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.
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?
I GOT 4.8×10^−19 C WHICH IS CORRECT
2.How many extra electrons are on this particular oil drop (given the presently known charge of the electron)?
To determine the number of extra electrons on the oil drop, we need to calculate the ratio of the charge of the drop to the charge of a single electron.
We already know the charge of the drop, which is 4.8×10^−19 C, and we also know the charge of a single electron, which is 1.6×10^−19 C.
To find the number of extra electrons, we can divide the charge of the drop by the charge of a single electron:
Number of extra electrons = Charge of the drop / Charge of a single electron
Number of extra electrons = 4.8×10^−19 C / 1.6×10^−19 C
Simplifying the expression gives:
Number of extra electrons = 3
Therefore, there are 3 extra electrons on this particular oil drop.