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To add to ACuriousMind's answer on the Liénard-Weichert potentials. You can put these formulas into an even more wonderfully descriptive form since you can derive Feynman's formula for the radiation from a moving charge: \vec{E} = ...

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The force does not change instantaneously, the correct way the electromagnetic field of (and thus the force exerted by) a moving electric charge is given by the Liénard-Wiechert potential, where one can see that the effect of the charge does not travel faster than light.

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The force is not propagated instantly, it takes time for the information to get from one point to another. You can treat that as istant if you are working with small enough distances and velocities but it's not. If you'll ever study field theory you'll meet retarded potentials, that are just this: the field propagates at the speed of light and it's no longer ...

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I believe b is the impact parameter. So b should be the perpendicular distance between the asymtote you drew and the fixed ion. The Wikipedia link has a picture that is pretty clear, although the picture in the link is illustrating Coulomb REPULSION, while your problem involves Coulomb attraction..

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The force is just the gradient of the potential energy. So it's not true that the energy is just force multiplied by $r$. It is the integral of the force w.r.t. $r$, which gives you the $1/r$ dependence. Edit: here "potential energy" and "interaction energy" are used interchangeably.

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It gives a quantity that is proportional to each of the charges involved independently, so that increasing either one increases the (magnitude of the) force. $k$ has units of its own, and these include a $\dfrac{1}{[C]^2}$ (inverse coulombs squared) term, so the units of the intermediate quantities in the equation don't lead to any non-physical results.

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Say you had two equal charges, $q$ a distance $x$ apart. $E=kq/r^2$ in general. You are right. The electric field can be defined as the force per unit charge. So a $100N/C$ field is one where 100 newtons of force would be exerted on a 1 Coulomb charged particle. so if we find the field due to the two particles right at the middle of $x$, at a point $x/2$ ...

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