How Approximate is Electrostatics? Throughout High school, when we learnt electrostatics, we used the conservation of energy using the equation for potential:$$V=\frac{q}{4\pi\epsilon_0r}$$for solving questions where 2 isolated opposite charges initially at rest are allowed to move under mutual attraction. If we had to find velocity at some separation $r$ we would do something like this:$$\frac{1}{2}mv^2=\frac{q_1 q_2}{4\pi\epsilon_0}(\frac{1}{r_1}-\frac{1}{r_2})$$
But only after learning electromagnetic induction, that it struck me that we have failed to consider that the charges accelerate and hence will produce an electromagnetic field. Which in turn exerts a force on the charges.
Is is alright to just ignore the extra forces that tend to act?
How much is electrostatics valid?
 A: The following derivation assumes that at no point in the journey are the charges moving at relativistic speeds: $v\ll c$. It might also be the case that this result appears even with the charges moving at relativistic speeds, but I do not consider that here.
The magnetic field at a point $\vec{r}$ produced by a charge $q$ at a point $\vec{r_0}$ moving with a velocity $\vec{v}$ is 
$$\vec{B}(\vec{r})=\frac{\mu_0 q}{4\pi}\frac{\vec{v}\times\hat{r'}}{r'}$$
where $\vec{r'}=\vec{r}-\vec{r_0}$, $r'=|\vec{r'}|$ and 
$$\hat{r'}=\frac{\vec{r'}}{r'}$$ The Lorentz force is 
$$\vec{F}=q\vec{E}+q\vec{v}\times\vec{B}$$ 
The second term on the right-hand side of the Lorentz force equation is the force exerted by a magnetic field on a charge $q$, and that is what we are interested in. Let us now suppose we have two charged particles, one of charge $q_1$ and the other of charge $q_2$, moving with velocities $\vec{v_1}$ and $\vec{v_2}$, respectively. The magnetic force exerted on $q_2$ by the magnetic field produced by $q_1$ is 
$$\vec{F}_{1\rightarrow 2}=\frac{\mu_0}{4\pi}\frac{q_1q_2}{r'^2}\left(\vec{v_2}\times(\vec{v_1}\times\hat{r'})\right)$$
and if you wish to calculate $\vec{F}_{2\rightarrow 1}$ you would simply send $1\rightarrow 2$ and $2\rightarrow 1$, and you would then have have to adjust your $\hat{r'}$. To be precise, $\hat{r'}\rightarrow -\hat{r'}$. In general, these forces do not obey Newton's third law, so they are not equal and opposite and you should calculate both. However, notice that if you start with two charges at rest that move only because of the force that one exerts on the other, then $\hat{r'}$ and $\vec{v_i}$ are collinear: they are attracted or repelled so $\vec{v}$ is along the straight line from one charge to the other, but this is exactly what we have defined $\hat{r'}$ to be. Cross-products of collinear vectors are zero, so the magnetic force produced is zero. 
The result is that in this idealized case, there is no need to consider magnetic forces, because they are zero. If you are interested in electrodynamics, you might consider purchasing or borrowing Introduction to Electrodynamics The Fourth Edition by David J. Griffiths. This is a standard university textbook for studying electromagnetism. My calculations are based off of the material in this textbook. 
