Conductors and Capacitors

Four topics on what metal does with charge and how two conductors store energy between them. Electrostatics with conductors, where zero field in the material forces charge to the surface but leaves a cavity's field very much alive, redistribution of charge, where a wire equalizes potential rather than charge and grounding sets V to zero rather than Q, capacitors, where C is fixed by geometry and every energy question forks on what the wiring holds still, and dielectrics, which multiply C by kappa and leave everything else to the constraint.

AP exam 10-15%4 topics

Topics

10.1

Electrostatics with Conductors

Redistribution of Charge between Conductors

Capacitors

Dielectrics

Lesson

Diagnostic

after diagnostic

Key forms For every problem in this unit

Inside the material

E = 0 everywhere in the metal, in electrostatic equilibrium. This is the axiom the rest follows from

Where the charge goes

all excess charge sits on a SURFACE. A Gaussian surface drawn inside the metal encloses zero net charge

The surface

an equipotential. E just outside is perpendicular to it, with magnitude E = σ / ε₀

Why that is twice the sheet value

a lone sheet gives σ / (2ε₀) on each side. Here the inside is forced to zero, so all the flux leaves one way

Uniform σ

an isolated sphere. On other conductors charge tends to crowd where the surface curves most sharply; the exact σ depends on the whole geometry and any outside fields. For two distant connected spheres, σ and E go as 1 / R

Empty cavity

E = 0 inside it, whatever fields sit outside. This is real shielding

Cavity holding charge q

cavity field is NOT zero. Wall carries −q, outer surface carries +q plus any net charge on the conductor

Shielding is one way

outside fields stay out. An enclosed charge still shows outside, through the induced +q, unless the conductor is grounded

Grounding

sets V = 0, NOT Q = 0. What flows, and which way, is whatever V = 0 requires given the neighbors

Connected conductors

equalize POTENTIAL. Two distant spheres: Q₁/R₁ = Q₂/R₂, so Q splits in proportion to R

Consequence at the surfaces

E = V / R at each sphere, so the SMALLER sphere ends with the stronger field and the higher σ

Isolated sphere

V = kQ / R, C = 4πε₀R = R / k

Definition

C = Q / V, a ratio fixed by GEOMETRY. It does not track the charge or voltage you apply

Parallel plates

C = ε₀A / d, or κε₀A / d with a dielectric filling the gap

Cylindrical and spherical

C = 2πε₀L / ln(b/a) · C = 4πε₀ab / (b − a)

Where the charge sits

+Q and −Q on the FACING inner surfaces. Outer faces stay neutral in the ideal case

Field in the gap

E = σ / ε₀ = V / d, uniform. Outside the plates the two contributions cancel, E = 0

Force on ONE plate

Q times the OTHER plate's field: F = Q² / (2ε₀A). A plate cannot push on itself

Series

same CHARGE on each. 1/C(eq) = 1/C₁ + 1/C₂. Voltage divides inversely with C, and C(eq) lands below the smaller

Parallel

same VOLTAGE on each. C(eq) = C₁ + C₂. Charge divides in proportion to C

Stored energy

U = Q²/2C = QV/2 = CV²/2. Identical at an instant, NOT interchangeable across a change

The clamp

battery attached: V is fixed, use CV²/2. Disconnected: Q is fixed, use Q²/2C. This choice sets the SIGN of the answer

Energy density

u = ε₀E² / 2, or κε₀E² / 2 in a dielectric. Times the gap volume Ad it reproduces CV²/2

Current and the gap

nothing crosses the gap. Current runs in the WIRES while the plates fill; a charged capacitor in DC is an open switch

Dielectric

C → κC always. What happens to E depends on what is held fixed (next two rows): the slab reduces the field of a GIVEN free charge by κ, reduced and finite, never zero. Bound charge σ(b) = σ(f)(1 − 1/κ)

Dielectric, battery attached

V fixed → Q rises by κ, E unchanged, U rises by κ

Dielectric, disconnected

Q fixed → V falls by κ, E falls by κ, U falls by κ

Partial fill

half the AREA: two capacitors in PARALLEL. Half the GAP: two in SERIES. Never average κ

Force on the slab

pulled INTO the gap. At fixed Q, U = Q²/2C falls as C grows, and F = −dU/dx

Unit 10 tools

Challenge bank

1 / 60

60 open-ended problems.

Read the question, work it out, then flip the card to compare your reasoning to the worked solution. Mark each card so you can return to the ones that still bite.

Question

Tap card to reveal explanation

Worked solution

Tap card to return to question

Cumulative assessment

Test the unit.

Twenty mixed items drawn from across all 4 topics, with guaranteed misconception-code coverage. Identifies which misconceptions still bite when you cannot see which topic the question came from.

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Check what stuck.

Units 8 through 10, drawn evenly so earlier units get the same share as this one. Twenty questions or a full 42-question section, your choice. Even coverage means this is a retention check rather than a score estimate.

20 or 42questions

13topics

50codes covered

Begin assessment →