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+### Statement
+
+$14.3.12.$ [Insert the problem statement]
+
+### Solution
+
+Rest frame (S')
+The charge is at rest at the origin:
+
+$\mathbf{E}' = \frac{q}{r'^3}\,\mathbf{r}' \quad (\text{CGS}),\qquad \mathbf{B}' = 0$
+
+ Transformation to the laboratory (S)
+
+The system S' moves with velocity $\mathbf{v} = \beta c\,\hat{\mathbf{z}}$
+relative to S. For a boost along z, the parallel and perpendicular components transform as:
+
+$E_z = E'_z, \quad \mathbf{E}_\perp = \gamma\,\mathbf{E}'_\perp, \qquad
+B_z = 0, \quad \mathbf{B}_\perp = \gamma\,\frac{\mathbf{v}}{c}\times\mathbf{E}'$
+
+Change of coordinates
+
+The positions are related by$ x' = x,\; y' = y,\; z' = \gamma(z - vt)$
+ The distance in S' as a function of the coordinates in S is:
+
+$r' = r\,\frac{\sqrt{1-\beta^2\sin^2\alpha}}{\sqrt{1-\beta^2}}$
+
+where$ \alpha$ is the angle between $\mathbf{v}$ and$ \mathbf{r} in S$
+
+ Electric field in S
+
+Substituting the expression for r' into $\mathbf{E}'$ and applying the transformation of components, one obtains a radial field from the instantaneous position of the charge:
+
+$\boxed{\mathbf{E} = \frac{q}{r^3}\,\frac{1-\beta^2}{\bigl(1-\beta^2\sin^2\alpha\bigr)^{3/2}}\;\mathbf{r}} \quad (\text{CGS})$
+
+In SI, q is replaced by $\dfrac{q}{4\pi\varepsilon_0}$
+
+Magnetic field in S
+
+$\boxed{\mathbf{B} = \frac{\mathbf{v}}{c}\times\mathbf{E}} \quad (\text{CGS}),\qquad
+\boxed{\mathbf{B} = \frac{1}{c^2}\,\mathbf{v}\times\mathbf{E}} \quad (\text{SI})$
+
+#### Answer
+
+[Insert a concise answer or boxed result]