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Consider a system of two electrically charged particles. Their charge doesn’t matter for my question but suppose they are both positive.

I am confused about what the correct way to calculate forces on particles is, based on information of the field, when those particles themselves contribute to the field and are at an infinite potential: What is wrong with the following argument?

Particle 1 will contribute to the electric field potential proportional to $\frac 1 {d}$, where $d$ is the distance from particle 1. But at particle 1’s position, the distance is $0$, so that the electric potential is infinite at that point. Hence, the gradient of the electric potential is undefined there, and we cannot compute a force on particle 1, even though we would want a force from particle 2 to be exerted.

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2 Answers 2

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When computing the force on particle 1, you only take into account the effect of particle 2. This is true for any number of charges: if you had several charges, the force on particle 1 would be the vector sum of the individual forces caused by particle 2, particle 3, particle 4, etc. In essence, particle 1 does not exert a force on itself due to its own electric field; all that matters is the field produced by the others.

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  • $\begingroup$ That's write. You might want to add the formula $F_{12} = q_1 \cdot E_2 \propto \frac{q_1 \cdot q_2}{r_{12}^2}$. $\endgroup$
    – Semoi
    Jun 15, 2020 at 17:32
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The potential you cite is defined as the work required to bring a unit of charge from infinity to within a distance,d, of a given point charge. If the distance is very small (as within an atomic nucleus), the energies involved can be quite large. The limits are still being explored.

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