Photon, EM wave I have been told that an EM wave generate from an acceleration of a charged particle
As the wave propagates away from the charge the field strength of the charge gets weaker, so does the field strength of that EM wave maintain as it propagates?
Is the field strength of that EM wave proportional to the field strength of that charged particle?
 A: The Lienard Wiechert fields describe the field around an arbitrarily moving source point charge. The velocity is $ \boldsymbol{\beta}_s $ and the acceleration is $ \boldsymbol{\dot \beta}_s $ The fields are given by:
$$\mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \varepsilon_0} \left(\frac{q(\mathbf{n}_s - \boldsymbol{\beta}_s)}{\gamma^2 (1 - \mathbf{n}_s \cdot \boldsymbol{\beta}_s)^3 |\mathbf{r} - \mathbf{r}_s|^2} + \frac{q \mathbf{n}_s \times \big((\mathbf{n}_s - \boldsymbol{\beta}_s) \times \dot{\boldsymbol{\beta}_s}\big)}{c(1 - \mathbf{n}_s \cdot \boldsymbol{\beta}_s)^3 |\mathbf{r} - \mathbf{r}_s|} \right)_{t_r}$$
and
$$\mathbf{B}(\mathbf{r}, t) = \frac{\mathbf{n}_s(t_r)}{c} \times \mathbf{E}(\mathbf{r}, t)$$
where $$\boldsymbol{\beta}_s(t) = \frac{\mathbf{v}_s(t)}{c}$$ $$\mathbf{n}_s(t) = \frac{\mathbf{r} - \mathbf{r}_s(t)}{|\mathbf{r} - \mathbf{r}_s(t)|}$$ and $$\gamma(t) = \frac{1}{\sqrt{1 - |\boldsymbol{\beta}_s(t)|^2}}$$ (the Lorentz factor)
Notice that the velocity term is proportional to $|\mathbf{r} - \mathbf{r}_s(t)|^{-2}$. So this term of the field falls off as $1/r^2$. Since the energy density of the fields is equal to the square of the field, that means the energy density falls off as $1/r^4$. Since the volume increases as $r^2$ that means that the total energy falls off as $1/r^2$. In other words, most of the electromagnetic energy from a non-accelerating charge stays localized near the charge.
In contrast, the acceleration term for the field is proportional to $|\mathbf{r} - \mathbf{r}_s(t)|^{-1}$. That means the energy density falls off as $1/r^2$. Combined with the volume increasing as $r^2$ means that the total energy is constant as $r$ increases. This is why we say that energy is radiated away from an accelerating charge.
Both terms are proportional to $q$, so they are indeed proportional to the strength of the charge.
