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15 August, 14:03

An electron is in a vacuum near the surface of the Earth. Where should a second electron be placed so that the net force on the first electron to the other electron and to gravity, is zero?

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  1. 15 August, 17:29
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    There are two forces acting on the first electron.

    1) Force of gravity, Fg = mass of the electron * g

    mass of the electron = 9.11 * 10 ^ - 31 kg

    g = 9.8 m/s^2

    Fg = 9.11 * 10^ - 31 kg * 9.8 m/s^2

    2) Electrostatic force due to the second electron, Fe

    Use Coulomb Law.

    Fe = Coulomb constant * [charge of the electron*charge of the electron] / [separation]^2

    Coulomb constant = 9.0*10^9 N*m^2 / C^2

    Charge of the electron = 1.6 * 10 ^-19 C

    Separation = d

    Fe = 9.0 * 10^9 N*m^2/C^2 * [1.6*10^-19C]^2 / d^2

    Condition: net force = 0 = = > Fe = Fg

    Given that the second electron will exert a repulsion force, it has to be below (closer to the earth than) the first electron to counteract the atractive force of the earth.

    9.11 * 10^ - 31 kg * 9.8 m/s^2 = 9.0 * 10^9 N*m^2/C^2 * [1.6*10^-19C]^2 / d^2

    From which you can solve for d.

    d = sqrt { 9.0 * 10^9 N*m^2/C^2 * [1.6*10^-19C]^2 / (9.11 * 10^ - 31 kg * 9.8 m/s^2) }

    d = 5.08 m

    Then the second electron must be placed 5.08 m below the first electron.
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