Millikan measured the electron\'s charge by observing tinychargedoil drops in an

Question

Millikan measured the electron's charge by observing tinychargedoil drops in an electric field. Each drop had a chargeimbalance ofonly a few electrons. The strength of the electricfield wasadjusted so that the electric and gravitational forces ona dropwould balance and the drop would be suspended in air. In thiswaythe charge on the drop could be calculated. The charge wasalwaysfound to be a small multiple of 1.6e-19 C. Find the charge onanoil drop weighing 2.75 10-14 N and suspended in a downward fieldofmagnitude 2.15 104 N/C.

Explanation / Answer

as the E is downward, charge must be negative so as to cancel force of gravity force due to gravity=2.75 * 10^-14= electrostatic force=Eq i.e. q= 2.75 * 10^-14/2.15 *10-4 =1.27*10^-18 q=ne= n* 1.6* 10^-19 so n*1.6 * 10^-19= 1.27*10^-18 n=7.93 so we can take n=8 Ans: 8 electons

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