calculation of charge of an oil drops

Physics is the one of the most important and interesting topic in the world. Physics study is the study about the very small particle to very large particle i.e. photon to universe. We don’t saw an electron but scientist shows that electron is charge particle and has charge 1.6*10-19columb. The name of scientist who obtain the charge of an electron is Millikan. He takes required instrument which is called Millikan’s instrument. He applied require potential difference from battery to neutralize the oil drops. When the oil drops dropped from the above glass plate then there produce three forces they are’
Viscous force= 6πήrvt
Where; ή=coefficient of viscosity;
v=terminal velocity;
r=radius of oil drop;
t= time period;
Up thrust (U)=4/3πr^3σg
Where; σ=density of air;
g=gravitational force;
Weight of an oil drop(w)=4/3πr^3ρg
Where; ρ=density of oil;
Then if the oil drop contains terminal velocity then upward force must be equal to downward force i.e.;
Upward force = downward force
6πήrvt +4/3πr^3σg =4/3πr^3ρg
In this relation we can find the value of r.
By finding the value of r, we can find the value of charge of an oil drop. i.e.,
When the current is applied through the apparatus, the relation become,
qE= 4/3πr^3σ1g-4/3πr^3ρ1g +6πήrvtType equation here.
By this equation q can be calculated and we obtain the value of e from this equation.
i.e. q=ne
Where, n=no of electron;
By this way charge of an electron can be calculated from Millikan.