An oil drop having a mass `4.8xx10^(-18)`g and a charge `2.4xx10^(-18)`C stands still between two charged horizontal plates separated by a distance of 1 cm If now the polarity of the plates is changed, instantaneous accleration of the drop is `(g=10ms^(-2)`

Options

A. `5ms^(-2)`
B. `10ms^(-2)`
C. `15ms^(-2)`
D. `20ms^(-2)`

Answer

B

Explanation

For an oil drop ,
Mass,`m=4.8xx10^(-18)g`
Charge ,`q=24.xx10^(-18)`
Distance ,d-1cm
The expression for instantaneous accelaration can be obtained as ,
F=ma
`implies a=(F)/(m)`
`implies a=(qE)/(m)(because" Force of charge ,"F=qE)`
`=(mg)/(m)(becauseqE=mg)a=gtherefore a=10ms^(-2)`