A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field $100V\ m^{- 1}.$ If the mass of the drop is $1.6 imes 10^{- 3}g,$ the number of electrons carried by the drop is $(g = 10ms^{- 2})$

Question

A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field $100V\ m^{- 1}.$ If the mass of the drop is $1.6 	imes 10^{- 3}g,$ the number of electrons carried by the drop is $(g = 10ms^{- 2})$

Dr. Abubacker · Accepted Answer

(d) For the drop to be stationary,
Force on the drop due to electric field = weight of the drop $qE = mg$
$q = \frac{1.6 	imes 10^{- 6} 	imes 10}{100} = 1.6 	imes 10^{- 7}C$
Number of electrons carried by the drop is
$= \frac{q}{e} = \frac{1.6 	imes 10^{- 7}C}{1.6 	imes 10^{- 19}C} = 10^{12}$

Q. A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field $100V\ m^{- 1}.$ If the mass of the drop is $1.6 \times 10^{- 3}g,$ the number of electrons carried by the drop is $(g = 10ms^{- 2})$

Explanation

Reviewed by Dr. Abubacker

Want more practice?