Reply To: ToyModelProblem

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#3486 Score: 0

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.