Question

In his famous 1909 experiment that demonstrated quantization of electric charge, R. A. Millikan suspended small oil drops in an electric field. With a field strength 0f 20 MN/C, what mass drop can be suspended when the drop carries a net charge of 10 elementary charges?

Answer 1

Answer:

$3.26198\times 10^{-12}\ kg$

Explanation:

E = Electric field = 20 MN/C

q = Charge of electron = $1.6\times 10^{-19}\ C$

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of drop

The electrical force will balance the weight

$Eq=mg\\\Rightarrow 20\times 10^{6}\times 10\times 1.6\times 10^{-19}=m\times 9.81\\\Rightarrow m=\dfrac{20\times 10^{6}\times 10\times 1.6\times 10^{-19}}{9.81}\\\Rightarrow m=3.26198\times 10^{-12}\ kg$

The mass that can be suspended is $3.26198\times 10^{-12}\ kg$