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Forum Index -> Modern Physics like the article? email it to a friend.  
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[Post New]31 Mar 2007 14:34:34 IST
    Subject: helppppppp
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shipra22_bhardwaj (304)

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n oil drop carrying a charge q has a mass m .it is  falling freely in air wid terminal speed v de electric feild required to make drop move upward wid de same speed is
1. mg/q
2.2mg/q
3.mgv/q2
4.2mgv/q
    
[Post New]1 Apr 2007 00:20:56 IST
    Subject: Re:helppppppp
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pritam (5)

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is the ans 2?
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[Post New]1 Apr 2007 19:19:31 IST
    Subject: Re:helppppppp
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shipra22_bhardwaj (304)

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ya but how
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[Post New]2 Apr 2007 14:42:08 IST
    Subject: helppppppp
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krishna.gopal (2058)

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When its falling down net force is zero. So drag = mg upwards. If it has to move with velocity v upwards, weight of mg and drag mg will act downwards So E = 2mg/q (This assumes there is no boyancy i.e. density of air is very very less than density of air)
Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi
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