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 :

m g / 4

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m g / A

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m g ε_{0} A / q

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m ε_{0} A / q

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Solution

The correct option is C $m g ε_{0} A / q$
Gravitational force on the drop of mass m is $F_{g} = m g$
Electric force on the drop is $F_{e} = Q E = Q (\frac{q}{A ϵ_{0}})$ where $Q =$ charge on drop and $E =$ electric field in between plates.
In equilibrium, $F_{g} = F_{e} \Rightarrow m g = Q (\frac{q}{A ϵ_{0}})$
$∴ Q = \frac{A ϵ_{0} m g}{q}$

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