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

To prevent the droplets from falling, the electric force must balance the gravitational force. The gravitational force on the droplet is: \[ F_g = mg = \rho Vg = \rho \left(\frac{4}{3} \pi r^3 \right) g, \] where \( \rho \) is the density of the oil, \( r \) is the radius of the droplet, and \( g \) is the acceleration due to gravity.
The electric force on the droplet is: \[ F_e = qE = q \frac{V_A - V_B}{d}, \] where \( q \) is the charge on the droplet and \( d \) is the separation between the plates.
Equating the forces: \[ F_e = F_g. \] Substituting the known values and solving for \( V_A \) and \( V_B \), we find the values to be 580 V and 100 V respectively.
Final Answer: 580 V and 100 V