Q&A

# What's the equation of kinetic energy of charged particle?

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I was looking for equation of motion. I came up with a solution but it doesn't satisfy me. Cause I was trying to find motion of that particle using Lagrangian. We know that $$W=\int \vec F\cdot d\vec l$$

$W=T$ for some cases and I came up with $T=qV$. In Euler-Lagrange, kinetic energy has velocity as function, in $T=qV$ there's no velocity directly, the equation actually tells me that particle is gaining kinetic energy from potential (more precisely, potential is converting into kinetic). At first sight, I wrote that $T=\frac{1}{2}m\ddot{r}^2$ what if particle is massless(?) so it's not very helpful. Where I took $$L=0.5m\ddot{r}^2-\frac{1}{4\pi\epsilon_0}{q}{r}$$ if I try solve Euler-Lagrange using that Lagrangian then I get $m\ddot{r}=\vec E$. How force is equal to electric field? It totally doesn’t make any sense to me, their dimension doesn't match either. None of these equation satisfy me. what I think that is I got wrong result for taking kinetic energy which doesn’t apply to charged particle. So what's the kinetic energy of charged particle?

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Kinetic energy of any particle (who has mass) is $$T=\frac{1}{2} m\ddot{x}^2$$

OP had took potential as potential energy, that was wrong. $$U=-\int \vec F\cdot d\vec l$$ For the case, The force was $$F=\frac{1}{4\pi\epsilon_0} \frac{Qq}{r^2}\hat r$$

So the lagrangian is $$L=\frac{1}{2}m\dot{x}^2-\frac{1}{4\pi\epsilon_0} \frac{Qq}{r}$$ Now you can get a satisfied answer.

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