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Consider two particles $A$ with mass $m$ and $B$ with mass $M$ ($M>> m$), and each particle has charge $Q$. Suppose they are at a distance $r$ ($r$ is of the order $10^{10}m$) apart at a time $t$. Due to electric force of attraction both particle comes closer to each other.

The force exerted on particle $B$ due to particle $A$ at a time $t$ will be due to the field exerted by $A$ at a time $t-\Delta t$, when particle $A$ was a distance of $r+\Delta r$ from the position of particle $B$ at time $t$ (this is because electric field travels at the speed of light and not instantaneously). Therefore the force experienced by particle $B$ will be $\dfrac{Q^2}{4\pi\epsilon_0(r+\Delta r)^2}$.

Similarly the force exerted on particle $A$ due to particle $B$ at a time $t$ will be due to the field exerted by $B$ at a time $t-\Delta t'$, when particle $B$ was a distance of $r+\Delta r'$ from the position of particle $A$ at time $t$. Hence the force experienced by particle $A$ will be $\dfrac{Q^2}{4\pi\epsilon_0(r+\Delta r')^2}$.

Force on the two particles is approximately equal (because $\Delta r$ ,$\Delta r'<<r$) therefore $\Delta r'<<\Delta r~$ because the mass of particle $B$ is much larger than the mass of particle $A$. Then this would lead to the conclusion that force exerted on particle $A$ due to particle $B$ is not equal to the force exerted on particle $B$ due to particle $A$ at time $t$. So does this situation violate Newtons Third Law?

Asher2211
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Yes! Electromagnetic forces violate Newton's Third Law. The reason is that the Third Law is essentially a statement about momentum conservation. Notice that ensuring that $$\sum_i \mathbf{F}_i = \mathbf{0}$$ means ensuring $$\frac{\mathrm{d}}{\mathrm{d} t} \mathbf{P}_{\text{tot}} = \mathbf{0}.$$

However, the electromagnetic field itself carries momentum as well. Hence, electromagnetic momentum also enters the equation and ends up spoiling the Third Law: there can be different variations of momenta in the different particles because the momentum in the electromagnetic field accounts for the difference and ensures momentum is overall conserved.

Níck Aguiar Alves
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If you want to take into account (non-infinite) speed of light you are definitely beyond the scope of Newton laws. So it definitely does not apply here.

vtm11
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Your, excellent, question as such is inconsistent as Newton mechanics can only be applied at speeds $v\ll c$. However, it is really about momentum conservation, which holds at any speed.

Indeed the Coulomb force is the retarded force. However, the two retarded forces obey Newton's first law, as the retarded time is the same for both particles. See the illuminating description in Griffith's Electrodynamics, Ch.8. What is not included in your question is the magnetic force. The momentum given by the Poynting vector does obey Newton's third law, but only if the field contributions are included. Besides non-radiative contributions, which in my opinion are paradoxical, Poynting's vector distribution also contain radiative contributions, which describe the momentum carried off by radiation. The latter of course violate the simple two particle picture of the original question.

my2cts
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