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Attractive Force of two charges in water

  1. May 20, 2010 #1
    Two test charges suspended in vacuum attract each other with a force of 1x10^-6 N. What will the attractive force be if the vacuum is replaced with water at 20 C? The permittivity of water at this temperature is 7.07x10^ -10 C^2/N-m^2.

    Fwater = (1/4pi*Ewater)(Q1Q2/r^2)

    Force is in Newtons
    Ewater is the permittivity of water
    Q1 and Q2 are the charges in Coulombs
    r is the separation in meters

    My attempt :(

    I am told that this can be simplified down to the following equation.

    Fwater = K / Fwater

    K is constant, calculated by rolling everything that doesn't change (4pi, the charges, the distance) into one number.

    K = 1/4pi*Ewater

    K = 1/1.256^-1*7.07 x 10^-10 C^2/N-m^2

    K = 1/8.88^-11 C^2/N-m^2

    K = 1.12^-10 C^2/N-m^2

    After finding K is this the correct formula?

    Fwater = (1.12^-10 C^2/N-m^2) / (7.07 x 10^-10 C^2/N-m^2)
    Last edited: May 20, 2010
  2. jcsd
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