Question

In Millikan's oil-drop experiment, one looks ata small oil drop held motionless between two plates. Take the voltage betwee 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 3.8x10^6m. Find the charge on the drop, in terms of electron units. You need to round your answer to the nearest integer

Answer 1

Answer:

The charge on the drop  is

q = 1.741 x 10 ⁻²¹ C

Explanation:

Electric field due to plates

Ef = V/d  

Ef = 2033 V / (2.0 * 10^-2 m )

Ef = 101650 V/m

So, we can write  

Ef * q = m*g

q = m*g / E f

The mass can be equal using the density and the volume so:

m = ρ * v

The volume can be find as:

v = 2.298 x 10 ⁻ ¹⁶ m³

q =  ρ * v * g / Ef

q =  81 x 10 ³ kg/ m³ * 2.2298 x 10 ⁻ ¹⁶ m³ * 9.8 m/s² / 101650 V/m

The charge on the drop  is

q = 1.741 x 10 ⁻²¹ C