Question

A helium nucleus is spherical with a radius of about 1.9 × 10−15 m. The nucleus contains 2 protons (and can additionally have 2 neutrons). Because of their motions the protons can be considered on average to be uniformly distributed throughout the nucleus. Based on the net flux at the surface of the nucleus, calculate the divergence of the electric field as electric flux per unit volume.

Answer 1

To be able to solve this problem it is necessary to take into account the concepts of Electric Flow

on a closed surface.

By definition we know that the flow across a closed surface is

$\Phi = \oint\vec{E}d\vec{A}$

And we know that the Coulumb's law says

$\vec{E} = \frac{q}{4\pi\epsilon_0r^3}\vec{r}$

We can replace, so

$\Phi = \oint \frac{q}{4\pi\epsilon_0r^3}\vec{r}d\vec{A}$

Solving the integral we reach the expression,

$\Phi = \frac{q}{\epsilon_0}$

The charge of a proton is $q=1.6*10^{-19}C$, but we have 2 protons, then

$q=(2)*1.6*10^{-19}C=3.2*10^{-19}C$

The constant of permittivity of free space $(\epsilon_0$) is $8.85 x 10^{-12}$H/m

Replacing we have that the flux is

$\Phi = \frac{q}{\epsilon_0}$

$\Phi = \frac{3.2*10^{-19}}{8.85 x 10^{-12}}$

$\Phi = 3.62*10^{-8}N m^2 C^{-1}$

We can also calculate the Volume of a Sphere, that is

$V=\frac{4}{3} \pi r^3 =\frac{4}{3}\pi (1.9*10^{-15})^3$

$V=2.8730*10^{-44}m^3$

Through this value is easy now calculate the dirvegence of the electric field, which is given by,

$\xi = \frac{\Phi}{V}$

$\xi = \frac{3.62*10^{-8}}{2.8730*10^{-44}}$

$\xi = 1.2*10^36m^{-1} C^{-1}$