What is their total energy when they are far apart

Two protons and an alpha particle are held at rest at the corners of an equilateral triangle whose side length is 8e-10 m. The particles are released and move apart. What is their total energy when they are far apart?

Could someone please give an explanation with equations and the answer? Thank you!

The system consisting of the two protons and the alpha particles has no external forces on it, and no non-conservative forces within it. Therefore, mechanical energy is conserved. So, the total energy when the particles are far apart is the same as the total energy when they are held in their triangle.

The electric potential energy of a system of particles is the sum of the electric potential energies of all the pairs. In this case, there are three pairs: the two protons, one proton with the alpha particle, and the other proton with the alpha particle.

With respect to zero potential energy at infinite separation, the electric potential energy of a pair of particles is

U = k(q1)(q2) / r

So the potential energies of the three pairs is

U(pp) = k(e)(e) / r

U(pα) = k(e)(2e) / r

U(pα) = k(e)(2e) / r

where k is the Coulomb constant, and r is the distance between the particles,
8×10^-10 m. The total energy is

U = k(e)(e) / r + k(e)(2e) / r + k(e)(2e) / r

U = ke² (1 + 2 + 2) / r = 5ke² / r

U = 5 (9.0×10^9 N∙m²/C²) (1.6×10^-19 C)² / (8×10^10 m) = 1.44×10^-18 J

You're welcome. I'm glad to help.

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