Unit 10: Conductors and Capacitors — AP Physics C E&M Practice

What Unit 10 Covers in AP Physics C E&M

Unit 10 applies the electrostatics of Units 8 and 9 to two central physical systems: conductors in electrostatic equilibrium and capacitors. The AP Physics C E&M course expects students to derive capacitance from geometry using Gauss's law and the definition of potential, not merely to recall formula values. Energy storage and dielectric effects round out the unit.

Conductors in Electrostatic Equilibrium

Properties of Conductors

In electrostatic equilibrium, the electric field inside a conductor is zero, any net charge resides on the surface, and the surface is an equipotential. These results follow directly from Gauss's law (a Gaussian surface just inside the conductor encloses zero charge) and the condition that no work is done moving charge within the conductor. AP FRQs frequently ask students to justify these properties using calculus-based arguments from Units 8 and 9.

Charge Distribution on Conductor Surfaces

Surface charge density on a conductor is not always uniform. At sharp points, charge density — and the adjacent electric field — is higher. AP-style problems may involve spherical conductors, conducting shells, or conductors with cavities, requiring application of Gauss's law to determine induced charge distributions.

Capacitance and Geometry Derivations

Defining Capacitance

Capacitance C = Q/V represents the charge stored per unit potential difference. While the parallel-plate result C = ε₀A/d is widely known, AP Physics C E&M expects students to derive capacitance for non-planar geometries by: (1) assuming a charge Q on the conductors, (2) using Gauss's law to find E, (3) integrating E to find the potential difference V, and (4) computing C = Q/V.

Parallel-Plate Capacitor

Using Gauss's law for a large conducting plate yields E = σ/ε₀ = Q/(ε₀A). Integrating E over the plate separation d gives V = Ed = Qd/(ε₀A), so C = ε₀A/d. This derivation must be reproducible step-by-step in an AP FRQ.

Cylindrical Capacitor

For a cylindrical capacitor (coaxial conductors of radii a and b), Gauss's law gives E = Q/(2πε₀Lr). Integrating from a to b: V = Q/(2πε₀L) · ln(b/a). Therefore C = 2πε₀L/ln(b/a). The natural logarithm arises from the 1/r field — a clear example of calculus producing a non-trivial geometry result.

Spherical Capacitor

For concentric spherical shells of radii a and b, Gauss's law gives E = Q/(4πε₀r²). Integrating: V = Q/(4πε₀) · (1/a − 1/b). Thus C = 4πε₀ab/(b−a). As b → ∞, this gives the capacitance of an isolated sphere: C = 4πε₀a.

Energy Stored in a Capacitor

The energy stored in a capacitor is derived by integrating the work done to move infinitesimal charge increments dq against the growing potential: U = ∫₀Q (q/C) dq = Q²/(2C) = ½CV² = ½QV. AP FRQs expect students to derive this result — not merely quote it — and to apply it in contexts involving changing capacitor geometry or inserted dielectrics.

Dielectrics

Inserting a dielectric material with dielectric constant κ increases capacitance to C = κε₀A/d (for parallel plates). The dielectric reduces the electric field inside by a factor of κ, which can be understood through the concept of polarisation reducing the net surface charge density. AP questions test the effect of dielectrics on E, V, Q, C, and U — distinguishing between constant-charge and constant-voltage scenarios.

Key FRQ Skills for Unit 10

1. Capacitance derivation: Given geometry, apply Gauss's law → E → integrate for V → C = Q/V in full algebraic form.
2. Energy storage: Derive U = ½CV² from the integral definition; apply to both isolated and connected capacitor systems.
3. Dielectric analysis: Correctly determine how inserting a dielectric at constant Q versus constant V affects each electrical quantity.
4. Conductor equilibrium: Use Gauss's law to determine charge distributions on conductors and cavities.

Common Mistakes in Unit 10

• Skipping the Gauss's law step and assuming E without derivation in FRQs
• Confusing energy formulas — U = Q²/2C applies at constant charge; U = ½CV² applies at constant voltage
• Incorrectly applying the logarithm form for cylindrical geometry instead of the reciprocal difference for spherical geometry
• Forgetting to re-derive C after a dielectric is inserted, instead of simply multiplying by κ without justification