SCIENCE · INVESTIGATE AND EXPLAIN

Integrate charging work instead of using a final-state rectangle

An initially uncharged ideal capacitor has voltage V(q) = q/C at an intermediate charge q.

Learning outcome: Compare two capacitances at fixed final voltage and two final voltages at fixed capacitance. Sketch both the charging line and accumulated-work curve. Then hold final Q fixed on paper and explain why changing C reverses the fixed-voltage energy trend.

Starting recap

Declare conductor boundaries, integrate potential or charging work, and distinguish controlled charge from voltage.

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Try

With C = 2 nF and final V = 4 V, Q = 8 nC and integrated work is 16 nJ. The final-state rectangle QV is 32 nJ. At fixed C, doubling V quadruples U and doubles Q.

Open the experiment

Predict

Why does final voltage multiplied by final charge overestimate reversible charging work from zero?

Your prediction and reason

Experiment

Compare two capacitances at fixed final voltage and two final voltages at fixed capacitance. Sketch both the charging line and accumulated-work curve. Then hold final Q fixed on paper and explain why changing C reverses the fixed-voltage energy trend.

Integrate the growing charging voltage

Worked teaching model · synthetic values · Ideal linear capacitor initially uncharged, quasistatic reversible charging. Differential external work is V(q)dq with V(q)=q/C. Fixed geometry and no leakage, dielectric loss or circuit transient modeled.

Capacitance: 2 nF Final voltage: 4 V

Reset keeps your prediction, captures and practice history.

Voltage during quasistatic charging

Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.

Voltage during quasistatic charging: numerical coordinates

accumulated charge (nC)	voltage (V)
0	0
0.133333	0.0666667
0.266667	0.133333
0.4	0.2
0.533333	0.266667
0.666667	0.333333
0.8	0.4
0.933333	0.466667
1.06667	0.533333
1.2	0.6
1.33333	0.666667
1.46667	0.733333
1.6	0.8
1.73333	0.866667
1.86667	0.933333
2	1
2.13333	1.06667
2.26667	1.13333
2.4	1.2
2.53333	1.26667
2.66667	1.33333
2.8	1.4
2.93333	1.46667
3.06667	1.53333
3.2	1.6
3.33333	1.66667
3.46667	1.73333
3.6	1.8
3.73333	1.86667
3.86667	1.93333
4	2
4.13333	2.06667
4.26667	2.13333
4.4	2.2
4.53333	2.26667
4.66667	2.33333
4.8	2.4
4.93333	2.46667
5.06667	2.53333
5.2	2.6
5.33333	2.66667
5.46667	2.73333
5.6	2.8
5.73333	2.86667
5.86667	2.93333
6	3
6.13333	3.06667
6.26667	3.13333
6.4	3.2
6.53333	3.26667
6.66667	3.33333
6.8	3.4
6.93333	3.46667
7.06667	3.53333
7.2	3.6
7.33333	3.66667
7.46667	3.73333
7.6	3.8
7.73333	3.86667
7.86667	3.93333
8	4

Accumulated reversible work

Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.

Accumulated reversible work: numerical coordinates

accumulated charge (nC)	work (nJ)
0	0
0.133333	0.00444444
0.266667	0.0177778
0.4	0.04
0.533333	0.0711111
0.666667	0.111111
0.8	0.16
0.933333	0.217778
1.06667	0.284444
1.2	0.36
1.33333	0.444444
1.46667	0.537778
1.6	0.64
1.73333	0.751111
1.86667	0.871111
2	1
2.13333	1.13778
2.26667	1.28444
2.4	1.44
2.53333	1.60444
2.66667	1.77778
2.8	1.96
2.93333	2.15111
3.06667	2.35111
3.2	2.56
3.33333	2.77778
3.46667	3.00444
3.6	3.24
3.73333	3.48444
3.86667	3.73778
4	4
4.13333	4.27111
4.26667	4.55111
4.4	4.84
4.53333	5.13778
4.66667	5.44444
4.8	5.76
4.93333	6.08444
5.06667	6.41778
5.2	6.76
5.33333	7.11111
5.46667	7.47111
5.6	7.84
5.73333	8.21778
5.86667	8.60444
6	9
6.13333	9.40444
6.26667	9.81778
6.4	10.24
6.53333	10.6711
6.66667	11.1111
6.8	11.56
6.93333	12.0178
7.06667	12.4844
7.2	12.96
7.33333	13.4444
7.46667	13.9378
7.6	14.44
7.73333	14.9511
7.86667	15.4711
8	16

Calculated values under the declared model
Quantity	Value	Unit
Final charge magnitude	8	nC
Integrated reversible charging work	16	nJ
Final-voltage rectangle Q times V	32	nJ
Energy per final unit charge	2	J/C

Charging voltage grows with accumulated charge, so final voltage times total charge overcounts reversible stored energy.

Integrating q/C from zero to Q gives Q²/(2C) = QV/2 = CV²/2. These equivalent final-state expressions do not justify applying final voltage to every infinitesimal charging step. An actual resistive charging circuit can transfer additional source energy; that later transient is a different account.

Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.

Charging voltage grows with accumulated charge, so final voltage times total charge overcounts reversible stored energy.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Accumulate charge in a linear capacitor

An initially uncharged ideal capacitor has voltage V(q) = q/C at an intermediate charge q. Quasistatic reversible external work is dW = V(q)dq. Integrating from 0 to Q gives U = Q²/(2C), equivalently QV/2 or CV²/2 at the final state.

Explain the missing half

Multiplying final V by total Q treats every increment as if it were transferred against the final voltage. In fact the voltage rises from zero. The area under the straight V-versus-q line is a triangle, not the full rectangle. This is not a rule that every source in every charging circuit supplies only half QV; later resistive transients require accounting for dissipated energy separately.

Connect to a spatial field account

For a vacuum parallel-plate gap, substitute C = epsilon0 A/d and V = Ed into CV²/2. This gives U = epsilon0 E² Ad/2, so the uniform field-energy density is epsilon0 E²/2. For E = 2000 N/C, A = 0.01 m² and d = 0.002 m, the energy is approximately 0.354168 nJ. Doubling E at the same geometry multiplies this by four. This local energy description is a derived extension, not a new storage mechanism in addition to U.

Worked original case

With C = 2 nF and final V = 4 V, Q = 8 nC and integrated work is 16 nJ. The final-state rectangle QV is 32 nJ. At fixed C, doubling V quadruples U and doubles Q.

Investigation

Changed-case practice

Try a new practice case

For C = 3 nF and final V = 6 V, what is stored energy in nJ?

Numerical result in the requested units

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 cases checked against the official framework on 30 October 2026, with OpenStax verification on 31 October 2026 against the current official framework, Unit 10 and OpenStax capacitance. These are virtual supplied models, not instructions to charge, ground, wire or handle real equipment. Public answers are practice; retained plans and self-reports do not establish independent mastery, actual laboratory participation or AP authorization.

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Which conductor boundary, geometric approximation or independent observation could change this conclusion?

Your explanation and limitations

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