Imagine a particle with charge $q$ at rest at the origin.
It is surrounded by a concentric spherical insulating shell, also at rest, with charge $Q$ and radius $R$.
At time $t=0$ I apply a constant horizontal acceleration $\mathbf{a}$ to the particle.
The electromagnetic field spreads out radially in all directions from the charge at the speed of light (Appendix 1).
The momentum density in the field is directed radially and also spreads out from the charge at the speed of light (Appendix 2).
As the momentum density at opposite field points cancel the total momentum in the field is always zero (Appendix 2).
At time $t=R/c$ the field reaches the charged shell and imparts a horizontal force on it (Appendix 3).
What balances this force if the rate of change of horizontal momentum in the field is always zero?
I think it has to be balanced by an opposite force on the particle that develops due to an advanced electromagnetic interaction from the spherical shell at time $t$ backwards in time to the particle at time $t_r$.
Appendix 1: EM field of an accelerated charge with velocity zero
Specializing the Lienard-Wiechert fields for an accelerating charge with zero velocity we find that the electric field is given by:
$$\mathbf{E}(\mathbf{r},t)=\frac{q}{4\pi\epsilon_0}\left(\frac{\mathbf{\hat r}}{r^2}+\frac{\mathbf{\hat r}\times(\mathbf{\hat r}\times \mathbf{a})}{c^2r}\right)_{t_r}$$
and the magnetic field is given by:
$$\mathbf{B}(\mathbf{r},t)=\frac{\mathbf{\hat r}(t_r)}{c}\times\mathbf{E}(\mathbf{r},t) $$
where $\mathbf{\hat r}$ is the unit vector in the direction of the field point and the retarded time $t_r$ at the charge $q$ is given by:
$$t_r=t-\frac{r(t_r)}{c}.$$
Appendix 2: Zero total momentum in EM field from an accelerated charge
The momentum density $\mathbf{g}(\mathbf{r},t)$ in the electromagnetic field is given by:
$$\mathbf{g}(\mathbf{r},t)=\epsilon_0\mathbf{E}(\mathbf{r},t)\times\mathbf{B}(\mathbf{r},t).$$
The momentum density is directed radially so that:
$$\mathbf{g}(\mathbf{r},t)=\frac{\epsilon_0E^2}{c}\mathbf{\hat r}$$
where the magnitude of the electric field $E$ is given by
$$E = -\frac{q a \sin \theta}{4\pi\epsilon_0c^2r}$$
and $\theta$ is the angle between acceleration $\mathbf{a}$ and radial direction $\mathbf{\hat r}$.
As $\sin(\pi-\theta)=\sin(\theta)$ then we have:
$$\mathbf{g}(-\mathbf{r},t)=-\mathbf{g}(\mathbf{r},t)$$
so that the momentum density at opposite field points cancel.
Thus the total momentum in the fields is always zero.
Appendix 3: Total force on charged spherical shell from accelerated charge
The electric field at the spherical shell consists of a static radial part and an acceleration part. The total force from the static radial part will cancel out in opposite pairs so that we only have to worry about the acceleration part.
The total horizontal force on the spherical shell with charge $Q$ and radius $R$ is given by:
$$F = \int_{sphere} E\sin \theta\ \sigma\ dA$$
where
$$E = -\frac{q a \sin \theta}{4\pi\epsilon_0c^2R}$$
$$\sigma=\frac{Q}{4\pi R^2}$$
$$dA = 2 \pi R^2 \sin \theta\ d\theta$$
Performing this integration over the sphere using:
$$\int_0^\pi \sin^3 \theta\ d\theta=\frac{4}{3}$$
We find that the total horizontal force is given by:
$$F = -\frac{2}{3}\frac{qQa}{4\pi\epsilon_0c^2R}.$$
