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Capacitance and energy for specified linear conductor geometry

Physical scope

Capacitance and stored electrostatic energy for a specified linear ideal-conductor geometry in a named inertial frame: with the additional bounded smooth ambient domain, a connected vacuum region, a grounded outer boundary and no volume source, the charge-potential map of the conductor bodies is linear with a symmetric positive definite capacitance matrix, the field energy equals one half of the pairing of potentials and charges, and for a single driven conductor the usual two-terminal relations and positivity of the capacitance hold. The item is conditional on the smooth solvability theorem and asserts no arbitrary-boundary or charge-data solvability, no material response, no transient charging dynamics, no radiation and no empirical claim.

Statement

Assume the Axiom of Choice for the cited unit-Dirichlet solvability and trace interfaces, and its Countable Choice consequence for surface and integral interfaces (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $(Ω; D_{1}, \dots, D_{m}; Ω_{v})$ be the conductor geometry of The ideal electrostatic conductor model, additionally with $Ω$ a bounded $C^{\infty}$ ambient domain and $Ω_{v}$ connected and with the outer boundary $\partial Ω$ grounded: the data on $\partial Ω$ are the constant Dirichlet datum $0$ ; let $ρ_{v} \equiv 0$ on $Ω_{v}$ (the pure capacitor configuration: no volume source), and let the framework constants be those of Framework-local SI units, the vacuum constants and the exact/measured status of their values. For each $j \in {1, \dots, m}$ let $φ_{j} \in C^{\infty} (\overline{Ω_{v}})$ be the unit-potential solution of Existence and regularity for the electrostatic boundary value problem(c) with the smooth data $φ_{j} ∣_{\partial D_{k}} = δ_{j k}, φ_{j} ∣_{\partial Ω} = 0,$ so that $- Δ φ_{j} = 0$ on $Ω_{v}$ with the Laplacian of The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field, and let $ν$ be the outward normal of $Ω_{v}$ (so on $\partial D_{j}$ the normal $ν$ points into the conductor). Define the capacitance matrix by $C_{j k} : = ε_{0} \int_{\partial D_{j}} \frac{\partial φ_{k}}{\partial ν} d S, j, k = 1, \dots, m,$ whose entries carry farads (F), with the unit potentials read as numerical volt values; these are flux integrals of continuous functions over the compact embedded hypersurfaces $\partial D_{j}$ (normal derivative of Classical normal derivative) (Surface integration on compact C1 hypersurfaces, The vacuum region around compact smooth conductors is a bounded smooth domain). Then: (a) [Linearity and matrix form] for every $V = (V_{1}, \dots, V_{m}) \in R^{m}$ the unique equilibrium potential with conductor potentials $V$ and grounded outer boundary is $φ^{V} = \sum_{j = 1}^{m} V_{j} φ_{j}$ , and the conductor charges of Surface charge and normal field on an ideal conductor(d) are $Q_{j} (V) = ε_{0} \int_{\partial D_{j}} \frac{\partial φ^{V}}{\partial ν} d S = \sum_{k = 1}^{m} C_{j k} V_{k}, i.e. Q = C V .$ (b) [Reciprocity] $C$ is symmetric: $C_{j k} = C_{k j}$ for all $j, k$ ; the symmetry is Green's symmetric pairing of the harmonic potentials $φ_{j}, φ_{k}$ . (c) [Energy identity] the field energy of the configuration is $U (V) = \frac{ε_{0}}{2} \int_{Ω_{v}} ∣ \nabla φ^{V} ∣^{2} d λ_{3} = \frac{1}{2} V^{T} C V = \frac{1}{2} \sum_{j = 1}^{m} V_{j} Q_{j} (V),$ and $U (V) > 0$ for every $V \neq 0$ : $C$ is symmetric positive definite. In particular $U (V) \geq 0$ , with equality exactly for $V = 0$ , so every nonzero conductor-potential data set stores positive field energy. (d) [One driven terminal] with only conductor $1$ driven and every other conductor grounded, $V = t e_{1}$ , where $e_{1} : = (1, 0, \dots, 0) \in R^{m}$ : $Q_{1} = C_{11} t$ , $Q_{j} = C_{j 1} t$ for $j \neq 1$ , and $U = \frac{1}{2} C_{11} t^{2} = \frac{1}{2} Q_{1} V_{1} = Q_{1}^{2} / (2 C_{11})$ ; the capacitance of the arrangement at conductor $1$ with respect to the grounded reference is the positive geometric number $C_{11} > 0$ , with $C_{11} = Q_{1} / V_{1}$ when $t = V_{1} \neq 0$ , and the energy identities $U = \frac{1}{2} Q V = Q^{2} / (2 C) = \frac{1}{2} C V^{2}$ hold with $C = C_{11}$ , $Q = Q_{1}$ , $V = V_{1}$ . By (c) the positivity holds for every diagonal entry, $C_{j j} > 0$ . (e) [Assembly reading] use the family $V_{j} (s) = s V_{j}$ , $φ_{s} = s φ^{V}$ , $Q_{j} (s) = s Q_{j} (V)$ for $0 \leq s \leq 1$ , with fixed geometry and grounded reference. Prescribe the quasistatic external charge-transfer work by $d W_{e x t} = \sum_{j} V_{j} (s) d Q_{j} (s)$ . Then $W_{e x t} = \int_{0}^{1} s d s \sum_{j} V_{j} Q_{j} (V) = U$ . The potentials vary during assembly; only the final potentials and geometry are fixed. This derives the sheet-charge accounting directly and does not apply the volume-source theorem Finite-source assembly energy equals the electric field energy to singular sheets. The family is an accounting prescription, not an asserted time-dependent Maxwell/material solution. (f) [Scope] the theorem assumes: the compact separated conductor geometry, the additional bounded smooth ambient domain, and connected vacuum region of the model, the grounded outer boundary $\partial Ω$ , zero volume source, and the existence and uniqueness of the unit-potential Dirichlet solutions (Existence and regularity for the electrostatic boundary value problem(c), Uniqueness and flux compatibility for the electrostatic boundary value problem(a)). It asserts no solvability for arbitrary boundary data (the existence theorem is used only for the constant data above), no prescribed-total-charge problem, no dielectric or material response, no finite-conductivity or transient charging dynamics, no radiation, and no measurement; the matrix $C$ is a geometric object of the ideal electrostatic model, and infinite parallel planes require their own explicitly declared idealization, outside this bounded geometry.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice as declared in the statement; the conductor geometry $(Ω; D_{1}, \dots, D_{m}; Ω_{v})$ of the model in the bounded smooth ambient branch with $Ω$ a bounded $C^{\infty}$ domain, $Ω_{v}$ connected, outer boundary grounded and $ρ_{v} \equiv 0$ ; the unit-potential solutions $φ_{j} \in C^{\infty} (\overline{Ω_{v}})$ with $- Δ φ_{j} = 0$ , $φ_{j} ∣_{\partial D_{k}} = δ_{j k}$ , $φ_{j} ∣_{\partial Ω} = 0$ ; the outward unit normal $ν$ of $Ω_{v}$ ; and the capacitance matrix $C_{j k} = ε_{0} \int_{\partial D_{j}} \partial φ_{k} / \partial ν d S$ .

[F1]

The model and its vacuum geometry: in the bounded smooth ambient branch of The ideal electrostatic conductor model, by The vacuum region around compact smooth conductors is a bounded smooth domain the vacuum region $Ω_{v}$ is a bounded $C^{\infty}$ domain (in particular a bounded $C^{1}$ domain), its boundary is the disjoint union $\partial Ω \cup \partial D_{1} \cup \dots \cup \partial D_{m}$ of nonempty compact embedded smooth hypersurfaces, and $\partial Ω$ is nonempty; the model requires $Ω_{v}$ connected and nonempty, carries the pure electrostatic regime and the fields of the framework, and its constants are those of Framework-local SI units, the vacuum constants and the exact/measured status of their values (The ideal electrostatic conductor model, The vacuum region around compact smooth conductors is a bounded smooth domain, The static and pure electrostatic regimes and their fixed-time reading, The electric and magnetic fields as fixed-time spatial vector fields).

[F2]

Unit-potential solvability and uniqueness: the datum equal to $δ_{j k}$ on $\partial D_{k}$ and $0$ on $\partial Ω$ is constant on each compact boundary piece, hence smooth on the disjoint union $\partial Ω_{v}$ ; with $ρ_{v} \equiv 0$ the conductor reading of Existence and regularity for the electrostatic boundary value problem(c) is a special case of its Dirichlet branch, so there is exactly one $φ_{j} \in C^{\infty} (\overline{Ω_{v}})$ with $- Δ φ_{j} = 0$ on $Ω_{v}$ and the prescribed boundary values; uniqueness in the class of $C^{\infty} (\overline{Ω_{v}})$ functions with identical boundary data is the Dirichlet uniqueness clause (Existence and regularity for the electrostatic boundary value problem(a),(c), Uniqueness and flux compatibility for the electrostatic boundary value problem(a), The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field, $C^{k}$ Euclidean maps and diffeomorphisms).

[F3]

Surface charges and fluxes: for an equilibrium solution of the model with the joint one-sided regularity, $Q_{j} = \int_{\partial D_{j}} σ d S = ε_{0} \int_{\partial D_{j}} \partial φ / \partial ν d S$ , where $ν$ is the outward normal of the vacuum; the integrals exist as integrals of the continuous flux over the compact embedded hypersurfaces $\partial D_{j}$ (Surface charge and normal field on an ideal conductor(d), Surface integration on compact C1 hypersurfaces, Classical normal derivative).

[F4]

Green's first identity on the bounded $C^{1}$ domain $Ω_{v}$ : for real $u \in C^{2} (\overline{Ω_{v}})$ and $v \in C^{1} (\overline{Ω_{v}})$ , $\int_{Ω_{v}} (v Δ u + D u \cdot D v) d λ_{3} = \int_{\partial Ω_{v}} v \partial_{ν} u d S$ with the outward normal $ν$ ; the boundary integral splits over the disjoint pieces $\partial Ω, \partial D_{1}, \dots, \partial D_{m}$ (First Green identity, [F1]).

[F5]

Positivity and constancy tools for (c): a continuous function on the compact $\overline{Ω_{v}}$ is bounded and Borel, so for $φ^{V} \in C^{\infty} (\overline{Ω_{v}})$ the function $∣ \nabla φ^{V} ∣^{2}$ is nonnegative, Borel, bounded and integrable over $Ω_{v}$ ; a nonnegative measurable $f$ with $\int f d λ_{3} = 0$ vanishes almost everywhere, and a continuous such $f$ on the connected open $Ω_{v}$ then vanishes everywhere (otherwise a positive value would stay bounded below on a small Euclidean ball of positive finite measure); a totally differentiable function on a nonempty connected open Euclidean set with vanishing differential is constant; a function constant on the connected open $Ω_{v}$ and continuous on $\overline{Ω_{v}}$ has that constant equal to its value $0$ on the nonempty boundary piece $\partial Ω$ (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere, Euclidean balls have positive finite Lebesgue measure, A differentiable map on a connected open Euclidean set has zero derivative exactly when it is constant, Continuous functions on Euclidean spaces are Borel measurable, Monotonicity and nonnegative homogeneity of the nonnegative integral, Integral over a measurable subset).

[F6]

Path-work reading and the elementary integral: for open $U \subseteq R^{n}$ , $ϕ \in C^{1} (U)$ and a piecewise $C^{1}$ path $γ : [a, b] \to U$ , $\int_{γ} \nabla ϕ \cdot d r = ϕ (γ (b)) - ϕ (γ (a))$ ; and $\int_{0}^{1} s d s = \frac{1}{2}$ by the one-dimensional fundamental theorem (The gradient theorem: the line integral of a gradient is the endpoint increment, The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ).

Proof

technique · direct

1.1F1F2F3given

Setup and well-definedness: by [F1] the vacuum region $Ω_{v}$ is a bounded $C^{1}$ domain (indeed $C^{\infty}$ ), connected, nonempty, with boundary the disjoint union of the nonempty compact pieces $\partial Ω, \partial D_{1}, \dots, \partial D_{m}$ , and with $\partial Ω$ nonempty. By [F2] each unit-potential solution $φ_{j} \in C^{\infty} (\overline{Ω_{v}})$ exists and is unique with $- Δ φ_{j} = 0$ and $φ_{j} ∣_{\partial D_{k}} = δ_{j k}$ , $φ_{j} ∣_{\partial Ω} = 0$ . The normal $ν$ is the outward unit normal of $Ω_{v}$ ; it points into the conductor on each $\partial D_{j}$ [F1]. Each flux $\int_{\partial D_{j}} \partial φ_{k} / \partial ν d S$ exists because $\partial φ_{k} / \partial ν$ is continuous on the compact embedded hypersurface $\partial D_{j}$ carrying finite surface measure [F3], so $C_{j k}$ is well defined.

2.1F2F3givenstep 1.1

Linearity and the charge-potential map (a): fix $V \in R^{m}$ and put $φ^{V} : = \sum_{j = 1}^{m} V_{j} φ_{j} \in C^{\infty} (\overline{Ω_{v}})$ . By linearity of the Laplacian, $- Δ φ^{V} = \sum_{j} V_{j} (- Δ φ_{j}) = 0$ on $Ω_{v}$ ; on $\partial D_{k}$ one has $φ^{V} = \sum_{j} V_{j} δ_{j k} = V_{k}$ , and on $\partial Ω$ , $φ^{V} = 0$ . By the uniqueness clause of [F2], $φ^{V}$ is the unique equilibrium potential with conductor potentials $V$ and grounded outer boundary, and the solution is linear in $V$ by construction. Its charges are given by [F3] and the definition of $C$ : for each $j$ , $Q_{j} (V) = ε_{0} \int_{\partial D_{j}} \partial φ^{V} / \partial ν d S = ε_{0} \sum_{k = 1}^{m} V_{k} \int_{\partial D_{j}} \partial φ_{k} / \partial ν d S = \sum_{k = 1}^{m} C_{j k} V_{k}$ , that is $Q = C V$ .

2.2F4givenstep 1.1

Reciprocity (b): apply Green's first identity [F4] on the bounded $C^{1}$ domain $Ω_{v}$ twice, first with $u = φ_{j}$ , $v = φ_{k}$ and then with $u = φ_{k}$ , $v = φ_{j}$ ; both functions lie in $C^{\infty} (\overline{Ω_{v}})$ and $Δ φ_{j} = Δ φ_{k} = 0$ , so subtracting the two identities gives $0 = \int_{\partial Ω_{v}} (φ_{j} \partial_{ν} φ_{k} - φ_{k} \partial_{ν} φ_{j}) d S$ . The boundary splits into $\partial Ω$ and the conductor surfaces [F4]; on $\partial Ω$ both traces of $φ_{j}, φ_{k}$ vanish, so the outer piece contributes $0$ , while on $\partial D_{ℓ}$ the traces are $δ_{j ℓ}$ and $δ_{k ℓ}$ , so the total reduces to $\int_{\partial D_{j}} \partial_{ν} φ_{k} d S - \int_{\partial D_{k}} \partial_{ν} φ_{j} d S$ . Multiplying by $ε_{0}$ gives $C_{j k} = C_{k j}$ for all $j, k$ ; the symmetry is exactly Green's antisymmetric pairing of the two harmonic potentials.

3.1F3F4F5givenstep 2.1step 2.2

Energy identity and positivity (c): apply Green's first identity [F4] once more with $u = v = φ^{V}$ : $\int_{Ω_{v}} (φ^{V} Δ φ^{V} + ∣ \nabla φ^{V} ∣^{2}) d λ_{3} = \int_{\partial Ω_{v}} φ^{V} \partial_{ν} φ^{V} d S$ . The Laplacian term vanishes by [step 2.1], the outer boundary contributes $0$ because $φ^{V}$ vanishes on $\partial Ω$ , and on $\partial D_{j}$ the trace of $φ^{V}$ is the constant $V_{j}$ , so $\int_{\partial Ω_{v}} φ^{V} \partial_{ν} φ^{V} d S = \sum_{j = 1}^{m} V_{j} \int_{\partial D_{j}} \partial_{ν} φ^{V} d S$ . By [F3] and [step 2.1] this equals $\sum_{j} V_{j} Q_{j} (V) / ε_{0} = \frac{1}{ε_{0}} \sum_{j, k} V_{j} C_{j k} V_{k} = \frac{1}{ε_{0}} V^{T} C V$ ; multiplying by $ε_{0} / 2$ gives the identity $U (V) = \frac{ε_{0}}{2} \int_{Ω_{v}} ∣ \nabla φ^{V} ∣^{2} d λ_{3} = \frac{1}{2} V^{T} C V = \frac{1}{2} \sum_{j} V_{j} Q_{j} (V)$ . The energy is nonnegative because the integrand is; if $U (V) = 0$ , then $\int_{Ω_{v}} ∣ \nabla φ^{V} ∣^{2} d λ_{3} = 0$ , so $∣ \nabla φ^{V} ∣^{2} = 0$ almost everywhere by [F5], hence everywhere by continuity on the connected open $Ω_{v}$ [F5]; thus $φ^{V}$ is constant on $Ω_{v}$ by [F5], and the constant is $0$ because $φ^{V} = 0$ on the nonempty boundary piece $\partial Ω$ and $φ^{V}$ is continuous on the closure [F5]; consequently $V_{j} = φ^{V} ∣_{\partial D_{j}} = 0$ for every $j$ , i.e. $V = 0$ . Hence $U (V) \geq 0$ with equality exactly at $V = 0$ , and since $C$ is symmetric by [step 2.2], the symmetric bilinear form $V \mapsto \frac{1}{2} V^{T} C V$ is positive definite.

4.1givenstep 2.1step 3.1

One driven terminal (d): take $V = t e_{1}$ with $t \in R$ . By [step 2.1], $φ^{V} = t φ_{1}$ , $Q_{1} = C_{11} t$ and $Q_{j} = C_{j 1} t$ for $j \neq 1$ ; by [step 3.1], $U = \frac{1}{2} C_{11} t^{2} = \frac{1}{2} Q_{1} V_{1}$ . For $t = 0$ all charges and the energy vanish while the unit flux defining $C_{11}$ is unchanged; for $t \neq 0$ one may divide by $t = V_{1}$ to obtain $C_{11} = Q_{1} / V_{1}$ and $U = Q_{1}^{2} / (2 C_{11})$ , using $C_{11} > 0$ from the positive definiteness of [step 3.1] (take $V = e_{1}$ : $U (e_{1}) = \frac{1}{2} C_{11} > 0$ ). Hence $C_{11}$ is the positive geometric capacitance of conductor $1$ with respect to the grounded reference, and the displayed identities $U = \frac{1}{2} Q V = Q^{2} / (2 C) = \frac{1}{2} C V^{2}$ hold with $C = C_{11}$ , $Q = Q_{1}$ , $V = V_{1}$ .

4.2F6givenstep 2.1step 3.1

Assembly reading (e): for $0 \leq s \leq 1$ put $V (s) : = s V$ , $φ_{s} : = φ^{s V} = s φ^{V}$ by [step 2.1], and $Q_{j} (s) : = Q_{j} (s V) = s Q_{j} (V)$ , so that $d Q_{j} (s) = Q_{j} (V) d s$ . The prescribed quasistatic external charge-transfer work is $d W_{e x t} = \sum_{j} V_{j} (s) d Q_{j} (s) = \sum_{j} (s V_{j}) Q_{j} (V) d s = s d s \sum_{j} V_{j} Q_{j} (V)$ , and summing over the conductors and integrating over $s \in [0, 1]$ gives $W_{e x t} = (\int_{0}^{1} s d s) \sum_{j} V_{j} Q_{j} (V) = \frac{1}{2} \sum_{j} V_{j} Q_{j} (V) = U$ by [F6] and [step 3.1]. The family is an accounting prescription on the fixed geometry with grounded reference: the potentials vary during assembly, only the final potentials and geometry are fixed, and the sheet-charge accounting is derived directly rather than from the volume-source theorem Finite-source assembly energy equals the electric field energy; the path-work identity of [F6] gives the endpoint potential-drop reading along any interior piecewise $C^{1}$ path, with boundary endpoints understood as limits of the continuous closure traces. This is (e).

5.1F1F2givenstep 2.1step 3.1step 4.1step 4.2∎

Scope (f): the theorem used the compact separated conductor geometry and bounded smooth ambient branch with connected vacuum region [F1], the grounded outer boundary, the zero volume source, and the existence and uniqueness of the unit-potential Dirichlet solutions [F2]; the existence theorem was applied only to the constant data $V$ and $0$ , so no solvability for arbitrary boundary data or for prescribed total charges is asserted, and no dielectric coefficient, material response, finite-conductivity or transient charging dynamics, radiation, measurement or apparatus claim enters [F1, F2]. The matrix $C$ is a geometric object of the ideal electrostatic model of The ideal electrostatic conductor model, and the infinite parallel-plane example is a separate idealization outside this bounded geometry; this completes (f) and the theorem.

Capacitance and energy for specified linear conductor geometry

Statement

Facts & Assumptions

Proof

Depends on

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