Question

fe ln he C charge in the y- and x- cdirections Problem 3 Separating Charges In the previous homework, we considered an alpha particle being accelerated towards a gold nucleus. One way to "create" alpha particles is by placing neutral Helium into a uniform, constant electric field and "tearing" (accelerating) the electrons away from the nucleus. What is the minimum electric ficld strength required to initially "tear" them away? The distance between the electrons and Helium nucleus are approximately 31pm apart. Hint Consider the electrons and nucleus as two point-charges. Consider the attractive force kegpine

Answer 1

Electric field strength required to tear the electrons away from the Helium nucleus is $E=\frac{1}{4\pi\epsilon_0}\frac{2e}{r^2}$

$2e=2*1.6*10^{-19}\,C$ is the charge on Helium nucleus.

Distance between the Helium nucleus and the electron (point where electric field is to determined) is $r=31\,pm=31*10^{-12}\,m$

$E=8.99*10^9*\frac{2*1.6*10^{-19}}{(31*10^{-12})^2}=3*10^{12}\,N/C$