Defend a charge-sharing transient across model boundaries

Separate the preparation from the sharing interval

Two capacitors are prepared at voltages 6 V and 0 V, then every source is removed before the modeled sharing interval. Corresponding selected plates exchange charge through a stipulated finite 2 kilohm resistor; the opposite plates have an ideal return connection. The system boundary includes both capacitors and the resistor. It does not include a battery maintaining 6 V during the transient. This is a virtual circuit specification, not a switching procedure.

Derive two state equations and their invariant

Define I as current from the first selected plate to the second. Then C1 dV1/dt = −I, C2 dV2/dt = I and I = (V1 − V2)/R. Adding the first two relations proves C1 V1 + C2 V2 is constant. At equilibrium, the common voltage is Vf = 6C1/(C1 + C2), a capacitance-weighted average, not generally 3 V. Equal final potential does not imply equal charge when the capacitances differ.

Derive the difference mode

Subtract the two voltage derivatives: d(V1 − V2)/dt = −(1/R)(1/C1 + 1/C2)(V1 − V2). Define Ceq = C1C2/(C1 + C2); the difference decays as 6 exp[−t/(R Ceq)]. Recover V1 = Vf + C2(V1 − V2)/(C1 + C2) and V2 = Vf − C1(V1 − V2)/(C1 + C2). The series-like Ceq controls this difference mode even though the final selected plates share one potential.

Explain the apparent missing energy

Initial energy is C1×6²/2. Final capacitor energy is (C1 + C2)Vf²/2, which is smaller. The difference is Ceq×6²/2 and equals the integral of I²R over the full transient. Charge has not disappeared; charge conservation and energy conservation describe different inventories. Changing positive R changes the timescale but not the ideal final dissipated amount. Taking R to exactly zero is a singular limit in a model that omitted inductance and radiation, not evidence for instantaneous loss without a mechanism.

Challenge the fixed-component assumption

If a dielectric changes during the interval, if a source remains attached, or if probe loading adds another branch, the equations above no longer describe the same boundary. A proposed follow-up must state the new work or charge source rather than force old conservation statements onto a changed system. Record a circuit diagram, a source inventory and a finite-time energy account before consulting public feedback.

Worked original case

C1 = 2 microfarad and C2 = 3 microfarad give Ceq = 1.2 microfarad, tau = 2.4 ms and Vf = 2.4 V. At two tau, V1 ≈ 2.887207 V and V2 ≈ 2.075195 V. Total selected-plate charge remains 12 microcoulomb. Final capacitor energy is 14.4 microjoule versus initial 36 microjoule; full-transient heat is 21.6 microjoule.

Investigation

Retain a first prediction, both state equations, charge invariant, difference-mode derivation and finite-time energy residual. Change one capacitance and independently recompute the limiting voltage. Then critique a claim that charge loss caused the energy decrease, and write a new-system prediction for a source that remains connected.

Changed-case practice

Try a new practice case

Initially C1 = 3 microfarad is at 8 V and C2 = 1 microfarad is uncharged. What is final common voltage after source-free sharing?

Numerical result in the requested units Check this practice result

This public, unsaved check is not a secure assessment. Explain the source integral, symmetry or charge boundary in the retained course evidence workspace.

Sources and boundaries

Original teaching checked 31 October 2026 against current official calculus framework and the relevant OpenStax discussion. No source exercises or diagrams are reproduced. All records and apparatus are virtual, supplied or generated; no real wiring, switching, magnetic braking or electrical measurement is requested. Public feedback and retained work do not establish independent mastery, supervised practical participation or AP authorization.