Chapter 19 - Electric Charges, Forces, and Fields - Problems and Conceptual Exercises - Page 684: 22

$r=31.8\ cm$

We know that $F=K\frac{q_1q_2}{r^2}$ $F=K\frac{q_1(\sigma A)}{r^2}$ $F=K\frac{q_1\sigma (4\pi R^2)}{r^2}$ This simplifies to: $r=\sqrt{\frac{4\pi kq\sigma R^2}{F}}$ We plug in the known values to obtain: $r=\sqrt{\frac{4\pi(8.99\times 10^9)(1.95\times 10^{-6})(12.1\times 10^{-6})(0.442)^2}{46.9\times 10^{-3}}}$ $r=0.318\ m$ $r=31.8\ cm$