Deriving the Coulomb force equation from the idea of photon exchange?
The classical Coulomb potential can be recovered in the non-relativistic limit of the tree-level Feynman diagram between two charged particles.
Applying the Born approximation to QM scattering, we find that the scattering amplitude for a process with interaction potential $V(x)$ is
$$\mathcal{A}(\lvert p \rangle \to \lvert p'\rangle) - 1 = 2\pi \delta(E_p - E_{p'})(-\mathrm{i})\int V(\vec r)\mathrm{e}^{-\mathrm{i}(\vec p - \vec p')\vec r}\mathrm{d}^3r$$
This is to be compared to the amplitude obtained from the Feynman diagram:
$$ \int \mathrm{e}^{\mathrm{i}k r_0}\langle p',k \rvert S \lvert p,k \rangle \frac{\mathrm{d}^3k}{(2\pi)^3}$$
where we look at the (connected) S-matrix entry for two electrons scattering off each other, treating one with "fixed" momentum as the source of the potential, and the other scattering off that potential. Using the Feynman rules to compute the S-matrix element, we obtain in the non-relativistic limit with $m_0 \gg \lvert \vec p \rvert$
$$ \langle p',k \rvert S \lvert p,k \rangle \rvert_{conn} = -\mathrm{i}\frac{e^2}{\lvert \vec p -\vec p'\rvert^2 - \mathrm{i}\epsilon}(2m)^2\delta(E_{p,k} - E_{p',k})(2\pi)^4\delta(\vec p - \vec p')$$
Comparing with the QM scattering, we have to discard the $(2m)^2$ as they arise due to differing normalizations of momentum eigenstate in QFT compared to QM and obtain:
$$ \int V(\vec r)\mathrm{e}^{-\mathrm{i}(\vec p - \vec p')\vec r}\mathrm{d}^3r = \frac{e^2}{\lvert \vec p -\vec p'\rvert^2 - \mathrm{i}\epsilon}$$
where Fourier transforming both sides, solving the integral and taking $\epsilon \to 0$ at the end will yield
$$ V(r) = \frac{e^2}{4\pi r}$$
as the Coulomb potential.
There is the (non-genetal) relation between the free energy of interacting of two currents $J^{a}, J^{b}$ and the propagator: $$ U = -\frac{1}{2} \int d^{4}xd^{4}y J^{a}(x) D_{ab}(x - y)J^{b}(y). $$ It's not general, but it realizes the simple example which can help you to understand how to get the expression for force.
The structure of field which causes structure of propagator helps us to get the expression for the force. For example, for intercation via scalar field (by setting $D_{ab}(x - y) = \frac{1}{p^{2} - m^{2}}$) after simple transformations for "point-like" currents $J(x) = \delta (\mathbf x - \mathbf x_{0})$ we can get $$ U = -\frac{1}{4 \pi |\mathbf r|}e^{-mr}. $$ For case $m = 0$ we get the Coulomb-like law of interaction.
Absolutely the same thing you may do with the case of EM field (in Feynman gauge $D_{\mu \nu} = -\frac{g_{\mu \nu}}{p^{2}}$).
If you need some explicit derivation I'll give it later.