Milikan Oil Drop Expt A Numerical Approach


An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 ×104 N/C in Millikan's oil drop experiment.
The density of the oil is 1.26 g/cm3 Estimate the radius of the drop

At equilibrium the gravitational force=electrostatic force
mg=qE
ie volume of drop×density×g=neE (since q=ne)
4/3πr3 ×ρg=neE
r3=(3×12×1.6×10-19×2.55 ×104) / (4× 3.14× 1.26×103× 9.81 )=0.94×10-18
. . .r= (0.94×10-18)1/3=9.81×10-7
So the radius of the drop is 9.81×10-7m

Comments

  1. Sai's Science and Maths class15 September 2014 at 08:48

    Plus 2 students attention please The crash course will commence in the month of November2014

    ReplyDelete

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