Charged Body Between Parallel Plates

When a charged particle is placed between two horizontal parallel plates connected to a battery, it experiences two opposing forces: gravity (downward) and electrostatic force (upward, due to the field). Under the right potential difference, the particle can be held in equilibrium — appearing to "float" in mid-air. This principle underlies the classic Millikan oil-drop experiment, used historically to determine the charge of an electron.

A condition in which a charged body remains stationary between two parallel plates because the upward electric force exactly balances the downward gravitational force.

Step 1: The electric field between two parallel plates carrying a potential difference V and a separation d is uniform:

Step 2: For the charged body to remain suspended, the upward electric force must balance the downward gravitational force:

Step 3: Substituting the expression for E:

Step 4: Solving for the required potential difference:

When the charged body is a sphere of radius r and density ρ that has gained or lost n electrons:

Mass of sphere: m = \[\frac{4}{3}\pi r^3\rho\]

Charge on sphere: q = ne

Substituting into equation (1):

• Misconception: "A negatively charged body always experiences a downward electric force."
  Correction: The direction of electric force depends on both the sign of the charge and the direction of the field, not the charge sign alone. A negative charge in a downward field experiences an upward force.
• Misconception: "The electric field between plates varies with position."
  Correction: The field is uniform throughout the region between infinite (or large, closely spaced) parallel plates; it does not depend on distance from either plate.

Given: mass m = 5 mg = 5 × 10−6 kg, charge q = 2 × 10−6 C, plate separation d = 5 cm = 0.05 m, g = 9.8 m/s².

Find: Required potential difference V.

Solution: V = \[\frac {mgd}{q}\] = \[\frac{(5\times10^{-6})(9.8)(0.05)}{2\times10^{-6}}\] ≈1.225 V

Answer: V ≈ 1.225V.

Given: An oil drop of mass 4.8 × 10⁻¹⁵ kg, density 900 kg/m³, suspended using the plate data above.

Find: Approximate number of excess electrons n.

Solution: Using m = \[\frac {4}{3}\]πr³ρ, radius r ≈ 1.08 µm. Applying equation (2) with the given plate voltage and separation yields n ≈ 3.

Answer: The drop carries approximately 3 excess electrons.