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
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
only an isolated sphere. Elsewhere charge crowds at high curvature: σ and E scale 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. Inside the slab E = E₀ / 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