Question

In Millikan’s oil drop experiment, an oil drop carrying a charge \( Q \) is held stationary by a potential difference 2400 V between the plates. To keep a drop of half the radius stationary, the potential difference had to be made 600 V. What is the charge on the second drop?

• A. \( \frac{Q}{4} \)
• B. \( \frac{Q}{2} \)
• C. \( Q \)
• D. \( \frac{3Q}{2} \)

Hint:

The force on an oil drop in Millikan’s experiment is proportional to the square of its radius, and the potential difference required to balance the forces depends on the square of the radius.

Answer 1

The Correct Option is B