0

There is a post where CR Drost calculates the interaction engery in the center of the sun using only the density $\rho$ of the sun at it's center.

Assuming that all the mass comes from protons we get that for a given density $\rho=150g/cm^3$ every $cm^3$ contains about $N^3=150g/m_{p}$ protons, where $m_p$ is the protons mass. He looks at how many protons sit on an edge of the 1 $cm^3$ cube. This equals $N$ protons per $cm$. This makes sense but the following does not.

He then calculates a voltage and multiplies it with the charge $e$ of the proton to get the interaction energy. The voltage is calculated the following way: $$ 65 V\approx \frac{1}{4\pi\epsilon_0}\frac{e N}{d},\quad d=1\, cm$$

I see that 65 $eV$ this is the electric potential energy made up by $N+1$ point charges with equal charge $e$ but what has this to do with interaction?

ty.
  • 219

1 Answers1

0

The electric potential voltage exerts a force on each charged proton. The energy is the work done to bring the protons together. This results in the multiplication of the voltage (due to all protons in the cube) by the proton charge with which it is interacting.

JMLCarter
  • 4,462
  • I don't know why this should make sense. Why should there be a force on the particle? If I look at a proton in the center of the sun, and a proton at some position $\vec{r}$ then there is a third proton at position $-\vec{r}$. So the forces will cancel each other out. The proton in the center should not be pushed around. Where am I wrong with this thinking? – ty. Oct 03 '17 at 17:57
  • Although the protons have no relative motion on average, one can theorise that during star formation they moved into proximity from a nebula. Regardless of the exact process of formation, it takes work to bring like charges together . That is where the energy comes from that pushes the charges apart (in this case unsuccessfully due to gravitational considerations) – JMLCarter Oct 04 '17 at 22:22