The electron potential energy of a single electron (in $H^+_2$ ) at any point

Question

In my book it says:

We can choose a convenient reference by noting that the coulomb force at infinite distance is zero. It makes sense for this case, then, to choose $r=\inf$ as the vacuum level $E_v$: $E_p(r=\inf)=E_v$ (In here, p means potential)

When the nuclei are allowed to approach each other, an electron would be influenced by both nuclei according to Coulomb's law. The electron potential energy of a single electron (in $H^+_2$ ) at any point is now

$E_p=E_v-\frac{q^2}{(4\pi\epsilon_0)r^2_1}-\frac{q^2}{(4\pi\epsilon_0)r^2_2} $

where $r_1$ and $r_2$ are the distances between the electron and each of the two nuclei, respectively.

But i can't understand. As i know, the potential energy's formula is $E_p=\int_C\vec{F}\bullet\vec{dr}$

And, by superposition, we can get $E_p=E_1+E_2=\int_C(\vec{F}_1+\vec{F}_2)\bullet\vec{dr}$

where $\vec{F_1}$ and $E_1$ are force and potential energy due to one charge. And $\vec{F_2}$,$E_2$ are due to other charge.

So i think, $E_1=E_v-\frac{q^2}{(4\pi\epsilon_0)r^2_1} $,$E_2=E_v-\frac{q^2}{(4\pi\epsilon_0)r^2_2} $

Thus, it think $E_p=2E_v-\frac{q^2}{(4\pi\epsilon_0)r^2_1}-\frac{q^2}{(4\pi\epsilon_0)r^2_2} $, not $E_p=E_v-\frac{q^2}{(4\pi\epsilon_0)r^2_1}-\frac{q^2}{(4\pi\epsilon_0)r^2_2} $

What am I misunderstanding?

Answer 1

(a) First, you can add any constant you want to the energy without changing the physics, so even if you decide to add $E_v$ to each of $E_1$ and $E_2$ so that $E_p=E_1+E_2=2E_v+\cdots$, you could always decide to subtract $E_v$ from $E_p$ to get the expression in the book. So in that sense, there is not really a right answer to this question.

(b) If you want to interpret what the book is saying physically, then the correct energy superposition you should do is with three sources, not two:

\begin{eqnarray} {\rm Potential\ Energy} &=& {\rm Energy\ of\ Charge\ 1} + {\rm Energy\ of\ Charge\ 2} + {\rm Vacuum\ Energy} \\ E_p &=& E_1 + E_2 + E_v \\ &=& -\frac{q^2}{4\pi \epsilon_0 r_1}-\frac{q^2}{4\pi \epsilon_0 r_2} + E_v \end{eqnarray}