- #1

Smacal1072

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I've been reading Griffiths E&M and Feynman Lectures (vol 2, E&M), and it made me think about a

*gedanken*that I'm trying to resolve. I'm pretty sure once I really peruse the chapters on the relativistic formulation it'll make sense, but I'm impatient

I have two identical positive charges, [itex]q_0[/itex] and [itex]q_1[/itex], on the x-axis.

[itex]q_a[/itex] is at [itex]-ax[/itex], travelling toward the origin at [itex]v_0[/itex].

[itex]q_b[/itex] is at [itex]+bx[/itex], travelling toward the origin at [itex]-v_0[/itex].

Since they are both positive charges, they will repulse each other due to the Coulomb field. But, from what I understand, the repulsive force is less than what they would experience in a static case, since at each moment in time it is the retarded (further away) position of the particles that influence each other. (I think the field in the x-axis is reduced by a factor of [itex]\gamma = \frac{1}{\sqrt{1-v^{2}_{0}/c^2}}[/itex])

This is what I don't understand: The faster the charges zoom toward each other, the less work they impart into the field potential energy (since it is reduced in the x-direction). If they return to their previous positions at a slower velocity, the return trip will result in them having more kinetic energy than they started with?