advanced27 min studyLesson 237 in Physics

Make capacitance a matrix with a reference

Connect reciprocity, grounding, floating conductors, and positive field energy.

On this page

What you’ll understand

  • Compute conductor charges from Maxwell capacitance coefficients.
  • Explain symmetry and the role of a reference conductor.
  • Distinguish full-matrix zero row sums from reduced grounded coefficients.

A single capacitance relates one voltage difference to equal and opposite charges. Several conductors require a matrix because changing one potential redistributes charge on all the others. The matrix is constrained by electrostatics: reciprocity makes it symmetric, and field energy supplies positivity. Its apparent singularity depends on whether the reference conductor is still included. Keeping the reference explicit prevents a common mistake: demanding zero row sums from a matrix whose ground row and column have already been removed.

Superposition becomes a charge map

Fix the conductor geometry and assume a linear dielectric or vacuum between them, with no independently prescribed volume charge. Solve once with each electrode at unit potential and the others at zero. Superposition combines these basis potentials into any voltage configuration. Integrating the normal displacement around each conductor then gives its total charge as a linear combination of all conductor voltages. These coefficients are Maxwell capacitance coefficients, not arbitrary circuit fitting parameters.

A diagonal coefficient is positive and off-diagonal coefficients are nonpositive in the usual connected electrostatic setting. Raising one electrode’s voltage draws opposite-sign charge onto others held fixed. This sign convention differs from listing positive two-terminal capacitances in a circuit diagram, where the minus signs are written explicitly through voltage differences. State which convention is in use before comparing numerical tables. Reciprocity follows from Green’s identity, and exchanges the source and observation electrodes.

Qi=∑jCijVj,Cij=Cji,U=12∑iQiVi=12VTCVQ_i=\sum_jC_{ij}V_j,\qquad C_{ij}=C_{ji},\qquad U=\frac12\sum_iQ_iV_i=\frac12\mathbf V^{\mathsf T}C\mathbf V
Qi is conductor charge in C, Vi is potential in V, Cij is Maxwell capacitance in F, and U is stored energy in J. Geometry and linear material properties remain fixed.↔ Scroll wide equations horizontally.

Grounding removes a null direction

Include an enclosing reference conductor, or the appropriate infinity reference, as an additional index. Raising all potentials together changes no potential differences or fields. The complete matrix therefore annihilates the uniform vector, so every row sums to zero and its field-energy quadratic form has a gauge null direction. This statement includes the reference in the sum. A common voltage shift cannot be applied to the finite electrodes while secretly holding infinity at zero.

Once the reference is grounded, delete its row and column to form a reduced matrix. This matrix is normally positive definite and invertible for independent electrodes; its row sums need not vanish. They measure coupling to the removed reference. A floating electrode instead has fixed total charge and unknown voltage, which must be solved from its matrix row. Floating does not mean zero voltage, and a neutral floating electrode need not have zero local surface charge density.

C~1=0,V0=0 ⟹ Q=CredV\widetilde C\mathbf 1=0,\qquad V_0=0\ \Longrightarrow\ \mathbf Q=C_{\rm red}\mathbf V
C̃ includes every electrode and the reference; Cred omits the grounded index 0. The complete matrix has zero row sums, while reduced row sums equal minus the omitted reference coefficients.↔ Scroll wide equations horizontally.

Energy supplies a physical audit

For a passive linear medium, field energy is the integral of one half electric field dotted with displacement. It is nonnegative even when individual products of a conductor’s charge and voltage are negative. The total quadratic form, rather than each separate term, is the invariant physical quantity. Checking symmetry, signs, dimensions, and the relevant positivity can expose a table that cannot describe the proposed electrostatic geometry.

The matrix lab demonstrates why a positive determinant alone is a weaker statement. Reset and set diagonal scale to 1 and shear to 0; predict determinant 1 and condition number 1. Reset and set scale to 1 and shear to 1; predict the same determinant but unequal singular values and worse conditioning. These predictions concern readouts for the selected final matrix A. At Reset the square is still the identity; Play or Step interpolates geometry between identity and A. Its triangular matrix is nonsymmetric when shear is nonzero, so it is not a capacitance matrix or a solver for conductor charge.

U=12∫E⋅D d3r≥0,D=ϵE(ϵ>0)U=\frac12\int\mathbf E\cdot\mathbf D\,d^3r\geq0,\qquad \mathbf D=\epsilon\mathbf E\quad(\epsilon>0)
E is in V/m, D in C/m², and ε in F/m. This energy expression assumes a static, linear, reversible dielectric with positive permittivity; the integral excludes perfect-conductor interiors where E=0.↔ Scroll wide equations horizontally.

Two electrodes and a grounded enclosure

The reduced Maxwell matrix is [[3,−1],[−1,2]] nF. Apply V₁=2 V and V₂=1 V relative to a grounded enclosure. Find both charges, the stored energy, and the omitted reference couplings.

  1. Multiply each row by the voltage vector: Q₁=3×2−1×1=5 nC and Q₂=−1×2+2×1=0 nC. Zero net charge on electrode 2 does not imply zero voltage or zero surface polarization.

  2. Compute U=½(Q₁V₁+Q₂V₂)=½(5×2+0×1)=5 nJ. The symmetric reduced matrix has positive leading diagonal and determinant 5 nF², consistent with a positive-definite two-electrode energy form.

  3. The reduced row sums are 2 nF and 1 nF. The omitted full-matrix entries are therefore C₁₀=−2 nF and C₂₀=−1 nF. The reference charge is −5 nC, balancing the finite-electrode charges in this enclosing-conductor model.

The result

The charges are 5 nC and 0 nC, and stored energy is 5 nJ. Nonzero reduced row sums encode coupling to the grounded enclosure.

Explore the model

Interactive lab

Interactive lab · matrix

Separate determinant from conditioning

Reset and set a=1, b=0; predict determinant 1, singular values 1 and 1. Reset and set a=1, b=1; predict determinant 1 with singular values about 1.618 and 0.618.

Separate determinant from conditioningThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forReadouts describe the selected final A=[[a,b],[0,1]]. At Reset the square is identity; Play or Step interpolates identity↔A. This matrix family illustrates geometry and conditioning, not a capacitance solver.

Check your understanding

Why can a reduced grounded capacitance matrix have nonzero row sums?

Try it yourself

Use the same reduced matrix, but hold electrode 1 at 3 V while electrode 2 floats with zero net charge. Find V₂, Q₁, and U.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Set the second row’s charge to zero, then use the first row and the quadratic energy.

Compare with the solution

The floating condition is −V₁+2V₂=0, hence V₂=1.5 V. Then Q₁=3×3−1.5=7.5 nC and Q₂=0. The energy is ½×7.5 nC×3 V=11.25 nJ. Grounding electrode 2 instead would impose V₂=0 and change the charges, so floating and grounded are distinct boundary conditions.

Take it with you

  • Maxwell coefficients map all electrode potentials to their charges.
  • Only a matrix including the reference must have zero row sums.
  • A floating conductor supplies a charge constraint rather than a prescribed voltage.

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