Question:
In Millikanâs 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?
In Millikanâs 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?
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The force on an oil drop in Millikanâs 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.
Updated On: Apr 22, 2026
- \( \frac{Q}{4} \)
- \( \frac{Q}{2} \)
- \( Q \)
- \( \frac{3Q}{2} \)
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The Correct Option is B
Solution and Explanation
Step 1: Use the relation between radius and electric force.
The force on an oil drop is given by: \[ F = \frac{V Q}{d} \] where \( V \) is the potential difference, \( Q \) is the charge, and \( d \) is the distance between the plates. The force on the drop is also related to the radius \( r \) by: \[ F \propto r^2 \] Since the radius is halved, the new force will be \( \frac{1}{4} \) of the original force.
Step 2: Relate the new potential difference.
The new potential difference is proportional to the change in force: \[ V_2 \propto V_1 \quad \text{and} \quad V_2 = \frac{V_1}{2} \] Thus, the charge on the second drop is \( \frac{Q}{2} \).
The force on an oil drop is given by: \[ F = \frac{V Q}{d} \] where \( V \) is the potential difference, \( Q \) is the charge, and \( d \) is the distance between the plates. The force on the drop is also related to the radius \( r \) by: \[ F \propto r^2 \] Since the radius is halved, the new force will be \( \frac{1}{4} \) of the original force.
Step 2: Relate the new potential difference.
The new potential difference is proportional to the change in force: \[ V_2 \propto V_1 \quad \text{and} \quad V_2 = \frac{V_1}{2} \] Thus, the charge on the second drop is \( \frac{Q}{2} \).
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