Question

a small object has an excess of 5.00 x 109 electrons.an electron is placed in the field near this object and has 2.307E-18 joules of PE,find answer

Answer 1

$$E=\frac{kNe^2}{r}\\2.307\cdot10^{-18}=\frac{8.99\cdot10^{9}(5\cdot10^{9})(1.6\cdot10^{-19})^2}{r}\\r=0.499\ m$$