6kg q1=2.5×10−6c q2=10−9c by solving it v=3.65 m/s

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