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The oil drop is negatively charged and weighs 8.5 10-15 N. The drop is suspended in an electric fi?
The oil drop shown below is negatively charged and weighs 8.5 10-15 N. The drop is suspended in an electric field intensity of 5.3 103 N/C.
(a) What is the charge on the drop?
(b) How many electrons does it carry?
1 Answer
- Anonymous9 years agoFavorite Answer
Known:
F= 8.5x10^-15N
E= 5.3x10^3 N/C
a) What is the charge on the drop?
E = F/q
So,
q = F/E
q= (8.5x10^-15N)/(5.3x10^3 N/C)
q= 1.6x10^-18C
(b) How many electrons does it carry?
q/qe
So,
q= 1.6x10^-18C and qe= 1.6x10^-19C
1.6x10^-18C/1.6x10^-19C
apx: 10 electrons