MCAT content library / Chemistry & physics / Capacitors: distinguish stored charge, voltage, and energy
Start with what the connections hold fixed, then use capacitance to predict charge, voltage and energy without mixing their roles.
About 15 minutes, including the worked examples and checks.
Useful foundations
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Concept: capacitance is not the same quantity as charge
A capacitor can change its stored energy while retaining the same charge, or change its charge while retaining the same voltage. The outcome depends on what the connections allow to change. Before substituting into an energy equation, identify that constraint. AAMC Category 4C includes ideal capacitors, their energy, combinations and dielectric effects.1
For an ideal linear capacitor, Q = CV. V is the potential difference between the two plates. Q is the magnitude of charge on one plate: the pair carries +Q and −Q, so its net charge is zero. “A capacitor stores 36 µC” therefore does not mean that the whole device has 36 µC of net charge or that Q should be doubled when using the equation.2
Capacitance is measured in farads:1 F =1 C/V. For fixed geometry and an unchanged linear insulating material, increasing Q also increases V, leaving C unchanged. The relationship does not say that a device’s capacitance doubles whenever a battery supplies twice the charge.
Pattern: geometry and material set the capacitance
For parallel plates whose separation is small relative to their dimensions, neglecting edge effects gives C = κε₀A/d. A is the overlapping area of one plate, d is the gap and κ is the relative permittivity of a material filling the entire gap. Vacuum has κ =1. Larger area or smaller separation increases C; doubling both A and d leaves it unchanged.24
For a separate numerical example, take vacuum plates with A =0.010 m² and d =1.00 mm =0.00100 m. Using ε₀ =8.85×10−12 F/m gives C =8.85×10−11 F =88.5 pF. Filling the unchanged gap with a linear dielectric of κ =4 raises this to 354 pF. Do not put a millimeter value directly into a formula using SI permittivity.
A dielectric polarizes: its bound charges rearrange or orient and partly oppose the field produced by a specified free plate charge. This does not create a conducting path across the gap. At fixed free Q, the net field and voltage decrease. If a battery instead holds V fixed across an unchanged gap, additional free charge arrives through the external circuit and the net field V/d stays fixed.5
Rule: choose an energy form that exposes the fixed quantity
The stored energy has three equivalent forms: U = ½CV² = Q²/(2C) = ½QV. They give the same answer for a state that satisfies Q = CV. The first is convenient when voltage is held fixed; the second is convenient when charge is held fixed. U is energy in joules, whereas C is capacitance in farads.3
Use two identical original teaching capacitors, each initially C₀ =3.0 µF at V₀ =12 V. Each has Q₀ =3.0×12 =36 µC and U₀ =½×3.0×12² =216 µJ. A dielectric with κ =4 will fully fill each gap without changing plate area or spacing. Assume a linear ideal dielectric and no leakage.
| State | C (µF) | V (V) | Q (µC) | U (µJ) |
|---|---|---|---|---|
| Initial, either capacitor | 3 | 12 | 36 | 216 |
| A: battery remains connected | 12 | 12 | 144 | 864 |
| B: battery disconnected first | 12 | 3 | 36 | 54 |
Battery connected: voltage is fixed
The battery maintains 12 V while capacitance rises to 12 µF. Charge rises to 144 µC and energy to 864 µJ. Both are four times their initial values. The additional 108 µC comes through the external leads; it does not move through the insulating dielectric.
The battery delivers electrical energy VΔQ =12×108 =1296 µJ during insertion. The capacitor’s stored energy rises by 648 µJ. In an ideal controlled insertion with no dissipative loss, the remaining 648 µJ is mechanical work done on the inserted material and can be extracted by the controlling agent. If that movement is not controlled, other forms of energy can appear. Counting only the capacitor while ignoring the battery would miss the energy input.
Battery disconnected first: charge is fixed
With no conducting path and no leakage, each plate keeps its original free charge magnitude 36 µC. Capacitance still rises to 12 µF, but voltage becomes Q/C =3 V and energy becomes Q²/(2C) =54 µJ. The dielectric change reduces this capacitor’s stored energy by 162 µJ. In the ideal controlled model, that decrease can supply mechanical work during insertion.4
These opposite energy changes do not contradict one another. The two devices are subject to different electrical constraints. “Adding a dielectric increases capacitance” is sufficient to determine C; it is insufficient to determine U until you know whether Q or V is being held fixed.
Application: series and parallel preserve different quantities
For initially uncharged ideal capacitors connected across a battery, parallel branches share the same voltage. Their plate charges add at each battery terminal, giving Ceq = C₁ + C₂. A series chain has equal plate-charge magnitudes, with no extra charge placed on its isolated intermediate connection. Its voltages add, giving 1/Ceq =1/C₁ +1/C₂. These rules differ from the corresponding resistor rules.6
Connect 3 µF and 6 µF in parallel across 12 V: Ceq =9 µF, the charges are 36 and 72 µC, and the battery-terminal charge magnitude is 108 µC. Reconnect those initially uncharged capacitors in series across 12 V: Ceq =2 µF and each has 24 µC. Their voltages are 24/3 =8 V and 24/6 =4 V, which sum to 12 V. The smaller capacitor has the larger voltage in that series pair.
The series pair stores½×2×12² =144 µJ, equal to½×3×8² +½×6×4² =96 +48 µJ. This provides a consistency check: the equivalent device and its individual components must agree about stored energy. Identify the actual shared nodes before classifying a connection; the page position of a symbol does not establish series or parallel.
For a changed geometry, dielectric or network, first recalculate C. Then apply the relevant fixed voltage or conserved charge. Finally compute energy and account for any battery input or mechanical work. Treating C, Q, V and U as separate quantities prevents a correct formula from being used with the wrong constraint.
ACTIVE PRACTICE · ELECTRIC CIRCUITS
Choose the Constraint Before Calculating the Energy
Start with the original example: an ideal capacitor has C₀ = 3.0 µF, V₀ = 12 V, Q₀ = 36 µC and U₀ = 216 µJ. Compare the two distinct ways of adding a dielectric to an unchanged plate geometry.
Capacitance µF; plate-charge magnitude µC; voltage V; stored energy µJ.
With the battery disconnected, charge remains fixed. Increasing C reduces V and U.
Ideal linear dielectric with unchanged plate dimensions, complete filling, and no leakage. Energy change can involve work and energy exchange with the battery or mechanical agent; it is not created without an energy source.
Check your reasoning
Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.
Check 1
A vacuum parallel-plate capacitor is charged to 9 V with capacitance 2 µF. Both leads are then disconnected, with no leakage. An external agent slowly doubles the plate separation while keeping the plate area unchanged and edge effects negligible. What happens to voltage and stored energy?
A. Voltage becomes 4.5 V; energy halves.
Review this choice. Doubling the gap halves capacitance. Since charge is fixed, both voltage Q/C and energy Q²/(2C) increase instead.
B. Voltage stays 9 V; energy stays unchanged.
Review this choice. The disconnected capacitor is not held at a fixed voltage. Its unchanged charge occupies a smaller capacitance.
C. Voltage becomes 18 V; energy doubles.
Correct. Initially Q=2×9=18 µC. The new capacitance is 1 µF, so V=18 V. Energy grows from 81 to 162 µJ; the external separating agent supplies the increase.
D. Voltage becomes 18 V; energy halves.
Review this choice. The voltage change is correct, but at fixed Q the energy is½QV. Doubling voltage doubles that energy.
The principle to carry forward
Disconnecting the battery fixes free plate charge in the ideal model; changing geometry can still change voltage and energy.
Check 2
Two identical capacitors begin at 2 µF and 6 V. A dielectric with κ =3 fully fills each unchanged gap. One remains connected to an ideal 6 V battery; the other is disconnected before insertion with no leakage. What is U_connected/U_disconnected after insertion?
A. 1/9
Review this choice. This reverses the ratio. The connected capacitor gains energy while the isolated one loses energy.
B. 1/3
Review this choice. This is the isolated capacitor’s final-to-initial energy ratio, not the requested ratio between the two final states.
C. 3
Review this choice. The connected capacitor’s energy triples, but the isolated capacitor’s energy also changes: it falls to one third of its initial value.
D. 9
Correct. Initial energy is 36 µJ. Connected: C=6 µF and V=6 V, so U=108 µJ. Disconnected: Q=12 µC and C=6 µF, so U=12 µJ. Their ratio is 9.
The principle to carry forward
With equal initial states and the same full-gap dielectric, the final-energy ratio is κ divided by 1/κ, or κ².
Check 3
An ideal 10 V battery connects to two parallel branches. One branch contains a single 3 µF capacitor. The other contains two 4 µF capacitors in series. All capacitors are initially uncharged and allowed to finish charging. What are the whole network’s equivalent capacitance and the charge magnitude on either 4 µF capacitor?
A. Equivalent 2 µF; individual charge 20 µC.
Review this choice. The series branch does have 2 µF and 20 µC, but this omits the parallel 3 µF branch from the whole-network capacitance.
B. Equivalent 5 µF; individual charge 20 µC.
Correct. The series pair gives(1/4+1/4)⁻¹=2 µF. In parallel with 3 µF, the network gives 5 µF. The series branch carries 2×10=20 µC, which is the magnitude on each of its capacitors.
C. Equivalent 5 µF; individual charge 50 µC.
Review this choice. The capacitance is correct, but 50 µC is the battery-terminal charge magnitude for the entire network. It is divided between the two branches.
D. Equivalent 11 µF; individual charge 110 µC.
Review this choice. Adding 3+4+4 treats all three capacitors as parallel. The two 4 µF components instead share a series branch.
The principle to carry forward
Reduce the series branch first, combine the parallel branches, then return to the relevant branch to find an individual charge.
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Sources
1. AAMC: What’s on the MCAT Exam? — Category 4C
2. OpenStax University Physics Volume 2: Capacitors and Capacitance
3. OpenStax University Physics Volume 2: Energy Stored in a Capacitor
4. OpenStax University Physics Volume 2: Capacitor with a Dielectric
5. OpenStax University Physics Volume 2: Molecular Model of a Dielectric
6. OpenStax University Physics Volume 2: Capacitors in Series and in Parallel