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

$4.7\times 10^{-10}m$

In the given scenario $F_{e}=F_c$ $\implies \frac{Ke^2}{r^2}=\frac{m_ev^2}{r}$ This simplifies to: $r=\frac{Ke^2}{m_eV^2}$ We plug in the known values to obtain: $r=\frac{8.99\times 10^9(1.6\times 10^{-19})^2}{9.11\times 10^{-31}(7.3\times 10^5)^2}$ $r=4.7\times 10^{-10}m$