Charge vs. Gravity: Calculating Electrostatic to Gravitational Force Ratio

[tony873004]

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

My effort:
$$\begin{array}{l} \frac{{\left( {\frac{{kq_{electron}^2 }}{{r^2 }}} \right)}}{{\left( {\frac{{GM_{electron}^2 }}{{r^2 }}} \right)}} \Rightarrow \\ \\ \frac{{kq_{electron}^2 }}{{GM_{electron}^2 }} = \frac{{8.988 \times 10^9 Nm^2 C^{ - 2} \times \left( { - 1.60217653 \times 10^{ - 19} C} \right)^2 }}{{6.97 \times 10^{ - 11} Nm^2 kg^{ - 2} \times \left( {9.10953826 \times 10^{ - 31} kg} \right)^2 }} = 3.9889 \times 10^{42} \\ \end{array}$$
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 textbooks 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
http://en.wikipedia.org/wiki/Fundamental_force
and 10^39 on
http://public.lanl.gov/alp/plasma/EM_forces.html
So I'm a bit less confident about my answer now.

 

[Andrew Mason]

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.

AM

 

[tony873004]

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