   Chapter 30, Problem 5PE

Chapter
Section
Textbook Problem

In Millikan's oil-drop experiment, one looks at a small oil drop held motionless between two plates. Take the voltage between the plates to be 2033 V, and the plate separation to be 2.00 cm. The oil drop (of density 0.81 g/cm3) has a diameter of 4.0 × 10 − 6 m. Find the charge on the drop, in terms of electron units.

To determine

The charge on the drop when a small oil drop held motionless between two plates.

Explanation

Given Data:

Voltage between the plates is2033V and the plate separation is2cm . The oil has a diameter of4×106m and its density is0.81g/cm3.

Formula Used:

In Millikan's oil drop experiment, the oil drop is held between the two charged plates and the force due to electric field and gravitational force are in equilibrium.

Thus, we have

FE=FG

Also we have

FE=qE , whereq= charge on oil droplet andE= electric field

AndE=Vd , whereV= voltage andd= separation between the plates.

AndFG=mg , whereg= acceleration due to gravity,m= Mass of droplet

Calculation:

We have

FE=FG

qVd

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