In the famous Millikan oil-drop experiment, tiny spherical droplets of oil are sprayed into a uniform vertical electric field. The drops get a very small charge (just a few electrons) due to friction with the atomizer as they are sprayed. The field is adjusted until the drop (which is viewed through a small telescope) is just balanced against gravity and therefore remains stationary. Using the measured value of the electric field, we can calculate the charge on the drop and from this calculate the charge e of the electron. In one apparatus the drops are 1.10 μm in diameter, and the oil has a density of 0.850 g/cm3.(a) If the drops are negatively charged, which way should the electric field point to hold them stationary (up or down)? (b) Why? (c) If a certain drop contains four excess electrons, what magnitude electric field is needed to hold it stationary? (d) You measure a balancing field of 5183 N/C for another drop. How many excess electrons are on this drop?

(b) The electric field point to hold them downward since the electric force balances the weight downward, Fe must act upward and for a negative charge electric force is opposite to the electric field

c and d are also answered in the picture above

What is Millikan oil drop experiment?

Millikan's oil-drop experiment is based on comparing applied electric force with changes in the motion of the oil drops.

In his experiment, he was able to determine the electric charge on each drop. He found that all of the drops had charges that were simple multiples of a single number which is the fundamental charge of the electron.

The simple conclusion drawn from millikan's oil drop experiment is that charge on an oil droplet is quantised and

= n x e

where;

n = number of electrons
e = charge on one electron.

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