Electrostatic energy

Electrostatic energy can be written as one half the charge-potential integral or epsilon-zero over two times the electric-field energy integral.

The local energy per unit volume associated with an electric field in vacuum is $u_E={ϵ}_0|E{|}^2/2$. Its volume integral gives the electric contribution to the electromagnetic energy. For a static field outside an isolated charged conducting sphere, integrating this density from its radius to infinity gives the electrostatic energy $Q^2/(8\pi {ϵ}_{0}R)$.

In the parallel-plate approximation, a central plate at potential $V$ between grounded plates at $z=\pm H$ forms two parallel capacitors of gaps $H+a$ and $H-a$. Summing their field energies gives the displayed formula. At fixed voltage it is minimized at $a=0$. Finite circular plates have fringe fields; the uniform-gap expression assumes those edge corrections are neglected.

A solid sphere of radius $R$ and uniform charge density has total electric charge $Q$. Gauss law gives the radial electric field $E_r=Qr/(4\pi {ϵ}_0{R}^3)$ inside and $E_r=Q/(4\pi {ϵ}_0{r}^2)$ outside. The electric potential vanishing at infinity is $Q(3-r^2/R^2)/(8\pi {ϵ}_{0}R)$ inside and $Q/(4\pi {ϵ}_{0}r)$ outside. Integrating the electrostatic energy density ${ϵ}_0{E}^2/2$ over both regions gives $U=3Q^2/(20\pi {ϵ}_{0}R)$. In a fixed-density family with $Q\propto Z$ and $R^3\propto Z$, the electrostatic energy scales as $Z^{5/3}$.

A first spherical shell with charge $q$ followed by shells of charges $2(-1{)}^{n+1}q$ leaves enclosed charge $(-1{)}^{n+1}q$ after each shell. Gauss's law therefore gives the same squared electric field $q^2/(16{\pi }^2{ϵ}_0^2{r}^4)$ in every gap outside the first radius. Integrating the electrostatic energy density over all space gives $W=q^2/(8\pi {ϵ}_0{r}_1)$, independent of the other shell radii.

A thin spherical shell of radius $a$ carrying total charge $Q$ has surface potentialand electrostatic self-energy

For distinct point charges $q_i$ at positions $r_i$, the electrostatic potential energy isWhen some charges are fixed, the terms depending on a movable charge form its potential energy in their electric potential.

For coaxial cylinders of radii $a<b$ and length $L\gg b$, carrying charges per unit length $\pm \lambda$,The field energy is $U={\lambda }^{2}L\log (b/a)/(4\pi {ϵ}_0)=QV/2$.

An isolated conducting sphere of radius $R$ initially at potential $V_0$ has charge $Q=4\pi {ϵ}_{0}R{V}_0$. Inserting it without contact into the hollow inner conductor of a long coaxial cylindrical capacitor conserves that charge. The neutral inner conductor acquires $-Q$ on its cavity wall and $+Q$ outside. With the outer conductor grounded, the annular electric field is $Q/(2\pi {ϵ}_{0}lr)$ and the inner potential is the displayed result. The original electric potential is not conserved. End effects are neglected; exact cavity fields are geometry-dependent.

A long coaxial cylindrical capacitor of length $L$ and radii $a,b$ has a dielectric-filled overlap of length $x$ and relative permittivity ${ϵ}_r$. Neglecting end and interface fringing, its filled and vacuum portions are in parallel because both share the same two conductors. The electric potential difference $V$ produces $E_r=V/[r\ln (b/a)]$ in either portion; the electric displacement field and electrode charge differ by the local permittivity. Adding their capacitances gives the title formula. The dielectric boundary is axial while the interior electric field is radial and tangential to that interface, consistent with the electrostatic interface boundary conditions.

At fixed applied voltage, the mechanical potential of a capacitor and its source includes $-C{V}^2/2$. Increasing the liquid overlap of a partially filled coaxial capacitor therefore pulls the liquid upward. Balancing $V^2(dC/dh)/2$ against the excess hydrostatic pressure times the annular area gives the title formula for a wide external reservoir. Its differential sensitivity is $2\Delta h/V$; it vanishes at zero bias and measures the magnitude, rather than the sign, of voltage.

A driven cylindrical conductor between two grounded coaxial conductors forms two coaxial cylindrical capacitors in parallel. For radii $a,\lambda a,2a$, the capacitance per unit length is the sum $C^{\prime }=2\pi {ϵ}_0[1/\log \lambda +1/\log (2/\lambda )]$. Writing $u=\log \lambda$ and $L=\log 2$, the function $1/u+1/(L-u)$ is strictly convex on $(0,L)$ and symmetric about $L/2$. Hence $C^{\prime }$ has its unique minimum at $\lambda =\sqrt{2}$, with value $8\pi {ϵ}_0/\log 2$. As $\lambda =1+\delta$ approaches the inner conductor, $C^{\prime }\sim 2\pi {ϵ}_0/\delta$, agreeing with the local parallel-plate capacitor approximation.