Question:
In Millikanâs oil drop experiment, a charged drop of mass \( 1.8 \times 10^{-13} \, \text{kg} \) is stationary between its plates. The distance between its plates is 0.9 cm, and the potential difference is 2000 V. The number of electrons in the drop is:
In Millikanâs oil drop experiment, a charged drop of mass \( 1.8 \times 10^{-13} \, \text{kg} \) is stationary between its plates. The distance between its plates is 0.9 cm, and the potential difference is 2000 V. The number of electrons in the drop is:
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In Millikanâs experiment, the charge on the oil drop is quantized, and the number of electrons can be found by dividing the total charge by the charge of one electron.
Updated On: Apr 22, 2026
- 500
- 50
- 5
- 10
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The Correct Option is B
Solution and Explanation
Step 1: Calculate the electric force.
The force acting on the oil drop is the electric force, given by: \[ F = qE \] where \( q \) is the charge on the drop, and \( E \) is the electric field between the plates. The electric field is: \[ E = \frac{V}{d} \] where \( V \) is the potential difference and \( d \) is the distance between the plates. Substituting the given values: \[ E = \frac{2000}{0.009} = 2.22 \times 10^5 \, \text{N/C} \]
Step 2: Set up the balance of forces.
The oil drop is stationary, so the electric force is balanced by the gravitational force: \[ mg = qE \] The gravitational force is \( mg = 1.8 \times 10^{-13} \times 9.8 = 1.764 \times 10^{-12} \, \text{N} \). Thus: \[ q \times 2.22 \times 10^5 = 1.764 \times 10^{-12} \] Solving for \( q \): \[ q = \frac{1.764 \times 10^{-12}}{2.22 \times 10^5} = 7.95 \times 10^{-18} \, \text{C} \]
Step 3: Find the number of electrons.
The charge on a single electron is \( e = 1.6 \times 10^{-19} \, \text{C} \). The number of electrons is: \[ \text{Number of electrons} = \frac{q}{e} = \frac{7.95 \times 10^{-18}}{1.6 \times 10^{-19}} = 50 \]
The force acting on the oil drop is the electric force, given by: \[ F = qE \] where \( q \) is the charge on the drop, and \( E \) is the electric field between the plates. The electric field is: \[ E = \frac{V}{d} \] where \( V \) is the potential difference and \( d \) is the distance between the plates. Substituting the given values: \[ E = \frac{2000}{0.009} = 2.22 \times 10^5 \, \text{N/C} \]
Step 2: Set up the balance of forces.
The oil drop is stationary, so the electric force is balanced by the gravitational force: \[ mg = qE \] The gravitational force is \( mg = 1.8 \times 10^{-13} \times 9.8 = 1.764 \times 10^{-12} \, \text{N} \). Thus: \[ q \times 2.22 \times 10^5 = 1.764 \times 10^{-12} \] Solving for \( q \): \[ q = \frac{1.764 \times 10^{-12}}{2.22 \times 10^5} = 7.95 \times 10^{-18} \, \text{C} \]
Step 3: Find the number of electrons.
The charge on a single electron is \( e = 1.6 \times 10^{-19} \, \text{C} \). The number of electrons is: \[ \text{Number of electrons} = \frac{q}{e} = \frac{7.95 \times 10^{-18}}{1.6 \times 10^{-19}} = 50 \]
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