Question

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 1.40 mm 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? Please explain your reasoning. (1)

b) If a certain drop contains four excess electrons, what magnitude electric field is needed to hold it stationary? Please express your answer to three significant figures. (2)

c) If you measure a balancing field of 5183 N/C for another drop. How many excess electrons are on this drop? Please express your answer as an integer. (2)

Answer 1

PART A:

The gravitational force is acting downwards and hence we requre a force directing upward to balance the gravitational force. We know that the droplet is negatively charged so the force on it is opposite to the electric field.

So, we require an electric field in downward direction so that the force on the droplet is in upward direction.

PART B:

The mass of one droplet is

The charge on the droplet is

The two forces are equal and opposite. i.e

Where

Putting the values of the known quantities in SI units we get

So, the required magnitude of electric field is

PART C:

We have

Putting the values

The magnitude of charge on an electron is

So, the number of electron is

So, the total number of excess electrons are