Question

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.

a) 4.8 × 10^(-19) C
b) 9.6 × 10^(-19) C
c) 1.6 × 10^(-19) C
d) 3.2 × 10^(-19) C

Answer 1

In Millikan's oil-drop experiment, the charge on an oil drop can be found using the formula q = mdrop / E. By calculating the mass of the oil drop and the electric field between the plates, we can determine the charge on the drop. In this case, the charge on the drop is 5.89 x 10^(-19) C.

Step-by-step explanation:

In Millikan's oil-drop experiment, the charge on the oil drop can be found using the formula mdrop = qE, where mdrop is the mass of the oil drop, q is the net charge of the oil drop, and E is the electric field between the plates. The charge on an electron, e, is 1.6 x 10^(-19) C. By rearranging the formula, we can solve for the charge on the drop:

q = mdrop / E

First, we need to calculate the mass of the oil drop using its density and diameter:

mass = volume x density

volume = (4/3) x π x (radius)^3

radius = diameter / 2

Using the given values, we find the mass of the oil drop to be 5.99 x 10^(-16) kg.

Next, we need to calculate the electric field between the plates:

electric field = voltage / distance

Using the given values, we find the electric field to be 10165 N/C.

Finally, we can calculate the charge on the drop:

q = (5.99 x 10^(-16) kg) / (10165 N/C) = 5.89 x 10^(-19) C