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 is `9 xx 10^(-15)kg`] (Neglect air resistance).

Explanation

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)`
So, qE= mg
`rArr` Charge, `q= (mg)/(E )= (9 xx 10^(-15) xx 10)/(27.5 xx 10^(4))= (900)/(275) xx 10^(-19)`
`rArr q = 3.2 xx 10^(-19)C`