Chapter 19 - Electric Charges, Forces, and Fields - Problems and Conceptual Exercises - Page 684: 21

$3.0\ electrons$

We know that $F=K\frac{q^2}{r}$ This can be rearranged as: $q=\sqrt{\frac{Fr^2}{K}}$ Now $N_e=\frac{q}{e}$ $\implies N_e=\frac{r}{e}\sqrt{\frac{F}{K}}$ We plug in the known values to obtain: $N_e=\frac{6.2\times 10^{-10}}{1.6\times 10^{-19}}\sqrt{\frac{5.4\times 10^{-9}}{8.99\times 10^9}}$ $N_e=3.0\ electrons$