Question

In an oil drop experiment, a drop with a weight of 1.9 x 10-14 N was suspended when the potential difference between 2 plates that were 63 mm apart was 780 V. What was the charge on the drop?

Answer 1

The charge on the oil drop in the Millikan's oil-drop experiment is 1.9 x 10-19 C.

Step-by-step explanation:

In Millikan's oil-drop experiment, the charge on the drop can be determined using the equation:

q = mg/E

Where:

• q is the charge on the drop
• m is the mass of the drop
• g is the acceleration due to gravity
• E is the electric field between the plates

Plugging in the given values, we have:

q = (1.9 x 10-14 N)/(780 V/0.063 m)

q = 1.9 x 10-19 C

Therefore, the charge on the drop is 1.9 x 10-19 C.