Question

A charged drop of radius $1.92mm$ is kept stationary by the application of an electric field of $1.65\times {10}^6$ N/C in Millikans oil drop experiment. The charge, if the density of oil is $920$ $Kg/m^3$, is

• A. $1.72\times {10}^{-18}$
• B. $16.2\times {10}^{-19}$
• C. $1.82\times {10}^{-17}$
• D. $1.92\times {10}^{-17}$

Answer 1

The correct option is B $16.2\times {10}^{-19}$

Mass of the drop $\frac{4\pi }{3}(1.92\times {10}^{-2}{)}^3\times 920kgs$
$=2.7275\times {10}^{-5}kgs$
Let charge be q,
So,
for equilibrium,
$\Rightarrow 2.7275\times {10}^{-11}\times g=1.65\times {10}^6\times q$
$\Rightarrow q=1.61999\times {10}^{-10}$
$\Rightarrow q=1.62\times {10}^{-10}$ Coulomb.