mg = qE
q =mg/E = 1•10^-14•9.8/2•10^5 =
= 4.9•10^-19 C
8 1/2 HRS ON THIS MASTERING PHYSICS WOW
n 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^-19 C-the charge of the electron. For this he won the Nobel Prize.
1. If a drop of mass 1.0 * 10^-14 kg remains stationary in an electric field of 2.0 * 10^5 N/C, what is the charge of this drop?
2. If a drop of mass 1.0 * 10^-14kg remains stationary in an electric field of 2.0 * 10^5; N/C}, what is the charge of this drop?
2 answers
thanks so much
How many extra electrons are on this particular oil drop (given the presently known charge of the electron)?
How many extra electrons are on this particular oil drop (given the presently known charge of the electron)?
2 answers