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26 March, 06:14

In the early 1900s, Robert Millikan used small charged droplets of oil, suspended in an electric field, to make the first quantitative measurements of the electron's charge. A 0.68-μm-diameter droplet of oil, having a charge of + e, is suspended in midair between two horizontal plates of a parallel-plate capacitor. The upward electric force on the droplet is exactly balanced by the downward force of gravity. The oil has a density of 860 kg/m3, and the capacitor plates are 4.0 mm apart. Part A What must the potential difference between the plates be to hold the droplet in equilibrium?

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  1. 26 March, 06:30
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    The gravitational force on the oil droplet is given by:

    F = mg

    F = gravitational force, m = mass, g = gravitational acceleration

    The mass of the droplet is given by:

    m = pV

    m = mass, p = density, V = volume

    The volume of the droplet is given by:

    V = 4π (d/2) ³/3

    V = volume, d = diameter

    Make some substitutions:

    F = 4pgπ (d/2) ³/3

    Given values:

    p = 860kg/m³, g = 9.81m/s², d = 0.68*10⁻⁶m

    Plug in and solve for F:

    F = 4 (860) (9.81) π (0.68*10⁻⁶/2) ³/3

    F = 1.389*10⁻¹⁵N

    The magnitude of the electric force is equal to the magnitude of the gravitational force. The electric force is given by:

    F = Eq

    F = electric force, E = electric field, q = droplet charge

    The electric field between the capacitor plates is given by:

    E = ΔV/d

    E = electric field, ΔV = potential difference, d = plate separation

    Make a substitution:

    F = ΔVq/d

    Given values:

    F = 1.389*10⁻¹⁵N, q = + e = 1.6*10⁻¹⁹C, d = 4.0*10⁻³m

    Plug in and solve for ΔV:

    1.389*10⁻¹⁵ = ΔV (1.6*10⁻¹⁹) / (4.0*10⁻³)

    ΔV = 34.73V

    Round ΔV to 2 significant figures:

    ΔV = 35V
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