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For steady inviscid flow, Bernoulli's equation along a streamline isWriting steady Euler flow in divergence form and integrating over $V$ givesby incompressibility and the divergence theorem.

Using $\rho ={\epsilon }_0\nabla \cdot E$, $E=-\nabla \varphi$, integration by parts, and $\varphi =0$ on the boundary givesPut $C=q/(4\pi {\epsilon }_0)$. Gauss's law gives the radial fieldTaking zero potential outside,The charge formula givesThe field formula gives the same result after integrating $4\pi r^2{E}_r^2$ over the two nonzero annuli.

Write $E=-{\hslash }^2{\kappa }^2/(2m)$. An even bound state is proportional to $\cos (kx)$ inside the well and to $e^{-\kappa |x|}$ outside, whereContinuity of the logarithmic derivative at $a$ givesFor sufficiently small $a$, $ka$ lies in the first monotone branch, so this equation has exactly one positive root and the even state is unique up to scale.

As $a\downarrow 0$, $\kappa \sim k^{2}a\to m/{\hslash }^2$. HenceThe wells converge distributionally to $V_0(x)=-\delta (x)$, whose normalized even bound state isIts derivative jump is precisely the one imposed by the delta potential.