Can we apply Coulomb's law for accelerating point charges when they are at rest?

Question

Can we apply Coulomb's law for accelerating point charges at the instant when they are at rest?

Answer 1

Not in general, accelerated charge produces different electric field than the Coulomb field. There is acceleration field, which has non-zero curl and thus is not a conservative field like the Coulomb field.

Only if the particle acceleration is small enough and we are interested in field close enough to the particle, then we can ignore the acceleration field component, because the Coulomb field is dominant. But in large enough distances the acceleration field always becomes dominant.

Answer 2

For an accelerating charge, even if it is temporarily at rest, you have to use retarded fields which depend on the past acceleration and velocity of the charge. This can be completely different than the Coulomb potential.

Answer 3

Classical electrodynamics gives the retarded electric field of a moving charge $q$ as (the primes indicating retarded values)

$$\vec{E}=\frac{q}{\lambda ^3\cdot d'^2}\cdot [\:\:(\vec{u'}-\frac{\vec{v'}}{c})(1-\frac{v'^2}{c^2}+\frac{\vec{d'}\vec{a'}}{c^2})\:\:-\frac{d'\vec{a'}}{c^2}\lambda\:\:]$$

where $\vec{d'}$ is the retarded vector from the charge to the observer (i.e. with the charge at the location as seen by the observer) , $\vec{u'}=\vec{d'}/\vec{d}$ the unit vector from the charge to the observer, $\vec{v'}$ and $\vec{a'}$ the retarded velocity and acceleration repectively, and

$$\lambda=1-\frac{\vec{u'}\vec{v'}}{c}$$

For $v'=0$ (as you assumed in your question) this turns into

$$\vec{E}=\frac{q} {d'^2}\cdot [\:\vec{u'} +\frac{\vec{u'}(\vec{d'}\vec{a'})}{c^2}\:\:-\frac{d'\vec{a'}}{c^2}\:\:]$$

The first term in the square brackets represents the usual Coloumb law (albeit relating to the retarded position) whereas the other two terms represent the acceleration terms. The latter add to zero however if $\vec{d'}$ and $\vec{a'}$ are parallel, so in this case we are just left with the usual form of the Coulomb law (for the retarded positon). If $\vec{d'}$ and $\vec{a'}$ are perpendicular on the other hand, the second acceleration term vanishes and we are just left with the acceleration field $-\frac{\vec{a'}}{c^2}\frac{q}{d'}$.

Having said this, the above is the usual result given by classical electrodynamics on the basis of the usual Liénard-Wiechert retarded potential. In a recent paper however, I have shown that the latter is incorrect as it violates charge conservation. The correct solution results just in a single acceleration term $-\frac{\vec{a'}}{c^2}\frac{q}{d'}$ in all situations, even if the velocity $\vec{v'}$ is not zero.