= Solution
On the lower plane outside the boss, write $s$ for distance from the boss centre along the plane, so $s>a$. Differentiate the potential normally before putting $z=0$:
$$
\left.\frac{\partial\Phi}{\partial z}\right|_{z=0}
=E_0\left(1-\frac{a^3}{s^3}\right).
$$
Thus the normal field is $E_z=-E_0(1-a^3/s^3)$ and
$$
\boxed{\sigma_{\rm plane}=-\frac{\varepsilon_0V}{d}
\left(1-\frac{a^3}{s^3}\right),\qquad s>a.}
$$
The source uses $r$ for this distance on the plane. Near the rim it tends to zero, while far from the boss it recovers the uniform parallel-plate surface charge.
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