In 1909 Robert Millikan was the first to find the charge of an electron in his n
ID: 1589358 • Letter: I
Question
In 1909 Robert Millikan was the first to find the charge of an electron in his now-famous oil drop experiment. In the experiment tiny oil drops are sprayed into a uniform electric field between a horizontal pair of oppositely charged plates. The drops are observed with a magnifying eyepiece, and the electric field is adjusted so that the upward force q E on some negatively charged oil drops is just sufficient to balance the downward force m g of gravity. Millikan accurately measured the charges on many oil drops and found the values to be whole-number multiples of 1.6 × 1019 C — the charge of the electron. For this he won the Nobel Prize. If a drop of mass 2.44898 × 1013 kg remains stationary in an electric field of 1.5 × 106 N/C, what is the charge on this drop? The acceleration due to gravity is 9.8 m/s2 . Answer in units of C. How many extra electrons are on this particu- lar oil drop (given the presently known charge of the electron)?
Explanation / Answer
Fg=m*a
Fe=E*q
deltaF=0
0= 2.44898(10-13)x(-9.8) + 1.5(106)xq
q= 2.44898(10-13)x(9.8)/1.5(106)
q=16x10-19C
this is the magnitude of the charge... but depending on the direction of the electric field, it may be positive or negative