Internal energy of three charged particles Three particles [latex] P_{1}, P_{2}, P_{3} [/latex] carry electric charges [latex] e_{1}, e_{2}, e_{3}[/latex] respectively. Find the internal potential energy Ψ.

In cgs/electrostatic units, the particles [latex] P_{1} \text { and } P_{2} [/latex] repel each other with the force [latex] h_{12}\left(r_{12}\right)=e_{1} e_{2}/\left(r_{12}\right)^{2}[/latex], where [latex]r_{12}[/latex] is the distance between [latex]P_{1} \text { and } P_{2} [/latex]. Their mutual potential energy is therefore [latex] H_{12}=-\int h_{12}\left(r_{12}\right) d r_{12}=-\int \frac{e_{1} e_{2}}{\left(r_{12}\right)^{2}} d r_{12}=\frac{e_{1} e_{2}}{r_{12}}.[/latex] The internal potential energy of the whole system is therefore [latex] \Psi=\frac{e_{1} e_{2}}{r_{12}}+\frac{e_{1} e_{3}}{r_{13}}+\frac{e_{2} e_{3}}{r_{23}}.[/latex]

Question 9.3: Internal energy of three charged particles Three particles P...

Classical Mechanics

Internal energy of three charged particles

Three particles $P_{1}, P_{2}, P_{3}$ carry electric charges $e_{1}, e_{2}, e_{3}$ respectively. Find the internal potential energy Ψ.

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In cgs/electrostatic units, the particles $P_{1} \text { and } P_{2}$ repel each other with the force $h_{12}\left(r_{12}\right)=e_{1} e_{2}/\left(r_{12}\right)^{2}$ , where $r_{12}$ is the distance between $P_{1} \text { and } P_{2}$ . Their mutual potential energy is therefore

$H_{12}=-\int h_{12}\left(r_{12}\right) d r_{12}=-\int \frac{e_{1} e_{2}}{\left(r_{12}\right)^{2}} d r_{12}=\frac{e_{1} e_{2}}{r_{12}}.$

The internal potential energy of the whole system is therefore

$\Psi=\frac{e_{1} e_{2}}{r_{12}}+\frac{e_{1} e_{3}}{r_{13}}+\frac{e_{2} e_{3}}{r_{23}}.$