Answer to Question #310549 in Electricity and Magnetism for Ase

Question

In a nucleus, two protons are 10^-15 m apart. a) What is their electrical potential energy? b) Knowing that they start from rest and are free to move, find their velocity when they are 4x10^-15 m apart.

Answer 1

(a)

$$E_{p1}=k\frac{e^2}{r_1}=9*10^9*\frac{(1.6*10^{-19})^2}{10^{-15}}=2.30*10^{-13}\:\rm J$$

(b)

$$E_{p2}=k\frac{e^2}{r_2}=9*10^9*\frac{(1.6*10^{-19})^2}{4*10^{-15}}=5.75*10^{-14}\:\rm J$$

$$v=\sqrt{2\Delta E_p/m}\\ =\sqrt{2*1.73*10^{-13}/1.67*10^{-27}}=1.0*10^7\:\rm m/s$$