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Spherical symmetry and the integral Gauss law give $4\pi r^2{E}_r(r)=Q/{ϵ}_0$ for $r>R$. Together with the stipulated zero interior field,The electric field jumps across the ideal charged surface; its value exactly on that surface is not needed for the volume integral. The electrostatic energy of a uniformly charged spherical shell isAs a check, the shell has constant electric potential $Q/(4\pi {ϵ}_{0}R)$, with zero reference at infinity, so $U=Q\varphi (R)/2$ gives the same result.

Use $\mathbf{E}=-\nabla \varphi$ and Gauss's law $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. The divergence theorem and the decay at infinity giveMultiplying by $1/2$ proves that the two expressions for the electrostatic energy agree.

The uniform volume charge density in the thick shell isSpherical symmetry and the integral form of Gauss's law giveThus $|\mathbf{E}|$ is zero inside, rises continuously through the charged region, reaches $Q/(4\pi {ϵ}_0{b}^2)$ at $b$, and then decays as $r^{-2}$. Taking $\varphi (\infty )=0$ and integrating $E=-d\varphi /dr$ givesThe electric potential is continuous, constant inside, decreases smoothly across the charge, and continues as $1/r$ outside.

As $b\to a$, the charge becomes a surface shell. ThenThe potential remains continuous, but the normal electric field jumps by $Q/(4\pi {ϵ}_0{a}^2)$ across the surface charge density.

The field-energy expression givesThe charge-potential expression gives the same result directly because the entire charge lies where the potential equals $\varphi (a)$:This is the electrostatic energy of a uniformly charged spherical shell.

Since $\varphi (a)=Q/(4\pi {ϵ}_{0}a)$ is proportional to $Q$,to first order. This is the work required to bring the additional charge $\delta Q$ from infinity to the potential created by the charge already present.