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Charge vs. gravity

  1. Sep 10, 2006 #1


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    Calculate the ratio of the electrostatic force to the gravitational force between two electrons.

    My effort:
    \frac{{\left( {\frac{{kq_{electron}^2 }}{{r^2 }}} \right)}}{{\left( {\frac{{GM_{electron}^2 }}{{r^2 }}} \right)}} \Rightarrow \\
    \frac{{kq_{electron}^2 }}{{GM_{electron}^2 }} = \frac{{8.988 \times 10^9 Nm^2 C^{ - 2} \times \left( { - 1.60217653 \times 10^{ - 19} C} \right)^2 }}{{6.97 \times 10^{ - 11} Nm^2 kg^{ - 2} \times \left( {9.10953826 \times 10^{ - 31} kg} \right)^2 }} = 3.9889 \times 10^{42} \\
    Sorry for small tex. This should be better:
    My units cancel nicely, so I assume I did it right. However, the book doesn't give the formula I used: F=kqq/r^2

    It gives
    F = Q\left( {E + v \times B} \right)

    Most text books give you the formulas you need to do the questions at the end of the chapter. So I figure I'd double check here.

    In Wikipedia I looked up the mass and charge of the electron. It gave
    9.109 3826(16) × 10−31 kg
    −1.602 176 53(14) × 10−19 C
    What do the (16) and (14) in these numbers mean?

    *** Edit
    I'm seeing an answer of 10^36 on
    and 10^39 on
    So I'm a bit less confident about my answer now.
    Last edited: Sep 10, 2006
  2. jcsd
  3. Sep 11, 2006 #2

    Andrew Mason

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    Close but your value for G is too high. G = 6.67x10^11 Nm^2/kg^2. I get 4.17x10^42
    Your approach and method is correct.

  4. Sep 11, 2006 #3


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    Thanks for catching that. I thought I had that memorized.

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