Question

An atomic nucleus has a charge of +40e. an electron is 10-9 m from the nucleus. what is the force on the electron?

Answer 1

The electron charge is equal to $-e=-1.6\cdot 10^{-19}C$. The atomic nucleus of the problem has a charge of $+40 e=40\cdot (1.6\cdot 10^{-19}C)=6.4\cdot 10^{-18}C$. The distance between the nucleus and the electron is $r=10^{-9}m$, so we can calculate the electrostatic (Coulomb) force between the two:
$F=k_e \frac{(-e)(+40e) }{r^2} =8.99\cdot 10^9 Nm^2C^{-2} \frac{(-1.6\cdot 10^{-19}C)(6.4\cdot 10^{-18}C)}{(10^{-9}m)^2} =$
$=-9.2 \cdot 10^{-9} N$
which is attractive, since the two charges have opposite sign.