AP · AP C:E&M · extended response
An isolated capacitor is not a constant-voltage capacitor
AP Physics C: Electricity and Magnetism · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Initial capacitor charging relation
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Stored charge / μC
Horizontal axis (x): Potential difference / V
Read the exact data as a table
| Potential difference / V | Stored charge / μC |
|---|---|
| 0 | 0 |
| 4 | 16 |
| 8 | 32 |
| 12 | 48 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Choose your AP course · School year 2026–27 · May 2027 exam
Choose a course to check this question’s reviewed parts and required depth. Its original AP label and topic codes do not establish suitability for the current syllabus.
Choose reviewed AP practice →A parallel-plate capacitor of capacitance 4.0 μF is charged to 12 V and then completely disconnected from its supply. Its plate separation is subsequently doubled with no leakage. The initial charge–voltage relation is supplied. Ignore edge effects and use C = ε₀A/d. Supplied data (Potential difference / V; Stored charge / μC): (0, 0), (4, 16), (8, 32), (12, 48).
- (a)
Find charge and stored energy before the change, then capacitance, voltage and energy after doubling the gap.
3 marks - (b)
Repeat the predictions if the 12 V supply instead remains connected and explain where energy goes.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Find charge and stored energy before the change, then capacitance, voltage and energy after doubling the gap.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Initial charge is CV = 4.0 μF×12 V = 48 μC. Initial energy is ½CV² = 288 μJ.
- 2
Doubling d halves capacitance to 2.0 μF. Disconnection fixes charge, so V = Q/C doubles to 24 V.
- 3
At fixed Q, U = Q²/(2C), so halving C doubles energy to 576 μJ. External mechanical work against attraction supplies the 288 μJ increase.
(b)
Repeat the predictions if the 12 V supply instead remains connected and explain where energy goes.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
The supply fixes potential difference rather than charge. With half the capacitance, Q = CV halves to 24 μC.
- 2
Stored field energy ½CV² also halves to 144 μJ, but this does not mean separating attracting plates releases mechanical work.
- 3
Quasistatic separation still needs positive mechanical work. Charge returns energy to the supply; the field energy, mechanical work and supply energy must all be included in the energy balance.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- Initial charge is CV = 4.0 μF×12 V = 48 μC. Initial energy is ½CV² = 288 μJ.
- Doubling d halves capacitance to 2.0 μF. Disconnection fixes charge, so V = Q/C doubles to 24 V.
- At fixed Q, U = Q²/(2C), so halving C doubles energy to 576 μJ. External mechanical work against attraction supplies the 288 μJ increase.
- The supply fixes potential difference rather than charge. With half the capacitance, Q = CV halves to 24 μC.
- Stored field energy ½CV² also halves to 144 μJ, but this does not mean separating attracting plates releases mechanical work.
- Quasistatic separation still needs positive mechanical work. Charge returns energy to the supply; the field energy, mechanical work and supply energy must all be included in the energy balance.
Private on this device