An oil drop of 12 excess electrons is held stationary under a constantelectric field of 2.55 × 104 NC–1 (Millikan’s oil drop experiment). Thedensity of the oil is 1.26 g cm–3. Estimate the radius of the drop.(g = 9.81 m s–2; e = 1.60 × 10–19 C).

Given: The magnitude of the electric field in Millikan’s oil drop experiment is 2.55× 10 4  N/C , the density of oil is 1.26 g/ cm 3 and the excess of electron on drop is 12.

The mass of the oil drop is given as,

m=V×ρ(1)

Where, the mass of the oil drop is m, the volume of the oil drop is V and the density is ρ.

The total charge density by electrons oil drop is given as,

q=ne(2)

Where, the number of excess of electrons is n and the charge on the electron is e.

The charge on an electron is given as,

e=1.6× 10 −19  C

the electric field intensity is E, the density of the oil is ρ,

By substituting the given values in equation (2), we get

q=12×1.6× 10 −19 =19.2× 10 −19  C

The volume of the oil drop is given as,

V= 4 3 π r 3 (3)

Where, the radius of the oil drop is r.

From equation (1) and (3), we get

m= 4 3 π r 3 ×ρ(4)

The weight of the oil drop is given by,

W=mg(5)

From equation (4) and (5), we get

W= 4 3 π r 3 ×ρ×g = 4 3 π r 3 ρg (6)

The force on a charge in electric field is given as,

F=Eq(7)

Where, the electric field is E and the charge is q.

The weight of the oil drop must be balanced by the force produced by the electric field.

From equation (6) and (7), we get

Eq= 4 3 π r 3 ρg r 3 = 3Eq 4πρg r= ( 3Eq 4πρg ) 1 3 (8)

By substituting the given values in equation (8), we get

r= ( 3×2.55× 10 4 ×19.2× 10 −19 4π×1.26× 10 3 ×9.81 ) 1 3 =9.82× 10 −7  m =9.82× 10 −4  mm

Thus, the radius of the oil drop is 9.82× 10 −4  mm.