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Coulomb's law

  1. Sep 2, 2005 #1
    Can someone explain this law to me in simple terms? I know that it is the electrostatic force between 2 charged objects in relation to the quantity and inversly related to the square of distance F=K q1 q2 / d^2 ..but what if you have atoms that are spearated by a certain distance and have extra electrons? What happens to the force if you double the charges?
    Thank you.
  2. jcsd
  3. Sep 2, 2005 #2
    [tex]q_1[/tex] and [tex]q_2[/tex] represents the net charge of the atoms. For example, if you have [tex]Fe^2^+[/tex] and [tex]Cl^-[/tex], then [tex]q_1[/tex] and [tex]q_2[/tex] would be 2 and 1, respectively. The second number would be positive because it accepts a negative charge by default, making the answer a positive number.
  4. Sep 3, 2005 #3
    so, the formula I typed above is correct and could be used with 2 oxygen atoms, 3 cm apart and each have 2 extra electrons ? Wouldn't I get a -2 charge?
    (-2)(-2) / 3cm ??
  5. Sep 3, 2005 #4
    well, atoms for the most part are neutral. if you had two oxygen atoms that somehow gained electrons and each had a negative 2 charge that were placed next to eachother, they'd fly apart. negative * negative = positive -- repulsive force. negative * posative = negative -- attractive force.

    doubling the charges of each one -- look at the equation.

    F = (1/4*pi*epsilon-zero)*(q1)(q2)/d^2

    say each charge is e (charge of an electron), and you double each one (2e), the magnitude of the force will increase by a factor of 4 (2*2).

    dealing with atoms is a bit tricky, but to simplify it enough to say that there is a uniform sphereical charge distribution at the location of each ion would work. two negative ions placed near eachother in a closed system would accelerate away from eachother on the line that they create.
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