Question

A proton is held at a fixed position. An electron is placed 0.95 m away and released. Find the electron's speed when it's 7.0 mm from the proton.

Answer 1

Conserving energies,

K.E_i + P.E_i = K.E_f + P.E_f ----------(1)

initial kinetic energy of electron K.E_i = 0

initial potential energy of electron   = $P.E._i=\frac{1}{4\pi\epsilon_0}\frac{-(1.6\times10^{-19})^2}{0.95}$

final kinetic energy of electron K.E_f = (1/2) mv²

final potential energy of electron   = $P.E._f=\frac{1}{4\pi\epsilon_0}\frac{-(1.6\times10^{-19})^2}{7\times10^{-3}}$

using equation (1) and mass of electron m = 9.1 x 10^-31 kg

$\frac{1}{4\pi\epsilon_0}\frac{-(1.6\times10^{-19})^2}{0.95}=\frac{1}{4\pi\epsilon_0}\frac{-(1.6\times10^{-19})^2}{7\times10^{-3}}+\frac{1}{2}mv^2$

speed of electron v = 267.8 m/s