$\overline{)\frac{{\mathbf{F}}_{\mathbf{elec}}}{{\mathbf{F}}_{\mathbf{gravity}}}}$

$\overline{){{\mathbf{F}}}_{{\mathbf{elec}}}{\mathbf{=}}\frac{{\mathbf{kq}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}}$

$\overline{){{\mathbf{F}}}_{{\mathbf{gravity}}}{\mathbf{=}}{\mathbf{G}}\frac{{\mathbf{m}}^{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}}$

The electrostatic force (not energy) of attraction between two oppositely charged objects is given by the equation F = k_{e}(Q_{1}Q_{2}/d^{2}) where k_{e} = 8.99 x 10^{9} N m^{2}/C^{2}, Q_{1} and Q_{2} are the charges of the two objects in Coulombs, and d is the distance separating the two objects in meters. The charges Q_{1} and Q_{2} will have the magnitude 1.60 x 10^{–19} C. How many times larger is the electrostatic force of attraction than the force of gravity?

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