= Solution
Write $\rho=|\mathbf x-\mathbf y|$. A rotationally symmetric <fundamental solution of the Laplace equation> is <harmonic> for $\rho>0$. The <Laplace equation in polar coordinates> reduces to $(\rho G')'=0$, so $G=A\log\rho+B$. Integrate $-\Delta G=\delta^{(2)}(\mathbf x-\mathbf y)$ over a disk of radius $\epsilon$ around the source. The <divergence theorem> gives
$$
1=-\int_{\partial D_\epsilon}\partial_\rho G\,ds=-2\pi A.
$$
Taking $B=0$ therefore gives
$$
\boxed{G(\mathbf x,\mathbf y)=-\frac1{2\pi}\log|\mathbf x-\mathbf y|.}
$$
Changing the reference length inside the logarithm merely adds a constant. Notice the sign convention: the potential equation in this question is $\Delta V=q\delta^{(2)}$, whereas this <Green function> satisfies $-\Delta G=\delta^{(2)}$. Thus the point-source potential used below is $-qG=+(q/2\pi)\log\rho$.
For the two insulating walls, successive <Euclidean reflections> in $x=b$ and $x=-b$ generate sources of strength $q$ at $(2nb,0)$ for every <integer> $n$. Only $n=0$ lies inside the physical strip. This is the <Neumann image lattice for a point source in a strip>. Set
$$
r_n=\sqrt{(x-2nb)^2+y^2}.
$$
The raw sum of logarithmic potentials diverges by additive constants. A well-defined potential is obtained by pairing opposite images and subtracting their constant reference terms:
$$
\boxed{V(x,y)=\frac q{2\pi}\left[\log r_0+\sum_{n=1}^\infty\left(\log r_n+\log r_{-n}-2\log(2nb)\right)\right]+C.}
$$
Each paired summand is $O(n^{-2})$ on compact subsets away from the source. The <series> therefore has local <uniform convergence>, as do the required derivative series. It gives
$$
\boxed{\frac{\partial V}{\partial y}=\frac q{2\pi}\sum_{n\in\mathbb Z}\frac{y}{(x-2nb)^2+y^2}.}
$$
In the corresponding $x$-derivative, the images pair as $n$ and $1-n$ at $x=b$, and as $n$ and $-1-n$ at $x=-b$. Their opposite horizontal separations make the derivatives cancel. Thus both homogeneous <Neumann boundary conditions> hold. The constructed solution is even in $y$; without a condition at infinity the side conditions alone would allow further homogeneous <harmonic functions>.
For fixed $y>0$, <uniform convergence> on $[-b,b]$ permits integration term by term. The $n$th term has integral
$$
\int_{-b}^b\frac{y\,dx}{(x-2nb)^2+y^2}=\arctan\frac{(2n+1)b}{y}-\arctan\frac{(2n-1)b}{y}.
$$
Here $\arctan$ is the ordinary real <inverse trigonometric function>, taking values in $(-\pi/2,\pi/2)$. Summing from $-N$ to $N$ makes a <telescoping series>:
$$
\int_{-b}^b\partial_yV\,dx=\frac q{2\pi}\lim_{N\to\infty}2\arctan\frac{(2N+1)b}{y}=\boxed{\frac q2}.
$$
The source flux splits equally between the upper and lower parts of the strip. The next two parts give the single-wall constructions that underlie the image lattice.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-2-2011-ib-neumann-images.png]
{title=Equal-sign image sources and the physical strip between two insulating boundaries}
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