JEE Challenger
More from Electric Charges and Fields
PhyMedAdv 2020 • Paper 2

Number of Electrons in Floating Oil Drop Millikan Experiment

As shown in the given figure, two large circular plates separated by 0.01 m0.01\text{ m} are attached to a voltage source via a switch. Charged oil droplets with a density of 900 kg m−3900\text{ kg m}^{-3} enter through a small aperture at the center of the upper disc. After reaching terminal velocity, the switch is closed, establishing a potential difference of 200 V200\text{ V} between the plates. Consequently, a specific oil droplet of radius 8×10−7 m8 \times 10^{-7}\text{ m} remains stationary and floats between the plates. Determine the total number of electrons on this drop. (Ignore buoyancy; take g=10 ms−2g = 10\text{ ms}^{-2} and elementary charge e=1.6×10−19 Ce = 1.6 \times 10^{-19}\text{ C}).

Question Diagram 1
Official Numerical Answer6

Topics & Concepts

Electric Charges and FieldsElectric FieldElectric Potential

Step-by-Step Solution

To find the total number of electrons present on the floating oil drop, we analyze the forces acting on the drop at equilibrium.

1. Forces Acting on the Oil Drop

When the switch is closed, an electric field is established between the circular plates. For the oil drop to remain stationary and float, the downward gravitational force (FgF_g) must be balanced by the upward electrostatic force (FeF_e).

  • Gravitational Force (FgF_g): Fg=mg=ρ⋅Vdrop⋅gF_g = m g = \rho \cdot V_{\text{drop}} \cdot g where:

    • Density of oil, ρ=900 kg m−3\rho = 900 \text{ kg m}^{-3}
    • Volume of the spherical drop, Vdrop=43πr3V_{\text{drop}} = \frac{4}{3} \pi r^3
    • Acceleration due to gravity, g=10 m s−2g = 10 \text{ m s}^{-2}
    • Radius of the drop, r=8×10−7 mr = 8 \times 10^{-7} \text{ m}

    Substituting the given values into the volume equation: Vdrop=43π(8×10−7)3=43π(512×10−21) m3=2048π3×10−21 m3V_{\text{drop}} = \frac{4}{3} \pi (8 \times 10^{-7})^3 = \frac{4}{3} \pi (512 \times 10^{-21}) \text{ m}^3 = \frac{2048 \pi}{3} \times 10^{-21} \text{ m}^3

    Thus, the mass of the drop is: m=ρ⋅Vdrop=900×(2048π3×10−21)=6.144π×10−16 kgm = \rho \cdot V_{\text{drop}} = 900 \times \left(\frac{2048 \pi}{3} \times 10^{-21}\right) = 6.144 \pi \times 10^{-16} \text{ kg}

    The gravitational force is: Fg=mg=(6.144π×10−16)×10=6.144π×10−15 NF_g = m g = (6.144 \pi \times 10^{-16}) \times 10 = 6.144 \pi \times 10^{-15} \text{ N}

  • Electrostatic Force (FeF_e): Fe=qE=(ne)EF_e = q E = (n e) E where:

    • nn is the total number of electrons
    • e=1.6×10−19 Ce = 1.6 \times 10^{-19} \text{ C} is the elementary charge
    • Electric field, E=Vd=200 V0.01 m=2×104 V m−1E = \frac{V}{d} = \frac{200 \text{ V}}{0.01 \text{ m}} = 2 \times 10^4 \text{ V m}^{-1}

    Substituting ee and EE: Fe=n×(1.6×10−19)×(2×104)=n×3.2×10−15 NF_e = n \times (1.6 \times 10^{-19}) \times (2 \times 10^4) = n \times 3.2 \times 10^{-15} \text{ N}

2. Condition for Equilibrium

Equating the upward electrostatic force to the downward gravitational force: Fe=FgF_e = F_g n×3.2×10−15=6.144π×10−15n \times 3.2 \times 10^{-15} = 6.144 \pi \times 10^{-15}

Solving for nn: n=6.144π3.2=1.92πn = \frac{6.144 \pi}{3.2} = 1.92 \pi

Using π≈3.1416\pi \approx 3.1416: n=1.92×3.1416≈6.03n = 1.92 \times 3.1416 \approx 6.03

Since the number of electrons must be an integer, rounding to the nearest whole number yields: n=6n = 6

Final Answer: The total number of electrons on the drop is 6.

PreviousNext