calculate the value of the characteristic times for the tiny oildrop in the millikan oildrop experiment when its terminal speed 6.1x10^-5 m/s.
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Terminal speed =5.8cm/s
Viscous force =3.9×10
−10
N
Radius of the given uncharged drop, r=2.0×10
−5
m
Density of the uncharged drop, ρ=1.2×10
−3
kgm
−3
Viscosity of air, η=1.8×10
−5
Pas
Density of air (ρ
0
) can be taken as zero in order to neglect buoyancy of air.
Acceleration due to gravity, g=9.8m/s
2
Terminal velocity (v) is given by the relation:
V=2r
2
×(ρ−ρ
0
)g/9η
=2×(2×10
−5
)
2
(1.2×10
3
−0)×9.8/(9×1.8×10
−5
)
=5.8×10
−2
m/s
=5.8cms
−1
Hence, the terminal speed of the drop is 5.8cms
−1
.
The viscous force on the drop is given by:
F=6πηrv
∴F=6×3.14×1.8×10
−5
×2×10
−5
×5.8×10
−2
=3.9×10
−10
N
Hence, the viscous force on the drop is 3.9×10
−10
N.
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