Chapter 19 - Electric Charges, Forces, and Fields - Problems and Conceptual Exercises - Page 684: 30

$240Km$

In the given scenario $F_e=F_g$ $\frac{Kqe}{r^2}=mg$ This can be rearranged as: $r=\sqrt{\frac{Kqe}{mg}}$ We plug in the known values to obtain: $r=\sqrt{\frac{8.99\times 10^9(0.35\times 10^{-9})(1.6\times 10^{-19})}{(9.109\times 10^{-31})(9.81)}}$ $r=2.4\times 10^5m= 240Km$