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Question

Two horizontal plates charged with +q and q charges are having an area of A. A charged drop of oil is suspended in equilibrium position between the plates, then the charge on the oil drop will be :

A
mg/4
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B
mg/A
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C
mgε0A/q
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D
mε0A/q
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Solution

The correct option is C mgε0A/q
Gravitational force on the drop of mass m is Fg=mg
Electric force on the drop is Fe=QE=Q(qAϵ0) where Q= charge on drop and E= electric field in between plates.
In equilibrium, Fg=Femg=Q(qAϵ0)
Q=Aϵ0mgq

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