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For a collection of fixed electrical conductors in linear electrostatics, the capacitance matrix relates conductor charges to potentials relative to a specified reference: $Q_i={\sum }_j{C}_{ij}{V}_j$. It includes mutual electrostatic effects omitted when each conductor is treated as an isolated capacitor. Its electrostatic energy is $U=\frac{1}{2}{\mathbf{V}}^{\mathsf{T}}\mathbf{CV}=\frac{1}{2}{\sum }_i{Q}_i{V}_i$. Geometry and separation are needed to determine its entries.

A capacitor stores electrostatic energy by separating electric charge between two electrical conductors. In the ideal two-terminal model their electric charges are $+Q$ and $-Q$, with a vacuum or insulating gap preventing direct transfer between them. Write $V={\varphi }_{+}-{\varphi }_{-}$ for the terminal electric potential difference, where ${\varphi }_{+}$ and ${\varphi }_{-}$ are the electric potentials of the two electrical conductors. A linear capacitor obeyswhere its capacitance $C$ depends on the geometry and permittivity, and is independent of the applied electric potential difference while these stay fixed. ^[ref].

For example, the parallel-plate capacitor below has capacitance $C=88.54\mathrm{pF}$. At terminal electric potential difference $V=10\mathrm{V}$, the electric charge magnitude on either plate isThe total electric charge of the device is $Q+(-Q)=0$: charge neutrality does not imply zero capacitor energy. This capacitor stores $4.427\mathrm{nJ}$ in its electric field. The schematic in parallel-plate capacitor shows the separation of electric charge; the plots in capacitor energy show how storage grows with terminal electric potential difference.

The useful distinction is between storing net electric charge and storing separated electric charge. A capacitor can release electrostatic energy into an electrical circuit; in an ideal RLC circuit, energy moves between the capacitor's electric field and the magnetic field associated with inductance. An ideal disconnected capacitor retains its separated electric charges and capacitor energy because there is no path for discharge.

The capacitor energy is the electrostatic energy stored when electric charge is separated. For a linear capacitor of fixed positive capacitance $C$, plate electric charge magnitude $Q$, and terminal electric potential difference $V=Q/C$, it isThe zero is the uncharged state. Thus the capacitor energy is positive for nonzero $Q$, even though the total electric charge of the two plates is zero. ^[ref].

For the parallel-plate capacitor example, capacitance $C=8.854\times 10^{-11}\mathrm{F}$ and terminal electric potential difference $V=10\mathrm{V}$ giveThe plate electric charge magnitude is $Q=0.8854\mathrm{nC}$. Doubling terminal electric potential difference with capacitance unchanged doubles the plate electric charge magnitude and quadruples the capacitor energy.

The electric charge is a linear function of terminal electric potential difference, while capacitor energy is a quadratic function. The second curve in each plot uses twice the capacitance: at a given terminal electric potential difference, both plate electric charge magnitude and capacitor energy double. These relations let us calculate how much energy an electrical circuit can receive when a capacitor discharges.

A brief charging argument explains the factor $1/2$. If the plate electric charge magnitude has reached $q$, the terminal electric potential difference is $q/C$. Transferring a further amount $dq$ of electric charge quasistatically requires work $dW=(q/C)dq$. With the geometry and permittivity fixed, the integral isUsing $Q=CV$ gives the other two forms. This integral accounts for the gradually rising terminal electric potential difference; it is the work stored in the capacitor's electric field, rather than a claim that all energy drawn from a fixed-voltage battery during charging ends up stored.

The field picture gives the same result. In the vacuum gap of a parallel-plate capacitor, the electric field has energy density $u={ϵ}_0{E}^2/2$. Multiplying by the gap volume $Ad$ gives the electrostatic energyFor the worked example, $u=4.427\times 10^{-4}J/{\mathrm{m}}^3$ and $Ad=10^{-5}{\mathrm{m}}^3$ reproduce $U=4.427\mathrm{nJ}$. The electrostatic energy resides in the electric field throughout the gap, despite overall charge neutrality.

When capacitance changes, specify the constraint. At fixed plate electric charge magnitude $Q$, increasing capacitance decreases capacitor energy $Q^2/(2C)$; doubling capacitance halves it. At fixed terminal electric potential difference $V$, increasing capacitance increases capacitor energy $C{V}^2/2$; doubling capacitance doubles it. In the latter case a connected battery supplies additional electric charge and energy, and there may also be mechanical work. Use the fixed-voltage capacitor energy balance to account for all transfers before inferring a force from changing capacitor energy.

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 $Q^2/(8\pi {ϵ}_{0}R)$.

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 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 $Z^{5/3}$.

For a quasistatic change of capacitance at fixed terminal electric potential difference, the capacitor energy is $U=C{V}^2/2$ and plate electric charge is $Q=CV$. Electrical work supplied to the electric field by the battery is $VdQ$, so mechanical work done on the capacitor is $d{W}_{\mathrm{mech}}=dU-VdQ=-V^{2}dC/2$. Increasing the separation of a parallel-plate capacitor makes $dC<0$: mechanical work is positive even though electrostatic energy decreases, because electrical energy returns to the source. Conversely, when $dC>0$, the battery supplies $V^{2}dC$: half increases capacitor energy and half is available as mechanical work done by the electric field. Differentiating capacitor energy alone at fixed terminal electric potential difference omits this exchange with the battery.

The polymer contour length and magnitude of total electric charge areUse a uniform charged-sphere model for the packed DNA, a homogeneous dielectric with $ϵ={ϵ}_r{ϵ}_0$, and initially neglect electrostatic screening. Gauss's law gives $E(r)=Qr/(4\pi ϵR^3)$ inside the sphere and $E(r)=Q/(4\pi ϵr^2)$ outside. Integrating the electrostatic energy both inside and outside,For an illustrative aqueous relative permittivity ${ϵ}_r=80$ and $R=20\mathrm{nm}$,The question does not specify permittivity; the result scales as $80/{ϵ}_r$. Using vacuum instead would increase it eightyfold and would not describe an aqueous interior.

Strictly, the work of packaging is a difference of electrostatic energies, rather than the packed self-energy alone. An extended charged rod also has self-energy. With a microscopic cutoff $a$ of order the DNA radius, the thin-rod approximation isOne derivation integrates the pair Coulomb potential along the rod, replacing separation by $\sqrt{z^2+a^2}$: the exact regularized expression is ${\lambda }^2[L\mathrm{arsinh}(L/a)-\sqrt{L^2+a^2}+a]/(4\pi ϵ)$. The diagram's $1\mathrm{nm}$ transverse label suggests $a\simeq 0.5\mathrm{nm}$, giving $U_{\mathrm{rod}}\simeq 6.25\times 10^6\mathrm{pN}\mathrm{nm}$. Hence this explicit reference-state estimate is $\boxed{\Delta U\simeq 1.32\times 10^8\mathrm{pN}\mathrm{nm}}$. The cutoff and precise initial conformation are additional model assumptions, but their correction is small compared with the unscreened sphere value.

The bare packed-sphere estimate exceeds $2.1\times 10^5\mathrm{pN}\mathrm{nm}$ by approximately $659$; subtracting the estimated extended-rod self-energy still leaves a factor of about $629$. The dominant missing effect is electrostatic screening: mobile positive counterions accompany the negatively charged DNA, greatly reducing long-range repulsion. Counterion condensation can further reduce its effective electric charge. Treating the genome's bare charge as uncompensated throughout and outside a $20\mathrm{nm}$ sphere therefore grossly overestimates its long-range electrostatic energy.

An optical tweezers measurement integrates an imposed force over a displacement. A reversible extraction or packaging comparison concerns a free energy difference relative to the unconfined DNA and ionic reservoirs. It can include entropy, bending and hydration contributions; dissipative pulling can add further work. Extraction can also be assisted by stored internal free energy, so the sign of externally supplied work requires the experimental convention. Here the supplied positive value is compared as an energy magnitude.

Fix the electric potential to vanish at infinity and assume a localized, nonsingular charge density of finite electrostatic energy. Assemble the charge by scaling ${\rho }_s=s\rho$ from $s=0$ to $1$. Linearity gives ${\varphi }_s=s\varphi$, so the work isThe factor $1/2$ prevents double counting of pair interactions. Using Gauss law, $\rho =-{ϵ}_0\Delta \varphi$, and integration by parts,The boundary integral at infinity vanishes for a localized distribution since $\varphi =O(r^{-1})$ and ${\partial }_r\varphi =O(r^{-2})$. Singular point-charge self-energies require a separate regularization and are not finite instances of this identity.

For the uniform solid sphere, spherical symmetry and Gauss law giveIntegrating inward from infinity and matching the continuous electric potential at $R$ givesThe value of the electric field at the centre is the zero vector.

Writing $Q=4\pi \rho R^3/3$, the electrostatic energy of a uniformly charged solid sphere follows either from the charge-potential integral or from the field integral including the exterior:For the nuclear scaling, $Q=Ze$ while constant volume per proton gives $R\propto Z^{1/3}$. Therefore the electric contribution scales as $Q^2/R\propto Z^{5/3}$ in this uniform-density model.

Take zero electric potential at infinity and put $K=(4\pi {\epsilon }_0{)}^{-1}$. By spherical symmetry and Gauss's law, the electric field is radial, with magnitude determined by the total enclosed electric charge. Thus, away from the idealized shell surfaces,Integrating $E_r=-d\Phi /dr$ from infinity, or adding the three spherical-shell potentials, givesThe expressions agree at each interface, so the potential is continuous everywhere, including the origin. The electric field has different interior and exterior limits at each infinitely thin charged surface, with jump equal to surface charge density divided by ${\epsilon }_0$. An ideal sheet has no single classical field value on the sheet itself; the displayed adjacent limits specify the field there. In conducting material of nonzero thickness the internal field is zero.

The electrostatic energy is the integral of ${\epsilon }_0|\mathbf{E}{|}^2/2$. There is no contribution from $r<a$, and spherical integration givesThe dependence on $b$ cancels because the magnitude of enclosed electric charge is $|q|$ on both sides of the middle shell, even though the field direction changes.

The differential Gauss law is $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. On a large ball $B_R$, the product rule and $\nabla \varphi =-\mathbf{E}$ giveThe divergence theorem consequently expresses the electrostatic energy asFor a localized charge distribution with the electric potential chosen to vanish at infinity, $\varphi =O(R^{-1})$ and $\mathbf{E}=O(R^{-2})$, so the surface term is $O(R^{-1})$ and tends to zero. More generally, it suffices to assume that this boundary integral vanishes. Passing to all space givesThe physical decay and potential reference are relevant: $\mathbf{E}\to 0$ by itself does not specify the additive constant in $\varphi$. Adding a constant $C$ changes the charge-based expression by $CQ/2$ when the total charge is $Q$, while leaving the field unchanged. The zero-at-infinity potential is the usual reference implicit in this all-space electrostatic energy formula.

The grounded conducting plane supplies the induced charge needed for a uniform field between the plates and zero field above the upper plate. Gauss's law gives $E_z=-\sigma /{ϵ}_0$ in the gap. With electric potential zero at the grounded plate,Integrating the electrostatic energy density ${ϵ}_0|\mathbf{E}{|}^2/2$ across the gap gives the capacitor energy per unit areaFor an isolated upper plate, charge is fixed, so increasing $d$ to $(1+\alpha )d$ gives exactly $\Delta u=\frac{1}{2}\sigma V\alpha$, with $V$ its initial value. The positive increase equals the mechanical work done against attraction.

At fixed voltage, the new charge density is $\sigma /(1+\alpha )$, and the energy becomes $u/(1+\alpha )$. Consequently $\Delta u=-\frac{1}{2}\sigma V\alpha +O({\alpha }^2)$. Charge $\sigma \alpha +O({\alpha }^2)$ per unit area flows back to the voltage source, which receives energy $\sigma V\alpha +O({\alpha }^2)$. The externally supplied mechanical work is still $\frac{1}{2}\sigma V\alpha +O({\alpha }^2)$: the field loses half this returned electrical energy and the mechanical work supplies the other half. Thus the fixed-voltage capacitor energy balance includes both field and source energy.

For a conductor in electrostatic equilibrium, the interior field is zero and the surface is equipotential. The tangential exterior field is zero. A thin Gaussian pillbox gives the normal jump $E_{\mathrm{out}}\cdot \mathbf{n}-E_{\mathrm{in}}\cdot \mathbf{n}=\sigma /{ϵ}_0$. Thus, in vacuum and with $\mathbf{n}$ pointing out of the conductor,These are the electrostatic boundary conditions at a conductor.

To obtain electrostatic energy in field form, integrate over the space outside the conductors, cutting it off first at a large sphere. From the Maxwell equations, $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$ and $\mathbf{E}=-\nabla \varphi$, soOn a conductor boundary, the outward normal of the integration region points into the conductor, so its flux contribution is $-{\varphi }_i{Q}_i/{ϵ}_0$. At the remote outer boundary the flux integral tends to zero: for a localized charge distribution, $\varphi =O(1/r)$ and $\mathbf{E}=O(1/r^2)$. The divergence theorem therefore givesSince the field vanishes inside conductors,The field integral is over all space, although the charges occupy a finite volume. For a finite outer integration boundary there is additionally the term $({ϵ}_0/2)\oint \varphi \mathbf{E}\cdot d\mathbf{S}$.

For the electrostatic energy of alternating-charge concentric shells, the total enclosed charge after shell $n$ isBy spherical symmetry and Gauss's law, the field is zero for $r<r_1$, and in every subsequent gap, including outside the last shell,Although its direction alternates, its squared magnitude is the same in every gap. ConsequentlyThe coefficient of $1/r_1$ is $q^2/(8\pi {ϵ}_0)$, independent of the other radii and of the number of shells.

Let the electrolyte permittivity be $\epsilon$ and let ${\kappa }_D=1/{\ell }_D$ be the inverse Debye–Hückel screening length. We reserve $k$ for the imposed lateral wavevector. In the Debye–Hückel approximation, linearization about an electrically neutral bulk electrolyte givesaway from fixed charges. For ionic species of valence $z_i$ and bulk number density $n_i$, ${\kappa }_D^2={\sum }_i{n}_i{z}_i^2{e}^2/(\epsilon k_{B}T)$; the approximation requires $|z_{i}e\varphi |\ll k_{B}T$. Substitution of a lateral cosine into this screened Poisson equation leaves a vertical decay constantThe screened sinusoidal surface-charge mode thus decays faster than a laterally uniform charge mode. A nonzero $k$ retains a finite decay length even in the zero-salt limit.

The exterior media and their boundary conditions are not specified in the paper. Specifying the two surface charge densities alone does not fix the normal derivative on the inside unless the outside response is also fixed. We first take the confined-gap idealization, with the prescribed charge supplying all the displacement flux into the electrolyte between the sheets. The electrostatic interface boundary conditions are thenThis is, for example, the limit of negligible exterior displacement admittance. The electrostatic potential solving these conditions and the Debye–Hückel approximation equation isDifferentiation checks the two surface signs directly. This solution allows a sine as well as a cosine lateral component when the phase is nonzero.

In linear screened electrostatics, the quadratic charging free energy is $\frac{1}{2}\int \sigma \varphi dA$. It is equivalently $\frac{\epsilon }{2}\int [(\nabla \varphi {)}^2+{\kappa }_D^2{\varphi }^2]dV$ with the stated boundary conditions. The second term represents the linearized ionic response; integrating only the bare electric field energy would omit it. If $\mathcal{A}$ denotes membrane area, the wavelength-averaged energy density isUsing $⟨{\cos }^2(kx){⟩}_x=1/2$ and $⟨\cos (kx)\cos (kx+\theta ){⟩}_x=\cos \theta /2$ gives the phase registration of screened charged sheets energyThe positive cosine coefficient shows that, for nonzero charge amplitude and finite separation,At the minimum the electrostatic potential simplifies to $\varphi =\alpha \cos (kx)\sinh (pz)/[\epsilon p\cosh (pd/2)]$. Positive and negative charge patches face opposite signs across the gap. The two charge patterns therefore favor a half-wavelength lateral offset rather than like-charge alignment.

For comparison, if identical electrolyte fills all of space on both sides of infinitesimally thin charge sheets, the electrostatic interface boundary conditions are continuity of $\varphi$, decay at infinity, and the charge-induced derivative jumpsSuperposing the two screened sinusoidal surface-charge modes gives, in particular between the sheets,Its full-space continuation replaces each vertical distance by the corresponding absolute distance. Evaluating the electrostatic potential on both sheets now yieldsThe constant term is the two isolated-sheet self energies, and the cosine term is their screened interaction. The energy depends on the exterior convention, but both stated physical idealizations give the same phase registration of screened charged sheets: unlike charge patches oppose each other. In the full-space convention the optimum interaction energy is $-{\alpha }^2{e}^{-pd}/(4\epsilon p)$, giving an attractive normal force density $-{\alpha }^2{e}^{-pd}/(4\epsilon )$. Phase sensitivity becomes exponentially weak for $pd\gg 1$; if $\alpha =0$ or the separation tends to infinity there is no selected phase.

One can also display how an exterior dielectric response interpolates between these cases. Let $Y_o\ge 0$ be its normal displacement admittance for this lateral Fourier mode: $Y_o=0$ for the confined-gap idealization, $Y_o=\epsilon p$ for identical exterior electrolyte, and $Y_o={\epsilon }_o|k|$ for an ion-free exterior dielectric. DefineThe symmetric and antisymmetric surface-charge combinations giveSince $D_s<D_a$, its cosine coefficient is positive. This makes the phase-minimizing conclusion robust while exposing the boundary information needed for an absolute electrostatic energy.

In electrostatics, Gauss's law is ${\oint }_{\partial \Omega }\mathbf{E}\cdot \mathbf{n}dS=Q_{\mathrm{enclosed}}/{ϵ}_0$, or locally $\nabla \cdot \mathbf{E}={\rho }_e/{ϵ}_0$, where ${\rho }_e$ is charge density and ${ϵ}_0$ is vacuum permittivity. Spherical symmetry makes the electric field radial, and Gauss's law givesChoose the additive constant of the electrostatic potential by $\varphi (\infty )=0$. Using $\mathbf{E}=-\nabla \varphi$ and continuity of potential across each charged shell givesThe field jumps at the shell surfaces, so the displayed formulas are for the requested open regions. The potential difference of the spherical capacitor is $\Delta \varphi =Q(1/a-1/b)/(4\pi {ϵ}_0)$. Thus its capacitance and electrostatic energy areThe energy follows either by charging work ${\int }_0^{Q}q/Cdq$, or directly from $W=({ϵ}_0/2)\int |\mathbf{E}{|}^2{d}^{3}x=Q^2(1/a-1/b)/(8\pi {ϵ}_0)$.

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.

Let $r$ be distance from the common centre. By Gauss's law, the electric field is radial and equalsTaking the electric potential to vanish at infinity and requiring continuity across each shell gives

The electrostatic energy is the integral of the electric-field energy density:Only the two nonzero-field regions contribute, soTherefore

Gauss's law in electrostatics statesCylindrical symmetry and a coaxial Gaussian cylinder giveThe field vanishes inside each perfect conductor, and outside the cable because the total enclosed charge per unit length is zero.

Choose the outer conductor's potential to be zero. Since $E=-\nabla V$,The capacitance per unit length is thereforeThe electrostatic energy per unit length isSubstitution of $C$ verifies $U=Q^2/(2C)$.

Isolation fixes charge, not voltage: $Q=C_0{V}_0$. The electrostatic energy is therefore $U(t)=Q^2/[2C(\delta (t))]$. Inverting the small-tilt capacitance givesThe leading variation has frequency $2{\omega }_0$, since tilts of opposite sign give the same capacitance. With $U_0=C_0{V}_0^2/2$, the result is $U_0[1-{\delta }_0^2/(6a^2)+({\delta }_0^2/(6a^2))\cos 2{\omega }_{0}t]$ to this order. Energy decreases as the gap becomes less uniform at fixed charge; the difference is exchanged mechanically with the oscillator.