A positively charged oil droplet remins stationary in the electric field between two horizontal plates separated by a distance of `1 cm` . The charge on the drop is `10^(-15) C` and mass of the droplet is `10^(-11) g`, the potential difference between the plates and if the polarity is reversed , the instantaneous of the droplet are

Explanation

As the droplet is at rest, its weight W= mg will be balanced by electric force F = qE
i.e., qE= mg

or `V = (mg d)/(q) V = ((10^(-14)) (9.8) (1xx 10^(-2)))/((9.8 xx 10^(-10)//3 xx 10^(9))) = 3 xx 10^(4)` volt
Now if the polarity of the plates is reversed, both electrical and gravitational force will act downward So, `F = mg + qE = 2mg` (as mg = qE)
And hence instantaneous acceleration of drop: `a = (F)/(m) = (2mg)/(m) = 2g= 19.6m//s^(2)`