ToyModelProblem

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  • #3463 Score: 0
    Brecht.Biesmans
    Participant

    The equation we have is composed of terms coming from the initial solution without the extra charge and an extra term stemming from the interaction of the extra charge with the dumbbell. Therefore, I would assume that a term is left out, namely the interaction of the extra charge with the outer charges which naturally would give rise to a positive contribution.

    #3486 Score: 0
    Stijn
    Participant

    I was thinking the same, but then I realized that taking that interaction into account doesn’t solve the paradox. The interaction with the outer charges will give a positive contribution to the energy, but the contribution will be smaller in magnitude than the interation with the dumbbell because the new charge in closer to the dumbbell than to the outer charges (l<d). So in the end the additional charge (-epsilon) will still lower the energy.

    #3560 Score: 0
    Brecht.Biesmans
    Participant

    I see what you mean and indeed think you are correct. having thought a bit about the problem, it might have something to do with the added extra charge. Since we are looking at descriptions of atoms and we know that they are normally neutrally charged, one can probably find the explanation in the total charge which is not zero anymore.

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