--- title: Electrostatic Energy and Pressure module: Electric Potential moduleNumber: 3 lessonNumber: 4 order: 304 summary: | Assembling a charge configuration takes work, and that work is stored, but where is it kept and how much is there? We total it two ways: as a sum over the charges, $U=\tfrac12\sum_i q_iV_i$, and as an integral over the field itself, $u_E=\tfrac12\varepsilon_0E^2$, energy the field carries in every region it fills. Differentiating the stored energy at fixed charge or at fixed voltage recovers the mechanical force on a conductor, and at a charged surface the same field scale appears as an outward electrostatic pressure. We work the parallel-plate case in full and mark where curvature and fringing make the pressure nonuniform. topics: [Electric Potential] draft: false sources: - book: Tipler & Mosca ref: "Ch. 23 — Electric Potential; §§23-5–23-6" --- ## Assembly energy of discrete charges Electrostatic energy is the external work required to assemble a charge configuration quasistatically from a reference state. For discrete charges, each pair contributes once: $$ U=\sum_{i=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \filldraw[draw=black,fill=black!8] (0.65,0.55) circle (0.16); \node[below] at (0.65,0.45) {$q_1$}; \filldraw[draw=black,fill=black!8] (4.45,0.55) circle (0.16); \node[below] at (4.45,0.45) {$q_2$}; \filldraw[draw=acc,fill=acc!10] (2.55,2.75) circle (0.16); \node[above] at (2.55,2.85) {$q_3$}; \draw[black,dashed] (0.65,0.55)--(4.45,0.55) node[midway,below] {$r_{12}$}; \draw[black,dashed] (0.65,0.55)--(2.55,2.75) node[midway,above left] {$r_{13}$}; \draw[black,dashed] (4.45,0.55)--(2.55,2.75) node[midway,above right] {$r_{23}$}; \end{tikzpicture} $$ > **Worked example.** Three charges sit at the corners of an equilateral triangle of > side $a=0.20\ \mathrm m$: $q_1=q_2=+2.0\ \mathrm{nC}$ and $q_3=-3.0\ \mathrm{nC}$. > Find the energy stored in assembling them. > > Every side has length $a$, so the pair sum is > > $$ > U=\frac{k}{a}\left(q_1q_2+q_1q_3+q_2q_3\right). > $$ > > With $q_1q_2=+4.0\times10^{-18}$ and $q_1q_3=q_2q_3=-6.0\times10^{-18}$ (units > $\mathrm{C^2}$), > > $$ > U=\frac{8.99\times10^9}{0.20}\,(4.0-6.0-6.0)\times10^{-18} > =-3.6\times10^{-7}\ \mathrm J. > $$ > > The total is negative: the two attractive pairs release more than the single > repulsive pair takes to build. A negative total does not mean every pair is > attractive, and it is the reason charge configurations dominated by opposite-sign > neighbors are bound. ## Pressure at conductor surfaces Electrostatic pressure is a surface-force statement. At the vacuum side of a conductor in electrostatic equilibrium, the tangential electric field is zero and the normal field satisfies $$ E_\perp=\frac{\sigma}{\varepsilon_0}, $$ where $\sigma$ is the surface charge density measured with the outward normal. A small patch of charge cannot use its entire exterior field to calculate its own force: half of the discontinuity is produced by that patch itself. The force comes from the average of the fields immediately on the two sides of the surface. Since the interior field vanishes, the external field acting on the patch has magnitude $E_\perp/2$. The outward traction is therefore $$ p=\frac{F}{A}=\sigma\frac{E_\perp}{2} =\frac{\sigma^2}{2\varepsilon_0} =\frac12\varepsilon_0E_\perp^2. $$ The energy-density calculation gives the same pressure magnitude. The surface-force construction also specifies the local direction. The vacuum field pulls a charged conductor surface outward along the exterior normal. On the inner face of the upper plate of a capacitor, that outward normal points down into the gap; the resulting surface traction draws the plate toward the lower plate. On the exterior face of an ideal broad capacitor, the field is nearly zero, so there is little balancing outward traction there. $$ % caption: Surface-force origin of electrostatic pressure. The conductor has zero interior field; a surface patch feels the average of the interior and exterior fields, giving outward traction $p=\sigma E_\perp/2=\varepsilon_0E_\perp^2/2$. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \fill[black] (0.45,0.35) rectangle (5.85,1.33); \draw[black,thick] (0.45,1.33)--(5.85,1.33); \foreach \x in {0.95,1.65,2.35,3.05,3.75,4.45,5.15} {\node at (\x,1.55) {$+$};} \foreach \x in {3.05,3.75,4.45,5.15} {\draw[->,thick] (\x,1.78)--(\x,2.85);} \node[above] at (4.10,2.85) {exterior E}; \draw[->,acc,very thick] (1.65,1.42)--(1.65,2.70); \node[acc,above] at (1.65,2.70) {outward pull}; \node[black!85] at (3.10,0.82) {conductor}; \end{tikzpicture} $$ > **Worked example.** A conductor surface carries charge density > $\sigma=1.0\ \mu\mathrm{C/m^2}$. Find the field just outside it and the outward > electrostatic pressure on the surface. > > The exterior normal field is set by the local surface charge: > > $$ > E_\perp=\frac{\sigma}{\varepsilon_0} > =\frac{1.0\times10^{-6}}{8.854\times10^{-12}}=1.1\times10^{5}\ \mathrm{V/m}. > $$ > > The patch feels the average of the interior ($0$) and exterior fields, giving > > $$ > p=\frac{\sigma^2}{2\varepsilon_0} > =\frac{(1.0\times10^{-6})^2}{2(8.854\times10^{-12})} > =5.6\times10^{-2}\ \mathrm{Pa}, > $$ > > directed outward along the normal. Using the field, $\tfrac12\varepsilon_0E_\perp^2$ > returns the same pressure, confirming the factor of one-half: a naive $\sigma > E_\perp$ would double it by letting the patch push on its own field. The same relation follows from the electric part of the Maxwell stress tensor, $$ \mathbf T=\varepsilon_0\left(\vec E\vec E- \frac12E^2\mathbf I\right). $$ On a closed surface in vacuum, integrating $\mathbf T\mathbin{\cdot}\hat n$ gives the net electric force on the enclosed matter. This method becomes valuable when field lines bend around edges or through apertures, because it does not require inventing a single scalar pressure for an asymmetric surface. A numerical field solution can be sampled on a surface that lies entirely in vacuum, then integrated to obtain force and torque without evaluating a divergent self-field at charge elements. A free isolated conducting sphere distinguishes local pressure from net force. Its charge is uniformly distributed, so the pressure has the same magnitude at every point and points radially outward. Opposite patches cancel in the vector sum: the sphere does not accelerate as a whole, although its material must resist a real tensile loading. For radius $R$ and charge $Q$, $$ \sigma=\frac{Q}{4\pi R^2}, \qquad p=\frac{Q^2}{32\pi^2\varepsilon_0R^4}. $$ The $R^{-4}$ scaling explains why electrostatic deformation becomes important for small highly charged droplets and fragile conductors. Material strength, surface tension, and charge emission set limits long before a literal mathematical surface could sustain arbitrarily large pressure. Near the central area of two parallel plates, the field is perpendicular to both metal surfaces and the pressure on each inner face is nearly uniform. The outer faces make an informative check on an idealization. If the plates are large and oppositely charged, the exterior fields cancel almost completely, while the two interior contributions add. A finite plate has fringing field at its perimeter; there, the stress has lateral components and the uniform-pressure estimate omits a small edge force. The error decreases as plate dimensions become large compared with separation. $$ % caption: Pressure on the two inner faces of an oppositely charged parallel-plate capacitor. The local tractions point into the gap, so the plates attract; exterior-face tractions are small in the broad-plate limit. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[black,very thick] (0.55,2.52)--(5.65,2.52); \draw[black,very thick] (0.55,0.55)--(5.65,0.55); \foreach \x in {1.15,2.05,2.95,3.85,4.75} {\node[above] at (\x,2.52) {$+$}; \draw[black,thick] ({\x-0.07},0.32)--({\x+0.07},0.32); \draw[->,black,thick] (\x,2.28)--(\x,0.82);} \draw[->,acc,very thick] (1.58,2.47)--(1.58,1.82); \draw[->,acc,very thick] (4.52,0.60)--(4.52,1.25); \draw[<->,black] (5.98,0.55)--(5.98,2.52) node[midway,right] {$d$}; \end{tikzpicture} $$ At a sharp point, the surface charge density and normal field become large. The continuum formula still describes the local tendency, but microscopic roughness, ionization of nearby gas, and charge leakage can intervene. A model that quotes a single plate-area pressure while ignoring a serrated electrode has discarded the geometry responsible for the largest fields. Energy derivatives based on a measured or computed capacitance retain the relevant shape information. **Incremental assembly and the role of the reference state.** An assembly calculation has an order, even though the final sum does not. Bring $q_1$ from the reference region first. No other charge is present, so the required external work is zero. With $q_1$ fixed, bring $q_2$ to its assigned position; the external work is $kq_1q_2/r_{12}$. The third step requires $$ W_3=q_3\left(k\frac{q_1}{r_{13}}+k\frac{q_2}{r_{23}}\right). $$ Adding the three steps produces the three pair terms in the displayed expression above. Reversing the order changes the intermediate work values but leaves their sum unchanged. Electrostatic force is conservative in the static configuration, so the work depends on the endpoint arrangement and the reference potential, rather than on the route used to carry a charge through empty space. A nearby wire, a grounded shield, or a dielectric changes the potential used at every step and must therefore be included in the physical system. The potential in an insertion step excludes the charge being inserted. Including its own point-charge potential would introduce an undefined self-term and would also count a force that cannot translate a charge against itself. For a finite charged body, self-energy is meaningful: the body must be built by bringing in small portions of its charge, each portion seeing the potential due to portions already present. That distinction separates the interaction energy between named charges from the energy required to create an extended charge distribution. A conductor illustrates a different charging process. Suppose an isolated conductor has capacitance $C$ and starts neutral. During an increment $\d Q$, the conductor already carries charge $Q'$ and its potential is $V(Q')=Q'/C$. The work is therefore accumulated as $$ U=\int_0^Q V(Q')\,\d Q' =\int_0^Q\frac{Q'}{C}\,\d Q' =\frac{Q^2}{2C}. $$ The one-half appears because the potential grows from zero to its final value as charge is deposited. It does not signal that half the energy has been discarded; it is the triangular area under a straight voltage--charge relation. This argument also applies to a linear capacitor connected to a slow charging source. If the capacitance changes while charge is supplied, both $V(Q,x)$ and the work done by the mechanical support must be tracked along the chosen path in state space. $$ % caption: Charging an isolated linear conductor. Its potential rises in proportion to deposited charge, so the triangular area under $V(Q)$ equals $Q^2/(2C)=QV/2$. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->,black] (0,0)--(5.65,0) node[right] {$Q$}; \draw[->,black] (0,0)--(0,3.25) node[above] {$V$}; \fill[acc!14] (0.55,0)--(4.95,0)--(4.95,2.55)--cycle; \draw[acc,very thick] (0.55,0.28)--(4.95,2.55); \draw[black,dashed] (4.95,0)--(4.95,2.55); \node[acc] at (3.15,0.80) {$U=\frac{QV}{2}$}; \node[below] at (4.95,0) {$Q_f$}; \node[left] at (0,2.55) {$\frac{Q_f}{C}$}; \end{tikzpicture} $$ A configuration near a grounded conductor has a reference region that includes the Earth and the charge reservoir that maintains ground. Image-charge constructions can calculate a force in simple geometries, but their energy requires care: a grounded metal can exchange charge with the reservoir, so its energy accounting differs from that of an isolated copy of the same shape. For grounded conductors, specify the controlled quantities and include reservoir work before differentiating for force. ## Energy stored in an electric field Continuous distributions have electrostatic energy expressible as an integral over field energy density in vacuum: $$ u_E=\frac12\varepsilon_0E^2, \qquad U=\int u_E\,\d V. $$ A parallel-plate capacitor has nearly uniform field between broad plates and weak exterior field when fringing is small. Its stored energy is concentrated primarily in the gap rather than in the metal, where equilibrium electric field vanishes. For plate area $A$, separation $d$, and uniform field magnitude $E$, the gap volume gives $$ U=\frac12\varepsilon_0E^2Ad. $$ Using $E=V/d$ and $C=\varepsilon_0A/d$ recovers the equivalent capacitor forms $$ U=\frac12CV^2=\frac{Q^2}{2C}=\frac12QV. $$ Each form holds a different control variable explicit. The $Q^2/(2C)$ form is convenient for an isolated charged capacitor. The $CV^2/2$ form is convenient when a voltage source fixes the terminal difference. Switching between them without stating whether charge or voltage is held fixed causes incorrect force calculations. $$ % caption: Field energy in a parallel-plate capacitor. The nearly uniform gap field has energy density $\varepsilon_0E^2/2$; the shaded gap holds the dominant stored energy in the broad-plate limit. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \fill[acc!12] (0.45,0.45) rectangle (5.35,2.35); \draw[black,very thick] (0.45,2.35)--(5.35,2.35); \draw[black,very thick] (0.45,0.45)--(5.35,0.45); \foreach \x in {1.00,1.85,2.70,3.55,4.40} {\draw[->,black,thick] (\x,2.08)--(\x,0.72);} \draw[<->,black,thick] (5.75,0.45)--(5.75,2.35) node[midway,right] {$d$}; \node[acc] at (5.05,1.40) {$u_E$}; \end{tikzpicture} $$ The field-energy integral has a model boundary. A classical point charge produces an energy density that diverges near its position, so its self-energy needs a finite source-size or more complete physical model. Energy differences for separated charges and macroscopic capacitor configurations remain well defined within the classical electrostatic approximation. **Continuous charge and finite source size.** The discrete pair sum approaches a continuous expression when a charge cloud is partitioned into small elements. An element carrying $\d q$ is placed in the potential due to all previously placed elements. In the limiting description, $$ U=\frac12\int \rho(\vec r)V(\vec r)\,\d^3 r. $$ The potential in this integral is the total physical potential after assembly, and the factor one-half keeps each interaction from appearing in both elements of a pair. For a smooth distribution, the formula is finite when the density remains finite over a nonzero volume. A delta-function point charge fails that test: its field grows too rapidly near the source for the field-energy integral to converge. The divergence identifies a limit of classical point-source electrostatics; it does not invalidate the finite interaction energy between separated charges. Gauss's law and $\vec E=-\nabla V$ connect the charge-potential expression to the field form. With fields that vanish sufficiently rapidly at a distant boundary, integration by parts gives $$ \frac12\int\rho V\,\d^3 r =\frac{\varepsilon_0}{2}\int V\,(\nabla\mathbin{\cdot}\vec E)\,\d^3 r =\frac{\varepsilon_0}{2}\int E^2\,\d^3 r. $$ The surface term deserves attention in laboratory geometries. A metal enclosure, an imposed voltage supply, or a truncated computational domain contributes a boundary condition. Extending the integration region through the surrounding conductors and returning to the physical reservoir gives a consistent total-energy calculation. Dropping a boundary term solely because a diagram has an edge can silently remove the work done by an external source. A uniformly charged insulating sphere of radius $R$ and total charge $Q$ has radial field $$ E(r)= \begin{cases} \dfrac{Qr}{4\pi\varepsilon_0R^3},&0\le r\le R,\\[5pt] \dfrac{Q}{4\pi\varepsilon_0r^2},&r\ge R. \end{cases} $$ The field rises linearly through the material, reaches its largest value at the surface, and then falls with inverse-square distance. Integrating $\varepsilon_0 E^2/2$ over both regions gives $$ U_{\rm uniform\ sphere}=\frac{3Q^2}{20\pi\varepsilon_0R} =\frac35\,\frac{kQ^2}{R}. $$ About one-sixth of this value lies inside the sphere and five-sixths lies outside; the exterior field occupies an unlimited volume even though its energy density falls rapidly. The result has a clear size dependence: compressing the same charge into a smaller sphere raises the energy needed to assemble it. $$ % caption: Radial field of a uniformly charged insulating sphere. The field rises linearly to the surface at $R$, then follows the inverse-square law; the energy density tracks $E^2$ on both sides. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->,black] (0,0)--(6.2,0) node[right] {$r$}; \draw[->,black] (0,0)--(0,3.35) node[above] {$E$}; \draw[acc,very thick] (0.35,0.22)--(2.55,2.55); \draw[acc,very thick] plot[smooth] coordinates {(2.55,2.55)(3.05,2.00)(3.65,1.45)(4.40,1.03)(5.25,0.72)(5.85,0.58)}; \draw[black,dashed] (2.55,0)--(2.55,2.55); \node[below] at (2.55,0) {$R$}; \node[acc,above left] at (1.45,1.45) {linear rise}; \node[acc,above] at (4.75,1.75) {inverse-square}; \end{tikzpicture} $$ An isolated conducting sphere makes the contrast sharp. Once electrostatic equilibrium is reached, excess charge resides on the surface and the electric field inside the metal is zero. Its capacitance is $C=4\pi\varepsilon_0R$, so charging it to $Q$ requires $$ U_{\rm conducting\ sphere}=\frac{Q^2}{2C} =\frac{Q^2}{8\pi\varepsilon_0R}. $$ All of that field energy lies outside the conductor. The lower value relative to a uniform insulating sphere of the same $Q$ and $R$ reflects the charge rearrangement allowed by conduction. No energy has vanished inside the metal; its equilibrium electric field is zero, and the stored electrostatic energy is carried by the external field configuration. Field energy can be concentrated in a finite dielectric-free gap without being uniform. A spherical capacitor has inner radius $a$, outer radius $b$, and charges $+Q$ and $-Q$ on its conductors. The field exists only for $a=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->,black] (0,0)--(5.65,0) node[right] {$d$}; \draw[->,black] (0,0)--(0,3.05) node[above] {$U$}; \draw[acc,thick] (0.40,0.45)--(5.15,2.60); \draw[black,thick] plot[smooth] coordinates {(0.40,2.62)(1.10,1.88)(2.00,1.32)(3.20,0.94)(4.30,0.72)(5.15,0.58)}; \node[acc,above left] at (5.15,2.55) {$U_Q$}; \node[black,above right] at (4.55,0.72) {$U_V$}; \end{tikzpicture} $$ The pressure expression is local for a smooth parallel gap. Sharp edges, fringing, nonlinear dielectrics, and nearby conductors change the field distribution and therefore the force. A force measurement can still be obtained from total energy or from the electromagnetic stress distribution, but a single uniform-gap formula cannot replace the full geometry. **Electrical constraints and the appropriate energy function.** A capacitor with geometry coordinate $x$ has two independently variable quantities: its charge and its geometry. For a reversible quasistatic change, the differential form of the field energy is $$ \d U=V\,\d Q-F_x\,\d x. $$ The term $V\,\d Q$ records electrical work delivered through the terminals. The sign of the second term follows the convention that positive $F_x$ does work while $x$ increases. When differentiating, hold the electrical variable specified by the physical constraint fixed. At fixed charge, $\d Q=0$ and the stored field energy is the mechanical potential: $$ F_x=-\left(\frac{\partial U}{\partial x}\right)_Q. $$ At fixed voltage, a source moves charge as the geometry changes. Subtracting the electrical term gives the constrained energy function $$ \mathcal G(U,Q,V)=U-QV, \qquad \d \mathcal G=-Q\,\d V-F_x\,\d x, $$ so that $$ F_x=-\left(\frac{\partial\mathcal G}{\partial x}\right)_V. $$ A linear capacitor has $U=Q^2/(2C)$ and $Q=CV$, which gives $\mathcal G=-CV^2/2$. The negative sign carries source accounting; it does not mean that a capacitor possesses negative field energy. The physical field energy remains $CV^2/2$, while the source-reservoir contribution changes by the additional amount needed to keep $V$ fixed. For parallel plates, take $x=d$ as the gap. With $C=\varepsilon_0A/d$, the two constrained calculations are $$ U_Q(d)=\frac{Q^2d}{2\varepsilon_0A}, \qquad \mathcal G_V(d)=-\frac{\varepsilon_0AV^2}{2d}. $$ They yield $$ F_d=-\frac{Q^2}{2\varepsilon_0A} =-\frac{\varepsilon_0AV^2}{2d^2} =-\frac12\varepsilon_0AE^2. $$ The minus sign states that electric force reduces the positive separation coordinate. The two forms agree when $Q=CV$ is imposed at the state being evaluated. Their derivations differ because one follows an isolated capacitor and the other follows a capacitor attached to a voltage source. Connecting those histories without the source term creates the familiar but false prediction of repulsion at fixed voltage. The field-pressure result can be checked directly. The uniform gap field has energy density $u_E=\varepsilon_0E^2/2$. Increasing the gap by $\d d$ at fixed area adds volume $A\,\d d$ of field. The energy change per added volume is $u_E$, and the magnitude of force per area is the same number. This local argument works because the field is approximately uniform and perpendicular to the plates; it becomes unreliable in a highly fringing geometry. An overlap capacitor provides a second geometry with a different force direction. Two broad plates remain separated by a fixed $d$, but their overlap length $x$ changes. If the plate width is $\ell$, the overlap area is $A=\ell x$ and $$ C(x)=\frac{\varepsilon_0\ell x}{d}. $$ At fixed voltage, the force toward increasing overlap is $$ F_x=\frac12V^2\frac{\d C}{\d x} =\frac{\varepsilon_0\ell V^2}{2d}. $$ There is no inverse-square dependence on $x$ in the broad-overlap approximation; each additional strip of overlap contributes the same incremental capacitance. At fixed charge, the energy form $Q^2/[2C(x)]$ produces the corresponding force after the derivative is taken at fixed $Q$. Edge fields alter the result near zero overlap, where the simple area formula no longer represents the full capacitance. $$ % caption: Variable-overlap capacitor for electrostatic actuators. With fixed gap $d$ and plate width $\ell$, extending the overlap $x$ adds a strip of area $\ell\,\mathrm dx$, so capacitance and the voltage-controlled force grow from the same geometric increment. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[black,very thick] (0.45,0.72)--(4.45,0.72); \draw[black,very thick] (1.62,2.05)--(5.62,2.05); \fill[acc!13] (1.62,0.72) rectangle (4.45,2.05); \draw[<->,black] (1.62,0.38)--(4.45,0.38) node[midway,below] {$x$}; \draw[<->,black,thick] (5.98,0.72)--(5.98,2.05) node[midway,right] {$d$}; \draw[->,acc,thick] (4.72,1.38)--(5.48,1.38) node[midway,above] {$F_x$}; \node[acc] at (3.03,1.36) {overlap area}; \end{tikzpicture} $$ ## Experimental and numerical use of energy methods Compare an energy calculation with an independently measured force. A translation stage can set a capacitor coordinate $x$ while a force sensor records the external force required to hold that coordinate. After subtracting the stage's mechanical baseline, the measured holding force has the opposite sign to the electric force. Repeating the measurement at fixed charge and at fixed voltage distinguishes the boundary conditions experimentally, provided the charge source or voltage source remains connected in the intended run. Capacitance data provide a second route. Measure $C(x)$ with a sufficiently small ac probe, then differentiate a smooth fit over the displacement interval of interest. At controlled voltage, the predicted force is $$ F_x=\frac12V^2\frac{\d C}{\d x}. $$ The derivative magnifies measurement noise. A two-point quotient using nearly identical coordinates can vary wildly even when each capacitance value is accurate. Several spacings, repeated sweeps in both directions, and a geometry-based fit make the slope more stable. The fit must include the region in which it will be used; extrapolating a parallel-plate expression into a fringing-dominated end position produces an apparent force law with no matching physical geometry. The electrical instrumentation has its own model. A capacitance bridge measures terminal response at a finite frequency, whereas the energy derivation assumes a quasistatic state. Series resistance, dielectric loss, leakage paths, and motion during the reading introduce phase shifts or transient currents. An instrument with a grounded shield can also change the surrounding capacitance. A report should state the drive amplitude, frequency, connection topology, temperature, and whether the electrodes were allowed to settle between position changes. The stored energy can be obtained from charge--voltage data without assuming a linear capacitor. For any reversible charging curve at fixed geometry, $$ U(Q)=\int_0^Q V(Q')\,\d Q'. $$ The graphical area under the measured curve gives the same quantity. A nonlinear dielectric or a geometry with position-dependent capacitance changes the curve; replacing it with $QV/2$ presumes a straight line through the origin. Integrating the measured curve preserves the actual response and makes hysteresis visible when the charging and discharging paths enclose a nonzero area. That enclosed area belongs to dissipative processes, not to recoverable electrostatic storage. Numerical electrostatics uses an analogous set of cross-checks. A solver returns potential values on a mesh, from which charge on an electrode can be found by a surface integral of the normal electric displacement. With one electrode held at voltage $V$ relative to another, compare $$ C_Q=\frac{Q}{V}, \qquad C_U=\frac{2U}{V^2}, \qquad U=\frac12\int\varepsilon_0E^2\,\d V. $$ Agreement between $C_Q$ and $C_U$ tests the potential solution, surface-charge calculation, and volume-energy integration at once. A mismatch often traces to coarse cells near narrow gaps, an outer boundary placed too close to a fringing region, or an incomplete inclusion of conducting surfaces. Refining only the wide empty region seldom improves the quantity that controls a small-gap force. Force from a numerical model can be checked in two independent ways. First, evaluate the constrained energy at $x-h$, $x$, and $x+h$, then use a central difference for the derivative. Second, integrate Maxwell stress over a vacuum surface enclosing one electrode. The coordinate increment must be small enough to represent local variation and large enough to exceed solver noise. Convergence of both estimates under mesh refinement and step-size variation provides stronger evidence than either result alone. The boundary of the modeled system determines whether a numerical energy supports a mechanical prediction. If a voltage source is represented merely by fixed potential nodes, its charge reservoir is implicit; use the fixed-voltage constrained function when taking a force derivative. If both conductors are isolated with fixed charges, integrate the field energy for that closed charge configuration. These are different experiments even when their instantaneous voltage and charge happen to match. ### Scale check: a broad vacuum capacitor > **Worked example.** A broad vacuum capacitor has plate area > $A=2.00\times10^{-2}\ \mathrm{m^2}$, gap $d=1.00\times10^{-3}\ \mathrm m$, and > terminal voltage $V=100\ \mathrm V$. Find its capacitance, charge, stored energy, > and the attractive force between the plates. > > In the broad-plate approximation (uniform gap field, fringing neglected), > > $$ > C=\frac{\varepsilon_0A}{d}=177\ \mathrm{pF},\qquad Q=CV=17.7\ \mathrm{nC}. > $$ > > The stored energy and the plate attraction are > > $$ > U=\tfrac12CV^2=8.85\times10^{-7}\ \mathrm J,\qquad > |F|=\frac{\varepsilon_0AV^2}{2d^2}=8.85\times10^{-4}\ \mathrm N. > $$ > > As a cross-check, the gap field is $E=V/d=1.00\times10^5\ \mathrm{V/m}$, the > electrostatic pressure is $\tfrac12\varepsilon_0E^2=4.43\times10^{-2}\ \mathrm{Pa}$, > and pressure times area returns the same $0.885\ \mathrm{mN}$. Recomputing $|F|$ > from $Q^2/(2\varepsilon_0A)$ at fixed charge gives the identical number. That pressure is small: a force sensor resolving several millinewtons would report mostly stage friction and gravity unless the geometry or voltage changes. The low value is why large-area electrostatic actuators rely on small gaps, many cells in parallel, or higher fields within material-breakdown limits. The fixed-charge and fixed-voltage force expressions agree here at a single operating point, but their source-work terms differ as the gap changes, so one history cannot be swapped for the other. ## Energy conservation during charging, discharge, and motion Energy bookkeeping becomes especially important when a capacitor is connected to a real circuit. Charge an initially uncharged linear capacitor through any positive resistance from an ideal constant-voltage source. The source delivers $$ W_{\rm source}=\int_0^Q V\,\d Q'=QV=CV^2. $$ At the final state, the capacitor stores $CV^2/2$. The remaining $CV^2/2$ appears as Joule heat in the resistance. The split does not depend on the resistance value; resistance changes the current history and the time required to settle. Taking the resistance toward zero does not eliminate the missing half within an ordinary classical circuit. Parasitic inductance, radiation, and source impedance then carry the transient energy until some dissipative mechanism absorbs it. Discharging the isolated capacitor through a resistor converts its entire initial field energy into heat. A voltage source connected during discharge changes this statement because the source can absorb energy, deliver energy, or do both during different intervals. Circuit diagrams that label a capacitor by a single voltage without identifying the source and switch state leave the energy destination undetermined. Electromechanical motion adds a third destination. A fixed-charge capacitor that moves from $x_1$ to $x_2$ changes its field energy by $$ \Delta U=-\int_{x_1}^{x_2}F_x\,\d x. $$ If no external circuit is attached, this energy change appears as mechanical work, kinetic energy, elastic energy, or dissipation in the mechanical support. At fixed voltage, the source also exchanges $\int V\,\d Q$, so field-energy change alone cannot determine mechanical work. The constrained function $\mathcal G$ accounts for that source exchange along a reversible voltage-controlled path. A rapid displacement introduces another scale. Charge redistribution through a finite lead resistance requires time, so a capacitor can behave closer to fixed charge during an abrupt motion even when its terminals are eventually attached to a voltage supply. Compare the mechanical motion time with the circuit time constant $RC$. Slow motion relative to $RC$ follows the voltage-controlled expression; fast motion can retain nearly constant charge for part of the trajectory. Intermediate motion requires simultaneous circuit and mechanical equations rather than an instant switch between the two static formulas. **Stability of voltage-controlled actuators.** Electrostatic force can compete with a restoring spring. Let the plate gap be $d$ and let a spring favor an unloaded gap $d_0$. With voltage held fixed, a suitable quasistatic potential for the combined system is $$ \mathcal G_{\rm total}(d)=\frac12k_s(d-d_0)^2 -\frac{\varepsilon_0AV^2}{2d}. $$ Equilibrium requires the derivative to vanish. Stability also requires positive curvature of this combined function. As the plates approach, the magnitude of the electrostatic attraction rises as $d^{-2}$ and its slope rises even faster. Beyond a geometry-dependent threshold, the spring cannot provide a neighboring stable equilibrium; the plates move together until a stop, contact, dielectric breakdown, or a nonideal force intervenes. Pull-in follows directly from the curvature of the combined energy function. The simple parallel-plate prediction treats the spring as linear, the voltage as perfectly regulated, and the overlap area as fixed. A compliant electrode can bend, changing both area and gap. A dielectric layer can add a series capacitance and a contact stop can prevent literal zero separation. Experimental pull-in data should therefore be compared with a model that includes the mechanical geometry, measured capacitance curve, and the electrical drive impedance. ## A disciplined solution method Start by drawing the conductors, dielectric regions, supports, and sources that remain connected during the proposed change. Mark each geometrical coordinate and its positive direction. The choice fixes the sign in $F_x\,\d x$ and prevents an attractive force from being reported as positive merely because a magnitude was calculated. Choose the controlled electrical quantity before differentiating. An isolated charged body calls for $U(Q,x)$ at fixed $Q$. A regulated source calls for $\mathcal G=U-QV$ at fixed $V$. A partially isolated circuit may require a charge constraint on one conductor and a voltage constraint on another; in that case write the full differential energy relation and solve for the source work explicitly. For mixed charge and voltage constraints, the circuit sketch must identify the controlled terminal variables. Obtain capacitance or field data from an appropriate geometry model. For broad parallel plates, $C=\varepsilon_0A/d$ follows from a nearly uniform central field. For coaxial, spherical, edge-dominated, or multi-electrode systems, derive or measure the relevant capacitance function. Check that the limiting dimensions have the expected behavior: increasing overlap should increase capacitance, increasing separation should reduce it, and removing a conductor should recover the simpler isolated configuration. Differentiate symbolically before inserting numbers. Units offer a fast filter: $\d U/\d x$ has units of newtons, $\varepsilon_0E^2/2$ has units of newtons per square metre, and $CV^2$ has units of joules. A result that depends on the sign of an area, gives a nonzero force where capacitance is independent of the coordinate, or predicts a force away from the region of stronger capacitance needs its assumptions revisited. Use a physical cross-check after the algebra. Electric force at fixed voltage tends to move a freely adjustable geometry toward larger capacitance because $\mathcal G=-CV^2/2$ decreases in that direction. At fixed charge, the same instantaneous tendency follows from reducing $Q^2/(2C)$. For a smooth conductor surface, compare the integrated force with the direction and scale of local pressure. For a circuit measurement, compare electrical input, field-energy change, mechanical work, and heat over the same interval. The following failure modes are worth testing deliberately in worked problems: | Modeling choice | Reliable check | | --- | --- | | Fixed charge | Disconnect the source before motion and verify $\d Q=0$. | | Fixed voltage | Include the source term through $\mathcal G=U-QV$. | | Uniform gap field | Compare plate width with gap and inspect the edge region. | | Surface pressure | Use the local normal field and integrate when curvature matters. | | Numerical derivative | Vary mesh spacing and displacement increment independently. | These checks test the electrical constraint, geometry approximation, pressure model, and numerical derivative independently. ## Several conductors and capacitance coefficients Many electrostatic devices have more than two conductors: a moving electrode may face two sensing pads, a shield may surround the assembly, and a reference plane may connect to a source reservoir. One scalar capacitance cannot represent every possible terminal condition. Choose a reference potential and relate the conductor charges to the conductor potentials through capacitance coefficients, $$ Q_i=\sum_j C_{ij}V_j. $$ For ordinary reciprocal electrostatics, the coefficient matrix is symmetric. Its diagonal entries are positive and its off-diagonal entries describe the way a raised potential on one conductor induces opposite charge on another. The energy is the quadratic form $$ U=\frac12\sum_iQ_iV_i =\frac12\sum_{i,j}C_{ij}V_iV_j. $$ The reference conductor belongs in the physical description. A distant enclosure, a grounded probe, or a voltage-source return can carry induced charge and can change both the matrix coefficients and the stored energy. If the reference at infinity is included as an additional conductor, charge conservation and the coefficient-sum relations become explicit; omitting it from a drawing does not remove its electrical role. Two isolated conductors provide the simplest matrix example. If their only relevant mutual capacitance is $C$, their charges satisfy $$ Q_1=C(V_1-V_2), \qquad Q_2=C(V_2-V_1), $$ and the energy depends only on the potential difference, $$ U=\frac12C(V_1-V_2)^2. $$ Grounding conductor 2 sets $V_2=0$ through a reservoir that supplies or removes its induced charge. Keeping conductor 2 electrically isolated instead holds its net charge fixed and changes the allowed relation between $V_1$, $V_2$, and motion. These alternatives look similar in a static field sketch yet lead to different energy derivatives when a conductor moves. The differential form extends directly to several terminals and several mechanical coordinates: $$ \d U=\sum_iV_i\,\d Q_i-\sum_\alpha F_\alpha\,\d x_\alpha. $$ When all listed voltages are held by ideal sources, form the constrained function $$ \mathcal G=U-\sum_iQ_iV_i, \qquad F_\alpha=-\left(\frac{\partial\mathcal G}{\partial x_\alpha}\right)_{V_1,V_2,\ldots}. $$ In a linear system, $sum_iQ_iV_i=2U$, so the constrained function is $-U$ after the voltage constraints have been imposed. The order matters. First express the charges and field energy for the actual geometry; then apply the terminal constraints; then differentiate. Substituting a fixed voltage too early into a fixed-charge expression hides the charge transferred by the source. Capacitive position sensors illustrate the matrix approach. A centered vane may have equal couplings to two pads. A small displacement raises one coefficient and lowers the other, allowing a differential bridge to reject common changes in temperature, cable capacitance, or uniform gap drift. A single-ended reading can still be distorted by a nearby grounded hand or shield because that object changes the complete coefficient matrix. Differential electronics reduce one class of error; they do not make the surrounding boundary irrelevant. Measurements of a capacitance matrix require controlled terminal conditions. Drive one terminal with a known small ac voltage while holding the others at their stated potentials, record the resulting terminal currents, and repeat for each independent drive. Reciprocity checks the matrix: after calibration and reference choice, it requires equal measured cross-couplings when source and response terminals are exchanged. Large disagreement points to wiring impedance, nonlinear materials, motion during acquisition, or an incomplete reference model. Force cross-talk can also be read from mixed derivatives. A geometry coordinate that changes both $C_{11}$ and $C_{12}$ can experience force when either pad is driven, and a two-voltage actuator can have a term proportional to $V_1V_2$. The sign follows the actual electrode polarities and geometry. Writing only the sum of individual plate pressures may miss these mutual terms when fields overlap. The same energy checks used for two conductors remain available. Compare charge obtained by surface integration with the matrix prediction, compare matrix energy with the volume integral of $\varepsilon_0E^2/2$, and compare coordinate derivatives with a stress-surface force. Agreement across those three calculations is especially valuable for a device whose motion occurs near multiple edges and shields. > **Worked example.** A grounded vane sits centered between two fixed sensing pads. > Near the center the two pad-to-vane capacitances fit straight lines, > $C_L(x)=C_0+\alpha x$ and $C_R(x)=C_0-\alpha x$, with $x$ toward the left pad and > $\alpha>0$. Drive the pads from ideal sources at $V_L$ and $V_R$ with the vane at > zero potential. Find the lateral force on the vane. > > Because both driven potentials are held fixed, the force follows from the > voltage-controlled function $\mathcal G$, with pad-to-pad coupling neglected: > > $$ > \mathcal G(x)=-\tfrac12C_L(x)V_L^2-\tfrac12C_R(x)V_R^2, > \qquad > F_x=-\frac{\partial\mathcal G}{\partial x}=\frac{\alpha}{2}\left(V_L^2-V_R^2\right). > $$ > > The force depends only on the difference of squared drives. Equal-magnitude drives > give zero first-order lateral force, even at opposite polarity, since the squares > match. The linear capacitance fit is local: it assumes an operating range far enough from contact that overlap, edge curvature, and dielectric shape vary smoothly, and a larger stroke needs the measured nonlinear functions instead. Zero net force does not mean zero local force: the pressure on each gap stays nonzero, and the two lateral tractions cancel. A differential readout is therefore sensitive to position without imposing a first-order lateral bias when the geometry and source amplitudes are balanced. Drive only the left pad and leave the right pad at the reference potential. The formula reduces to $F_x=\alpha V_L^2/2$, directed toward the left pad. The energy interpretation is immediate: leftward motion increases the driven-pad capacitance, so the voltage-controlled constrained energy decreases. Reversing the coordinate definition reverses the sign of $\alpha$ and of the reported force component while leaving the physical attraction unchanged. The measured electrical signal can be analyzed separately from the mechanical force. With a small ac excitation superposed on a dc bias, the pad current contains a term proportional to the time derivative of capacitance and a term proportional to voltage change. Holding the vane still during a capacitance calibration removes the motional contribution. Holding the source amplitude fixed during a slow position sweep allows the slope $\alpha$ to be fitted from the antisymmetric change in the two capacitances. A force measurement with the same bias then checks the derivative of the energy model. Parasitic capacitance changes the interpretation of an absolute measurement. Suppose each pad has an added fixed capacitance to a cable shield. That addition changes the measured total terminal capacitance but has zero derivative with respect to the vane coordinate, so it does not contribute to the intended lateral force. A calibration based only on absolute capacitance can therefore exaggerate the actuator coupling. Differencing the two position sweeps or fitting their slopes separates the position-dependent part from fixed lead and instrument contributions. The ideal symmetric result also defines diagnostic tests. A nonzero force at equal drive magnitudes can arise from unequal slopes, unequal nominal gaps, a tilted vane, or coupling through a third conductor. Interchanging the two drive channels should reverse the measured force if the coordinate convention and wiring are correct. Rotating the device by one hundred eighty degrees while retaining the laboratory coordinate can distinguish gravity or support friction from an electrostatic bias. Use those reversals as experimental tests of the matrix prediction and the stated geometry. At fixed pad charge, the same geometry uses a different energy function. The source terms disappear after disconnection, pad potentials change as the vane moves, and the force generally gains position dependence through the inverse capacitances. The voltage-controlled formula cannot be recycled by replacing $V$ with an initial value. The terminal condition must accompany every quoted force sensitivity, especially in microelectromechanical sensors where switching times can be comparable to mechanical motion times. This example links the scalar and matrix viewpoints. The broad-plate capacitor uses one coordinate and one capacitance. The three-conductor vane uses several terminal voltages and coordinate-dependent coefficients. In both cases, an energy derivative becomes a force only after the electrical constraints, reference conductor, and geometric coordinate have been named explicitly.