Question

Two metal plates (A, B) are kept horizontally with separation of \( \frac{12}{\pi} \) cm, with plate A on the top.} An atomizer jet sprays oil (density \( 1.5 \, \text{g/cm}^3 \)) droplets of radius 1 mm horizontally. All oil droplets carry a charge 5 nC. The potentials \( V_A \) and \( V_B \) are required on plates A and B respectively in order to ensure the droplets do not descend. The values of \( V_A \) and \( V_B \) are _______. (Neglect the air resistance to the droplets and take \( g = 10 \, \text{m/s}^2 \))}

• A. 100 V and 580 V
• B. 580 V and 100 V
• C. 600 V and 400 V
• D. 0 V and -200 V

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem is based on the Millikan oil-drop experiment principle.
For the charged droplets to not descend, the upward electrostatic force must exactly balance the downward gravitational force.
Since the charge on the droplet is positive (\(5 \text{ nC}\)), the electric field must point upwards. Therefore, the lower plate (B) must be at a higher potential than the upper plate (A).
Step 2: Key Formula or Approach:
Force balance: \(qE = mg\).
Mass of a spherical droplet: \(m = \frac{4}{3}\pi r^3 \rho\).
Electric field between parallel plates: \(E = \frac{V_B - V_A}{d}\).
Step 3: Detailed Explanation:
First, calculate the mass of a single oil droplet. Convert units to standard SI:
\(r = 1 \text{ mm} = 10^{-3} \text{ m}\)
\(\rho = 1.5 \text{ g/cm}^3 = 1500 \text{ kg/m}^3\)
\[ m = \frac{4}{3} \pi (10^{-3})^3 (1500) = \frac{4}{3} \pi (10^{-9}) (1500) = 2000 \pi \times 10^{-9} = 2\pi \times 10^{-6} \text{ kg} \] Now, equate the electric force to the gravitational force to find the required electric field \(E\):
\[ qE = mg \] \[ (5 \times 10^{-9}) E = (2\pi \times 10^{-6}) (10) \] \[ (5 \times 10^{-9}) E = 2\pi \times 10^{-5} \] \[ E = \frac{2\pi \times 10^{-5}}{5 \times 10^{-9}} = 0.4\pi \times 10^4 = 4000\pi \text{ V/m} \] Next, find the potential difference required across the plates. The distance \(d = \frac{12}{\pi} \text{ cm} = \frac{0.12}{\pi} \text{ m}\).
\[ V_B - V_A = E \times d = 4000\pi \times \frac{0.12}{\pi} = 480 \text{ V} \] The potential of plate B must be 480 V higher than plate A.
Now, review the given options to find the pair with a difference of +480 V:
Option (A): \(V_B - V_A = 580 - 100 = 480 \text{ V}\). (Matches)
Option (B): \(V_B - V_A = 100 - 580 = -480 \text{ V}\).
Option (C): \(V_B - V_A = 400 - 60 = 340 \text{ V}\).
Option (D): \(V_B - V_A = -200 - 0 = -200 \text{ V}\).
Step 4: Final Answer:
The values of \(V_A\) and \(V_B\) are \(100\text{ V and } 580\text{ V}\).