An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 * 10^4 N/C in Millikan`s oil drop experiment. The density of the oil is 1.26 g/cm^(-3). Estimate the radius of the drop.
Expert's answer
1
No1
First, we find charge of the drop. 12 electrons will create charge of
q=12⋅(−1.6)⋅10−19C≈1.92⋅10−18C
Next we find electrostatic force, acting on drop
F=qE=1.92⋅10−18C⋅2.55⋅104N/C≈4.896⋅10−14N
In Miliken experiment electrostatic force is equal to gravitational, so we can find size of drop