Question

What is the magnitude of electrical force of attraction between an copper nucleus (29 protons) and its innermost electron if the distance between them is 1.0 x 10-12 m?

Answer 1

The charge of the copper nucleus is 29 times the charge of one proton:
$Q=29 q= 29 \cdot 1.6 \cdot 10^{-19}C=4.64 \cdot 10^{-18}C$
the charge of the electron is
$e=-1.6 \cdot 10^{-19}C$
and their separation is
$r=1.0 \cdot 10^{-12} m$

The magnitude of the electrostatic force between them is given by:
$F=k_e \frac{Qe}{r^2}$
where $k_e$ is the Coulomb's constant. If we substitute the numbers, we find (we can ignore the negative sign of the electron charge, since we are interested only in the magnitude of the force)
$F=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(4.64 \cdot 10^{-18}C)(1.6 \cdot 10^{-19}C)}{(1.0 \cdot 10^{-12} m)^2}=6.68 \cdot 10^{-3} N$