Question

In a Millikan's oil drop experiment, a drop is observed to fall with terminal speed V= 1.4 mm/s when no electric field is applied. Then vertical electric field of 4.9 x 105 V/m is switched on and droplet is observed to continue to move downward at a new lower terminal speed V1 = 1.21 mm/s. How many quantum unit of charge does the droplet possess? (Given density of oil = 750 kgm-3and ηair-1.8 x 10-5kg/m

Answer 1

Answer:

23V

Explanation:

Correct option is

C

23V

In absence of electric field (i.e. ε=0)

Mg=6πnv

D1=6πnrv⟶(1)

In presence of electric field 

Mg+QE=6πn(2V)⟶(2)

When electric field D reduced to E/2

Mg+Q(E/2)=6πn(V1)

D3=6πnr(2V)⟶(3)

After solving (1), (2) and (3)

We get V1=23V.