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 1.10 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?

Answer 1

Concept:- here we find the mass of one droplet using the density, then we equate the electric force to weight considering the equilibrium of drops to find the electric field and number of electrons,