Capacitance and parallel plates

Charge a pair of conductors with +Q and −Q and a potential difference ΔV opens up between them. Capacitance is defined as the ratio C = Q/ΔV, measured in farads, one farad being one coulomb per volt. The whole point of the definition is that the ratio is constant. Electrostatics is linear: every field is proportional to the charges that make it, so tripling Q triples E everywhere, triples the integral of E along any path between the conductors, and so triples ΔV. Q and ΔV move together and their ratio does not move at all. What the ratio does depend on is shape, size and separation, so capacitance is a geometric property you can compute before the capacitor is wired to anything. The farad is an enormous unit: laboratory capacitors run in picofarads and nanofarads, and two plates of 1 cm² held 0.1 mm apart come to only 8.9 pF.

The plates are conductors, so in equilibrium the field inside the metal is zero and the charge sits on the facing surfaces with density σ = Q/A. Put a small pillbox across one of those surfaces, one face buried in the metal and one in the gap. No flux leaves the buried face, and the sides carry none because E is perpendicular to the plate, so Gauss's law reads E A_box = σ A_box/ε₀, giving E = σ/ε₀. Superposition says the same from the other direction: each charged sheet alone makes a field σ/2ε₀ pointing away from positive charge, and the two sheets reinforce between the plates while cancelling outside, leaving σ/ε₀ in the gap and zero beyond it. Read that result carefully. E depends on the charge per unit area and on nothing else — not on d, not on where along the plate you stand. With σ = 0.16 μC m⁻² the field is 1.8 × 10⁴ V m⁻¹ whether the gap is half a millimetre or five.

Potential difference is the line integral of the field, ΔV = ∫E⋅dl taken from one plate to the other, which in magnitude is the work per coulomb needed to carry a test charge across. In a uniform field parallel to that path the integral is trivial: ΔV = Ed. Now assemble the pieces. The field is Q/(ε₀A), so ΔV = Qd/(ε₀A), and the definition returns C = Q/ΔV = ε₀A/d. Q has cancelled, exactly as the linearity argument promised, and what survives is ε₀ and two pieces of geometry. Note which geometry. A is the area the plates actually share, not the area of the larger one, and d is the gap they share it across. Slide one plate sideways until only half of it faces the other and the capacitance halves, even though neither plate has changed size.

C = ε₀A/d makes two separate claims. Capacitance is proportional to area, because at the same ΔV a wider plate spreads the same field over more surface and so holds more charge. And it is inversely proportional to the gap, because at fixed ΔV a narrower gap means a steeper potential drop, a larger E, and therefore a larger σ. The second lever is the strong one and it costs almost no material, which is why real capacitors are thin gaps between large rolled or stacked films rather than big plates held far apart. The scale explains the design. To reach 1 F across a 1.0 mm air gap you would need A = Cd/ε₀ = 1.13 × 10⁸ m², a square plate 10.6 km on a side. Closing the gap is not free either: E = ΔV/d climbs as d falls, and every insulator has a field it stops withstanding, which is where the shrinking has to stop.

Real plates are finite, and near the rim the field lines bow outward into the surrounding space instead of crossing straight over. Two things follow. The region just outside the capacitor is not field-free, so a nearby conductor can feel it. And the true capacitance is larger than ε₀A/d, because at a given ΔV those extra paths hold extra charge on the plates: fringing always adds, never subtracts. How much it adds is set by the ratio of gap to plate size. For coaxial discs of radius R, Kirchhoff's edge correction multiplies ε₀A/d by 1 + (d/πR)(ln(16πR/d) − 1). At R = 0.10 m with d = 1.0 mm — a gap one two-hundredth of the plate width — that factor is 1.024, so the ideal formula reads 2.4% low; shrink the gap to 0.25 mm on a disc of radius 9.5 cm and the shortfall falls to 0.74%. The rule of thumb that follows is blunter than it looks. A gap one hundredth of the plate width already leaves ε₀A/d about four percent low, and you have to reach about a five-hundredth of it before the formula is good to one percent. Let d approach the plate width and it is not a model at all. The rim is also where E is strongest, which is where the air gives way first.

Turn the formula around and the capacitor becomes an instrument. Differentiating C = ε₀A/d at fixed area gives dC/dd = −C/d, so ΔC/C ≈ −Δd/d: a gap change of one part in ten thousand shows as a capacitance change of one part in ten thousand. Those are small quantities — a 2.0 cm² pair at a 50 μm gap is 35.4 pF, and a 10 nm shift of one plate moves it by 7.1 fF — which is why capacitive gauges are read with bridges rather than multimeters. What you read out depends on the circuit constraint. Hold ΔV fixed with a battery and it is the charge that moves, Q = ε₀AΔV/d, so the signal is a current. Disconnect the source instead and the charge is frozen, so E = Q/(ε₀A) is frozen with it and ΔV = Ed follows the gap in strict proportion — a displacement sensor linear in d, not merely linear to first order, which is how a condenser microphone turns the motion of a diaphragm into a voltage. Slide the plates sideways rather than apart and the shared area varies instead: that is a tuning capacitor, and a capacitive touch sensor.