From left to right, electron $e_1$, $e_2$ and proton $p_1$, $p_2$. $r_0=0.529nm$

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The total energy is sum of energy require to bring each particle to its place. Take the place of $e_1$ is zero reference. Here is my doing

$$E_{e_1} = \frac{e^2}{4\pi\epsilon_0}(\frac{-1}{-r_0}+\frac{1}{-2r_0}+\frac{-1}{-3r_0})$$ $$E_{e_2} = \frac{e^2}{4\pi\epsilon_0}(\frac{1}{2r_0}+\frac{-1}{r_0}+\frac{-1}{-r_0})=\frac{e^2}{8\pi\epsilon_0r_0}$$ $$E_{p_1} = \frac{e^2}{4\pi\epsilon_0}(\frac{-1}{r_0}+\frac{-1}{-r_0}+\frac{-1}{-2r_0})=\frac{-e^2}{8\pi\epsilon_0r_0}$$ $$E_{p_2} = \frac{e^2}{4\pi\epsilon_0}(\frac{-1}{3r_0}+\frac{1}{2r_0}+\frac{-1}{r_0})$$

then the total energy will be zero. What did I do wrong? Does the distance don't require direction and always positive value or each pair of particle just need to measure once which mean

$$E_{e_1} = \frac{e^2}{4\pi\epsilon_0}(\frac{-1}{-r_0}+\frac{1}{-2r_0}+\frac{-1}{-3r_0})$$ $$E_{e_2} = \frac{e^2}{4\pi\epsilon_0}(\frac{-1}{r_0}+\frac{-1}{-r_0})=0$$ $$E_{p_1} = \frac{e^2}{4\pi\epsilon_0}\frac{-1}{-2r_0}=\frac{-e^2}{8\pi\epsilon_0r_0}$$

  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/193746/2451 $\endgroup$ – Qmechanic Jul 12 '15 at 21:25
  • $\begingroup$ Just to clarify, are you trying to find the ground state of the H2 molecule (which would involve some quantum mechanics), or are you trying to solve a 4-body electrostatics problem inspired by an H2 molecule? $\endgroup$ – ragnar Jul 14 '15 at 6:07
  • $\begingroup$ Quantum physics is taught in later lessons so I think it's just 4-body electrostatic problem. $\endgroup$ – aukxn Jul 14 '15 at 6:25

The right way to derive the total energy is to put the particles one by one.

In a space centered at $e_1$ and having nothing else, the energy to bring $p_1$ from far away to a distance $r_0$ is $E_{e_1,p_1}$, this you know how to calculate.

The energy to put $e_2$ at its position will have two terms coming from $e_1$ and $p_1$. By superposition, this energy will be the sum of the two individual contributions $E_{e_2,e_1} + E_{e_2,p_1}$.

Similarly the energy to put $p_2$ at that position will have the form $E_{p_2,e_1} + E_{p_2,p_1} + E_{p_2,e_2}$.

Finally, the total energy is the overall sum of these terms. I get $\frac{-7}{3} \frac{e^2}{4 \pi \epsilon_0 r_0}.$

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