Solution
= Solution
For the grounded wall, use an image charge $-q$ at $(2b,0)$. Define the <Euclidean distances> $r=\sqrt{x^2+y^2}$ and $r_* =\sqrt{(x-2b)^2+y^2}$. The <method of images> gives
$$
\boxed{V_D(x,y)=\frac q{2\pi}\log\frac r{r_*},\qquad x<b.}
$$
The real source has the required <Dirac delta function> coefficient, and the image is outside the domain. On $x=b$, $r=r_*$, so $V_D=0$ exactly. This verifies the <Dirichlet boundary condition> rather than only its derivative.