Problem 19.25

19.25
The electric field at a distance r away from the center of a charge distribution is given by
$E = \frac{kq}{r^{2}}$
So, what is the radius? Well, the volume of the nucleus is 208 times the volume of one proton. So, the radius is
r_nucleus = (208)^1/3 r_proton
Plugging that into the equation for the electric field we find
$E = \frac{(9 \times 10^{9})(82 protons)(1.6 \times 10^{-19})} {\left((208)^{1/3} r_{proton}\right)^{2}}$
E = 2.33 N/C away from the nucleus.