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Millikan`s oil drop experiment attempts to measure the charge on a single electron,e,by measuring the charge of tiny oil drops suspended in an electrostatic field.It is assumed that the charge on the oil drop is due to just a small number of excess electrons. The charges 3.90×10^-19C, 6.50×10^-19C and 9.1×10^-19C are measured on three drops of oil. The charge of an electron is deduced to be,

Millikan`s oil drop experiment attempts to measure the charge on a single electron,e,by measuring the charge of tiny oil drops suspended in an electrostatic field.It is assumed that the charge on the oil drop is due to just a small number of excess electrons. The charges 3.90×10^-19C, 6.50×10^-19C and 9.1×10^-19C are measured on three drops of oil. The charge of an electron is deduced to be,

Grade:12

2 Answers

Vikas TU
14149 Points
6 years ago
As the resultant of the 3 of the charges :
Q1 = 3.9 x 10^-19 Q2 = 6.50 x 10^-19 Q3 = 9.1 x 10^-19

The resultant charge is 19 x10^-19 C
Mohd Ashraf
13 Points
6 years ago
The resultant charge will be the H.C.F. of the given chargesH.C.F. of 3.9, 6.5 and 9.1 is 1.3So the resultant charge will be 1.3*10^-19

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