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

Options

A. `1V , 9.8 m//sec^(2)`
B. `1 V , 0 m//sec^(2)`
C. `1V ,19.6 m//sec^(2)`
D. `2V ,19.6 m//sec^(2)`

Answer

3

Explanation

`V : p.d`. Between plates
`q(V)/(d) = mg`
`(10^(-15) V)/(10^(-2)) = 10^(-11) xx 10^(-3) xx 10`
`V = 1 volt`
If polartiy of plates are reversed , acceleration of ball
`a = (mg + (qV)/(d))/(m) = g + (qV)/(md) = g + g = 2g = 20 m//sec^(2)`