Equilibrium of Charged Oil Droplets in Uniform Electric Field

To ensure that the oil droplets do not descend, the downward gravitational force acting on each droplet must be balanced by an upward electrostatic force.

1. Mass of the Oil Droplet: Given:

• Radius of droplet, $r = 1\text{ mm} = 10^{-3}\text{ m}$
• Density of oil, $\rho = 1.5\text{ g/cm}^3 = 1500\text{ kg/m}^3$

The volume of a spherical droplet is: $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (10^{-3})^3 = \frac{4\pi}{3} \times 10^{-9}\text{ m}^3$

The mass $m$ of the droplet is: $m = \rho \cdot V = 1500 \times \frac{4\pi}{3} \times 10^{-9}\text{ kg} = 2\pi \times 10^{-6}\text{ kg}$

2. Gravitational Force: Taking $g = 10\text{ m/s}^2$, the downward gravitational force $F_g$ is: $F_g = mg = 2\pi \times 10^{-6} \times 10 = 2\pi \times 10^{-5}\text{ N}$

3. Electric Field Required for Equilibrium: The oil droplet carries a positive charge $q = 5\text{ nC} = 5 \times 10^{-9}\text{ C}$. To create an upward electrostatic force $F_e = qE$, the electric field $E$ must be directed vertically upwards.

Equating electrostatic force to gravitational force: $F_e = F_g \implies qE = mg$

$E = \frac{mg}{q} = \frac{2\pi \times 10^{-5}\text{ N}}{5 \times 10^{-9}\text{ C}} = 4000\pi\text{ V/m}$

4. Potential Difference Between Plates: Since Plate A is on top and Plate B is at the bottom, an upward electric field requires Plate B to be at a higher potential than Plate A ($V_B > V_A$).

The separation between the plates is $d = \frac{12}{\pi}\text{ cm} = \frac{12}{\pi} \times 10^{-2}\text{ m}$.

The required potential difference $V_B - V_A$ is: $V_B - V_A = E \cdot d = 4000\pi \times \left(\frac{12}{\pi} \times 10^{-2}\right) = 480\text{ V}$

5. Checking the Options:

• For Option A: $V_A = 100\text{ V}$ and $V_B = 580\text{ V}$ $V_B - V_A = 580\text{ V} - 100\text{ V} = 480\text{ V}$

This satisfies the required condition for equilibrium.

Correct Option: A