A 2D electrostatic (i.e. harmonic potential) boundary value problem is shown in the figure. The solid lines are conductors (all are parallel), the two conductors with potential $V$ are infinitely long, while the grounded (i.e. with potential $0$) conductors are semi-infinite. $\Delta,l$ are the relevant length parameters. The electrostatic potential $\varphi(x,y)$ satisfies the Laplace equation
$$(\partial_x^2+\partial_y^2)\varphi(x,y)=0,$$
subject to the boundary condition shown in figure.
Question: could the potential at the center of the figure $\varphi(0,0)$ be analytically solved?
Firstly, due to symmetry, we can consider the problem in $0<y<l$ and impose the condition $\varphi_y = 0$ on $y=0$ in the gap. This will result in a mixed boundary value problem. One way to deal with mixed boundary value problems is to map them to regions where each part of the boundary is mapped to a different boundary, and then applying a standard PDE method like separation of variables, e.g.
For simplicity, scale $x,y$ by a factor of $2l/\pi$ so that the upper boundary is at $y=\pi/2$ and the gap occurs in the interval $(-a,a)$, where $$ a = \frac{\Delta}{2} \frac{\pi}{2l} = \frac{\Delta\pi}{4l}. $$ Define $\varphi(x,y) = \frac{2}{\pi} V[y + u(x,y)]$, so that $u$ is harmonic with boundary conditions $$u(x, \pi/2) = 0, \qquad u(x, 0) = 0 \ (|x|>a), \qquad u_y(x,0) = -1 \ (|x| < a). $$
Now for some conformal mapping. Let $z=x + iy$. Transform the strip $0<y<\pi/2$ to the upper half-plane using the transformation $\omega = \tanh(z)$, where $x \in (-a,a)$ becomes $\omega \in (-A,A)$, $A = \tanh a$. We can then transform the upper half-plane to the semi-infinite strip $-\pi/2 < \xi < \pi/2$, $\eta > 0$, using $$ \zeta = \xi + i \eta = \sin^{-1}(\omega/A) = \sin^{-1}(\tanh(z)/A)$$ or $$z = \tanh^{-1}(A\sin\zeta).$$ Under this transformation, the gap $x \in (-a,a)$ has been transformed to $\xi \in (-\pi/2, \pi/2)$. The boundary condition $u_y=1$ is affected by the mapping: $$ u_\eta(\xi,0) = -\Im(w_\zeta) = -\Im(w_z z_\zeta) = -\Im\left(w_z \frac{A\cos\zeta}{1-A^2\sin^2\zeta} \right) = -\frac{A\cos\xi}{1-A^2\sin^2\xi} $$ (here $w$ is the analytic function with real part $u$). The other boundary conditions have been transformed to the conditions $u(\pm \pi/2, y) = 0$.
Now the solution to Laplace's equation in the $\zeta$ domain is found using separation of variables: $$ u(\xi,\eta) = \sum_{n=1}^\infty -\frac{B_n}{2n-1} e^{-(2n-1)\eta} \cos((2n-1)\xi) $$ where $B_n$ are the Fourier coefficients of the boundary condition: $$ B_n = -\frac{4}{\pi}\int_0^{\pi/2} \frac{A\cos\xi}{1-A^2\sin^2\xi}\cos((2n-1)\xi) \, d\xi. $$ This seems somewhat intractable though. If particularly interested in what happens at $\zeta = 0$ (corresponding to $x=y=0$), then \begin{align*} u(0,0) &= \sum_{n=1}^\infty -\frac{B_n}{(2n-1)} \\ &= \sum_{n=1}^\infty \frac{4}{(2n-1)\pi}\int_0^{\pi/2} \frac{A\cos\xi}{1-A^2\sin^2\xi}\cos((2n-1)\xi) \, d\xi \\ &= \frac{4}{\pi} \int_0^{\pi/2} \frac{A\cos\xi}{1-A^2\sin^2\xi} \left[\sum_{n=1}^\infty \frac{\cos((2n-1)\xi)}{2n-1}\right] \, d\xi \\ &= \frac{1}{\pi}\int_0^{\pi/2} \frac{A\cos\xi}{1-A^2\sin^2\xi} \cdot \log\left(\frac{1+\cos\xi}{1-\cos\xi}\right) \, d\xi \\ &= -\frac{2}{\pi}\int_0^{\pi/2} \frac{\tanh^{-1}(A\sin\xi)}{\sin\xi} \, d\xi \end{align*} (the last of these from integration by parts). This integral seems hard but is equal to $$ u(0,0) = \sin^{-1} A $$ (the only way I could see to establish this was to differentiate under the integral sign with respect to $A$ to obtain the derivative of the inverse sine). Thus returning to the original variables: $$ \varphi(0,0) = \frac{2V}{\pi}\sin^{-1}\left[\tanh\left(\frac{\Delta\pi}{4 l}\right)\right]. $$
The relative simplicity of $u(0,0)$ suggests to me that there might be a simple closed form solution for $u$ in the $\zeta$ plane, but it's not immediately obvious to me what it is. That would be a more elegant way of going about it if it were true.