$\overline{)\mathbf{F}\mathbf{=}{\mathbf{k}}_{\mathbf{e}}\frac{{\mathbf{Q}}_{\mathbf{1}}\mathbf{\xb7}{\mathbf{Q}}_{\mathbf{2}}}{{\mathbf{d}}^{\mathbf{2}}}}$

1 pm = 10^{-12} m

$\mathbf{d}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{\times}{\mathbf{10}}^{\mathbf{2}}\mathbf{}\overline{)\mathbf{pm}}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathbf{}\mathbf{m}}{\mathbf{1}\mathbf{}\overline{)\mathbf{pm}}}$

**d = 1.3x10 ^{-10} m**

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. What is the electrostatic force of attraction (in Newtons) between an electron and a proton that are separated by 1.3 x 10^{2} pm?

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