AP Physics C: Electricity and Magnetism, AP exams

Unit 10: Conductors and Capacitors

Unit 10 covers the behavior of conductors in electrostatic equilibrium and the devices that exploit it. Charge resides on conductor surfaces, the interior field vanishes, and charge redistributes when conductors are connected. Capacitance, stored energy, and dielectrics complete the unit.

Position: 3 of 6 units
Topics covered: 5 topics in this guide

What is covered

Topics inside this unit.

These topics describe the scope students most often review for this part of AP Physics C: Electricity and Magnetism. Treat them as a study checklist rather than an exhaustive list of what a single exam form can sample.

Electrostatics with conductors: Why the field inside a conductor in equilibrium is zero and how surface charge arranges itself.
Redistribution of charge between conductors: What happens when two charged conductors are connected, and the role of potential in the final state.
Capacitance: Definition of capacitance and its calculation for parallel-plate, cylindrical, and spherical geometries.
Energy stored in capacitors: The several equivalent expressions for stored energy, and energy density in the field.
Dielectrics: How an inserted dielectric changes capacitance, field, and stored energy under fixed charge or fixed voltage.

Study tips

How to practice Conductors and Capacitors.

Section-specific guidance that follows from how this material is usually tested.

Work dielectric problems twice: once with the battery connected and once disconnected, since the answers differ.
Derive parallel-plate capacitance from Gauss's law so the result is reconstructable.
Track which quantity is held constant in every capacitor problem before deciding what changes.

Where this fits

Unit 10: Conductors and Capacitors is 3 of 6 units in the AP Physics C: Electricity and Magnetism guide. Use the full guide to balance it against everything else you have to review.

Back to the AP Physics C: Electricity and Magnetism guide Official exam information

Sample Conductors and Capacitors questions

Practice questions, not yet expert-reviewed

Original practice questions written to the published framework for this section. No expert has reviewed them, and they are not taken from any official exam. Try each one before you open the answer.

Question 1Foundational
A solid metal sphere carries excess charge and has reached electrostatic equilibrium. Where does the excess charge reside, and what is the electric field inside the metal?
1. AThe charge spreads uniformly through the volume and the field inside the metal is zero
2. BThe charge lies entirely on the outer surface and the field inside the metal is zero
3. CThe charge lies entirely on the outer surface and the field inside the metal is nonzero
4. DThe charge collects at the center and the field inside the metal is zero
Show answer
Answer: B. The charge lies entirely on the outer surface and the field inside the metal is zero
If any field remained inside the conductor, free electrons would keep moving and equilibrium would not have been reached, so the interior field must be zero. Gauss's law applied to a surface just inside the metal then gives zero enclosed charge, which forces the excess charge onto the outer surface.
Key insight: Zero interior field is the definition of electrostatic equilibrium in a conductor, and surface charge is its consequence.
Question 2Foundational
A parallel-plate capacitor has plates of area 0.020 m^2 separated by 1.0 x 10^-3 m of vacuum. Using the permittivity of free space 8.85 x 10^-12 F/m, what is its capacitance?
1. A1.8 x 10^-16 F
2. B1.8 x 10^-13 F
3. C1.8 x 10^-10 F
4. D1.8 x 10^-7 F
Show answer
Answer: C. 1.8 x 10^-10 F
Deriving the field from Gauss's law and integrating it across the gap gives C = epsilon0(A/d) = (8.85 x 10^-12)(0.020)/(1.0 x 10^-3) = 1.8 x 10^-10 F. The 1.8 x 10^-16 F choice multiplies by the separation instead of dividing by it.
Key insight: Capacitance depends only on geometry and the material between the plates, never on the charge stored.
Question 3Intermediate
A 5.0 x 10^-6 F capacitor is charged to a potential difference of 200 V. How much energy is stored in the capacitor?
1. A1.0 x 10^-3 J
2. B0.050 J
3. C0.10 J
4. D0.20 J
Show answer
Answer: C. 0.10 J
Stored energy is (1/2)CV^2 = (1/2)(5.0 x 10^-6)(200)^2 = 0.10 J. The 0.20 J choice omits the factor of one half, and 1.0 x 10^-3 J is the stored charge CV in coulombs rather than an energy.
Key insight: The one-half factor comes from integrating V dq while the capacitor charges, since V climbs from zero as charge accumulates.
Question 4Intermediate
A parallel-plate capacitor is charged by a battery and then disconnected from it, so the charge on the plates is fixed. A dielectric slab of dielectric constant 3 is then slid in to fill the space between the plates. What happens to the potential difference across the plates?
1. AIt falls to one third of its original value
2. BIt falls to one ninth of its original value
3. CIt stays the same
4. DIt rises to three times its original value
Show answer
Answer: A. It falls to one third of its original value
The dielectric multiplies the capacitance by 3 while the isolated charge stays fixed, so V = Q/C drops to one third. If the battery had stayed connected, V would instead be held fixed and the charge would triple, which is why identifying the constrained quantity comes first.
Key insight: Ask what is held constant, charge or voltage, before deciding what a dielectric changes.
Question 5Advanced
Two isolated conducting spheres, one of radius R and one of radius 2R, each carry the same charge Q and are far apart. They are then joined by a long thin conducting wire and allowed to reach equilibrium. What is the ratio of the final charge on the larger sphere to the final charge on the smaller sphere?
1. A1 to 2
2. B1 to 1
3. C2 to 1
4. D4 to 1
Show answer
Answer: C. 2 to 1
Connected conductors share a common potential, and for an isolated sphere V = kq/r, so q is proportional to the radius and the larger sphere ends with twice the charge of the smaller one. The 4 to 1 choice scales charge with surface area, but it is potential, not area, that must match.
Key insight: Charge redistributes until potentials are equal, so the larger sphere holds more charge at a lower surface field.

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Unit 10: Conductors and Capacitors