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A charged oil drop is suspended in a uni...

A charged oil drop is suspended in a uniform electric field `27.5 xx 10^(4)V//m`, so that it neither rises nor falls. Find the charge on the drop. Take, mass of the drop `=9 xx 10^(-15)kg`. (Neglect air resistance)

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Given, `E= 27.5 xx 10^(4) V//m, m= 9 xx 10^(-15)kg, g= 10m//s^(2)` (say) and q=?
In equilibrium, `F_(e)= F_(g)`
`therefore qE= mg`
`rArr q= (mg)/(E )= (9 xx 10^(-15) xx 10)/(27.5 xx 10^(4))`
`=(900)/(275) xx 10^(-19)`
`therefore q= 3.2 xx 10^(-19)C`
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