Millikan measured the electron’s charge by observing tiny charged oil drops in a
ID: 1686481 • Letter: M
Question
Millikan measured the electron’s charge by observingtiny charged oil drops in an electric field. Each drop had
a charge imbalance of only a few electrons. The strength
of the electric field was adjusted so that the electric and
gravitational forces on a drop would balance and the drop
would be suspended in air. In this way the charge on the
drop could be calculated. The charge was always found
to be a small multiple of 1.60 x 10^-19 C. Find the charge
on an oil drop weighing 1.00 x 10^-14 N and sus pended
in a downward field of magnitude 2.08 x 10^4 N/C.
Explanation / Answer
The charge of electron e = 1.60 x 10^-19 C weight W = 1.00 x 10^-14 N downward electric field E = 2.08 x 10^4 N/C. given force on particle due to electrcifield = weight Eq = W from this charge of particle q = W / E = 0.6933 * 10 ^ -9 C = (0.6933 * 10 ^ -9 C/1.6* 10^-19 C ) e = 0.4333* 10^ 10 * e