Question

18. In Millikan's experience, a microscopic oil droplet, with 2 mm radius, is introduced between the plates of a plan capacitor, whose spacing is 5 cm. The oil density is 0.78 g/cm³. With the plates initially discharged, the droplet is observed to fall reaching, due to air resistance, a constant terminal speed v. When applying across the plates a potential difference of 40 kV with the electric field oriented upward, the falling speed doubles. What is the sign and the total charge in electron charge units?

Answer 1

Since the electric field is upward and after applying electric field the falling speed doubles which means the electric force on oil droplet is in the downward direction which is opposite to the direction of the electric field applied. Therefore the charge is negative.

Terminal velocity formula is given as

Since after applying the electric field, terminal velocity doubles which means there is an increase in acceleration of the oil droplet.

acceleration = g + acceleration due to electric field =

Since the square of terminal velocity is directly proportional to acceleration in this case Therefore

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