Question

In 1909, Robert Millikan performed an experiment involving tiny, charged drops of oil. The drops were charged because they had picked up extra electrons. Millikan was able to measure the charge on each drop in coulombs. Here is an example of what his data may have looked like.

Drop Charge (C)
A 3.20\times 10^{-19}
B 4.80\times 10^{-19}
C 8.00\times 10^{-19}
D 9.60\times 10^{-19}

Based on the given data, how many extra electrons did drop C contain?

Answer 1

What's the charge on the electron? I think 1.6 x 10^-19
So 8 x 10^-19/1.6 x 10^-19 = ??

Answer 2

answer: 5 electrons