In 1909, Robert Millikan was the first to find the charge of an electron
in his now-famous oil-drop experiment. In that experiment, tiny oil
drops were sprayed into a uniform electric field between a horizontal
pair of oppositely charged plates.The drops were observed with a
magnifying eyepiece, and the electric field was adjusted so that the
upward force on some negatively charged oil drops was just sufficient to
balance the downward force of gravity. That is, when suspended, upward
force qE just equaled mg. Millikan accurately measured the charges on
many oil drops and found the values to be whole number multiples of
1.6×10−19C the charge of the electron. For this, he won
the Nobel prize. If a drop of mass 1.08×10−14kg remains stationary in an electric field of 1.68×105NC−1, then the charge of this drop is :
In 1909, Robert Millikan was the first to find the charge of an electron
in his now-famous oil-drop experiment. In that experiment, tiny oil
drops were sprayed into a uniform electric field between a horizontal
pair of oppositely charged plates.The drops were observed with a
magnifying eyepiece, and the electric field was adjusted so that the
upward force on some negatively charged oil drops was just sufficient to
balance the downward force of gravity. That is, when suspended, upward
force qE just equaled mg. Millikan accurately measured the charges on
many oil drops and found the values to be whole number multiples of
1.6×10−19C the charge of the electron. For this, he won
the Nobel prize. If a drop of mass 1.08×10−14kg remains stationary in an electric field of 1.68×105NC−1, then the charge of this drop is :
The correct option is A 6.40×10−19C
Given that electrostatic force is just balancing gravitational force.
qE=mg q×1.68×105=1.08×10−14g
Charge of drop, q=6.40×10−19C
Given that electrostatic force is just balancing gravitational force.
qE=mg
