Solution
= Solution
With no free surface charge, tangential $E$ and normal $D$ are continuous. The image ansatz gives
$$
\frac{q'}q=\frac{1-\epsilon_r}{1+\epsilon_r},\qquad \frac{q''}q=\frac{2\epsilon_r}{1+\epsilon_r}.
$$
At $z=0^-$, with $R=(r^2+d^2)^{1/2}$,
$$
E_r=\frac{2q}{1+\epsilon_r}\frac{r}{4\pi\epsilon_0R^3},\qquad E_z=\frac{2\epsilon_rq}{1+\epsilon_r}\frac{d}{4\pi\epsilon_0R^3}.
$$