An oil drop having charge #q# and apparent weight #W# falls with a terminal speed #v# in the absence of electric field. When electric field is switched on it starts moving up with terminal speed #v#. The strength of the electric field is ??

1 Answer
A08
Nov 2, 2017

The drag force acting upon the oil drop is calculated from Stoke's law and is given as

#F_d=6pietarv# ......(1)

The apparent weight for a perfectly spherical body is given by
True weight #-# Upthrust (Weight of fluid displaced by the body)

#F_barg=W# .....(2)

At terminal velocity the oil drop is not accelerating as the net force acting on it must be zero

#F_d-F_barg=0#

Hence equating (1) and (2) we get

#W=6pietarv_(Tdarr)# ......(3)
where #r#-radius of oil drop, #η#-viscosity of air, #v_(Tdarr)#-terminal velocity.

When electric field #E# is switched on the oil drop starts moving up with terminal speed #v_(Tuarr)#.

Electric force will act upwards, net gravity and drag forces acts downwards. Net force acting on the oil drop is again zero.

#F_e-F'_d-F_barg=0#
#=>F_e=F'_d+F_barg=0#

where #F'_d# is the new drag force under the action of electric field.

Inserting various values above expression becomes

#qE=6pietarv_(Tuarr)+W#

Given #v_(Tuarr)and v_(Tdarr)=v#

#:.qE=W+W#
#=>E=(2W)/q# ......(4)

.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

Apparent weight in terms of other variables.

#F_barg=4/3pir^3g(rho-rho_(air))=W# .....(2)

Hence equating (1) and (2) we get

#r^2=(9etav_(Tdarr))/(2g(rho-rho_(air))# ......(3)
where #r#-radius of oil drop, #η#-viscosity of air, #v_(Tdarr)#-terminal velocity, #g#- acceleration due to gravity, #ρ#-density of oil and #ρ_(air)#-density of air

Related questions
Impact of this question
3300 views around the world
You can reuse this answer
Creative Commons License