Account for capacitor energy under different constraints

Choose what is fixed before predicting energy

For this linear capacitor, stored field energy can be written U = C V²/2 = Q²/(2C) = QV/2, where U is in joules, C in farads, V in volts and Q in coulombs. These are equivalent at the same state because Q = CV. A picojoule (pJ) is 10⁻¹² J.

At fixed voltage, U increases with C. At fixed charge, U decreases as C increases. There is no contradiction: these are different experiments. OpenStax: energy stored in a capacitor develops the equivalent energy expressions and the relation to the electric field.

Start from a shared reference

At 30 cm² overlap and 3 mm gap, both displayed scenarios have C = 8.854 pF, V = 6 V and Q = 53.124 pC. Stored energy is (8.854 × 10⁻¹²)(6²)/2 = 159.372 pJ.

Increase the gap to 4 mm without changing area. Predict the two outcomes before inspecting the common-scale energy bars:

• Connected to 6 V: C falls to 6.6405 pF and Q to 39.843 pC. Stored energy falls to 119.529 pJ, a change of −39.843 pJ.
• Isolated: Q remains 53.124 pC, voltage rises to 8 V and stored energy rises to 212.496 pJ, a change of +53.124 pJ.

The work table always compares the selected geometry with the 30 cm² / 3 mm reference. It does not use whichever setting you happened to view last. If comparing two other settings, subtract their energies and work totals consistently.

Where does the extra isolated energy come from?

Oppositely charged plates attract. In the isolated idealization, pulling them farther apart quasistatically requires external work. With no source attached, no friction and no kinetic-energy change, W external = ΔU. For the reference-to-4 mm change, the external agent supplies +53.124 pJ. The retained charge does not mean energy must stay fixed.

The calculation assumes a slow, controlled geometry change. It does not show plate acceleration, a force sensor, work lost to friction, or a discharge. At fixed area the isolated field remains constant, while the field-filled volume grows; this provides a second way to understand why energy increases in that comparison.

Include the source in the connected balance

At fixed 6 V, changing geometry transfers charge through the ideal source. For this quasistatic comparison, W source = V(Q final − Q reference), positive when the source transfers energy into the capacitor system. The capacitor’s balance is ΔU = W source + W external.

For the reference-to-4 mm change, charge change is 39.843 − 53.124 = −13.281 pC. Source work is therefore −79.686 pJ: energy is returned to the source. Since ΔU = −39.843 pJ, external work is ΔU − W source = +39.843 pJ. Pulling the plates apart still requires work, even though the stored field energy decreases. The source receives the released field energy plus the external work.

At the reference geometry all three work differences are zero. The stored energy itself is not zero. Returning to that setting means evaluating the reference again, not simulating an irreversible reset.

Avoid confusing geometry work with charging from zero

The factor one half in U is important because capacitor voltage rises as charge is accumulated at fixed geometry. The work calculation above instead holds source voltage fixed while geometry changes quasistatically. It is not a model of suddenly connecting an uncharged capacitor to a battery. A real charging transient can dissipate energy in resistance; that process needs a time-dependent circuit model.

The ideal geometry-work balance also assumes no leakage, no dielectric changes and no edge-field correction. Changing overlap area is an idealized geometry comparison; the diagram does not simulate moving machinery. The energy bars share a 0–800 pJ scale so both scenarios can be compared without automatic rescaling.

Transfer to another capacitance

A capacitor initially has C = 10 pF and V = 4 V. Compare reducing capacitance to 5 pF while either keeping it connected to 4 V or isolating it first. Calculate final energy and external work for each quasistatic case, using the initial state as reference.

Check the capacitor-energy transfer

Initially Q = 40 pC and U = 80 pJ. Connected: final Q = 20 pC and U = 40 pJ, so ΔU = −40 pJ. Source work = 4 × (20 − 40) = −80 pJ; external work = +40 pJ. Isolated: Q stays 40 pC, V rises to 8 V and U becomes 160 pJ; external work = ΔU = +80 pJ. Both cases require positive external work, but their stored-energy changes have opposite signs.

Optional field-energy check

For a uniform vacuum field, energy density is ε₀E²/2 in J/m³. Multiplying by field volume A d gives U = ε₀E² A d/2. Substituting E = V/d and C = ε₀ A/d yields C V²/2. This calculation shares the model's uniform-field assumption; agreement is not an independent measurement of edge effects.