1.

A charged oil drop of mass 9.75 xx 10^(-15) kg and charge 30 xx 10^(-16) C is suspended in a uniform electric field existing between two parallel plates. The field between the plates is (take g = 10 ms^(-2))

Answer»

`3.25 V m^(-1)`
`300 V m^(-1)`
`325 V m^(-1)`
`32.5 V m^(-1)`

Solution :`E = (MG)/(Q) = (9.75 xx 10^(-15) xx 10)/(30 xx 10^(-16)) = 32.5 VM^(-1)`


Discussion

No Comment Found

Related InterviewSolutions