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

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?

• In your energy formulas, you meant to write $1/r$, not $1/r^2$, right? Commented Mar 17 at 15:39
• This would be clearer if you used the standard $T$ ($V$) for kinetic (potential) energy.
– JEB
Commented Mar 17 at 18:41

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