Capacitors with Dielectrics

Take a pair of electrodes, measure C₀ with vacuum between them, slide in a material that fills the gap completely, and measure again. The ratio defines the dielectric constant, K = C/C₀, equivalently a permittivity ε = Kε₀. It is dimensionless and never below 1: vacuum is exactly 1, dry air 1.0006, paper about 3.5, mica about 5.4, liquid water about 80 at 20 °C. The mechanism is the polarization of the previous topic. Bound surface charge of density σ_b = σ_f(1 − 1/K) — here 0.706 σ_f — appears on the slab faces opposite in sign to the free charge it faces, so for a given free charge the interior field is E₀/K, the potential difference across the same gap is smaller by K, and C = Q/ΔV is therefore larger by K. For plates of area A = 0.0200 m² at separation d = 1.00 mm, C₀ = ε₀A/d = 177 pF; a K = 3.40 polymer film filling the gap makes it 602 pF. K is constant only for a linear material, below breakdown, at the frequency in use: water's 80 collapses at microwave frequencies.

Charge that empty 177 pF capacitor to ΔV₀ = 100 V, then disconnect the leads. The free charge Q = C₀ΔV₀ = 17.7 nC is stranded: nothing can add to it or drain it. Push the slab in and the field between the plates falls from 1.00 × 10⁵ V m⁻¹ to E₀/K = 2.94 × 10⁴ V m⁻¹, so ΔV = Ed drops to 29.4 V while Q holds. Capacitance is the ratio of the two, 17.7 nC / 29.4 V = 602 pF, as promised. For energy, use the form written in the quantity being held: U = Q²/(2C), so U = U₀/K = 0.885 µJ / 3.40 = 0.260 µJ. The capacitor is 0.625 µJ poorer than before. No source took it, because there is no source attached. The field did that much work on the slab while drawing it in — which is a prediction about a force, and it is checked below.

Repeat the insertion with the 100 V source still across the plates. Now ΔV is pinned, so E = ΔV/d stays at 1.00 × 10⁵ V m⁻¹: the field is not reduced at all. It cannot be, because the gap and the terminal voltage fix it. What changes is the free charge, which rises until it offsets the bound charge and restores that field, giving Q = KQ₀ = 60.2 nC and C = Q/ΔV = 602 pF — the same capacitance as before, since C depends only on geometry and material, never on the circuit. Energy now takes the form written in the constant, U = ½CΔV², so U = KU₀ = 3.01 µJ, a gain of 2.13 µJ. Meanwhile the source pushed ΔQ = 42.5 nC through 100 V and delivered W = ΔV ΔQ = 4.25 µJ. Only half of it reached the capacitor. In general, at constant ΔV the stored energy grows by ½ΔV²ΔC while the source supplies ΔV²ΔC, exactly twice as much.

K multiplies capacitance only when the dielectric fills the gap; anything less is a network to be reduced first. Slide the slab in lengthwise a depth x along plates of length L = 0.200 m and width w = 0.100 m. The filled and empty regions share the same pair of plates and therefore the same ΔV, so they sit in parallel: C(x) = ε₀w[(L − x) + Kx]/d = ε₀w[L + (K − 1)x]/d. That is linear in x, running from 177 pF at x = 0 to 602 pF at x = L, and passing 390 pF half way. A thin slab of thickness t = 0.400 mm laid flat across the whole area behaves oppositely. It and the remaining 0.600 mm of air carry the same free charge and their voltages add, so they are in series: C = ε₀A/(d − t + t/K) = 247 pF. Two-fifths of the gap filled with a K = 3.40 material buys a factor of only 1.39, because in a series pair the smaller capacitance — the air layer — dominates.

Take the isolated case first. The capacitor is the entire system, so F = −(dU/dx) at fixed Q = +[Q²/(2C²)] dC/dx = ½ΔV² dC/dx, positive because C grows as the slab enters. With the source connected, −(dU/dx) at fixed ΔV returns the wrong sign, because the source is now part of the system: add its contribution ΔV dQ = ΔV² dC to the ledger and the result flips back to F = +½ΔV² dC/dx. One expression, both constraints, always inward. For these plates dC/dx = ε₀w(K − 1)/d = 2.13 nF m⁻¹, so at 100 V the pull is ½(100 V)²(2.13 nF m⁻¹) = 1.06 × 10⁻⁵ N. At fixed voltage that force is constant over the whole stroke; at fixed charge it starts there and decays to 0.92 µN as ΔV sinks to 29.4 V. The pull does not come from the uniform interior field, which exerts no net force on a neutral polarized slab, but from the non-uniform fringing field at the plate edge.

Fixed voltage, full stroke: the source delivers 4.25 µJ, the capacitor stores 2.13 µJ, and the electrical force does 1.06 × 10⁻⁵ N × 0.200 m = 2.13 µJ of work on the slab. The three numbers close, and the force found from dC/dx reproduces the energy split independently. Guide the slab in slowly and your hand must pull backwards and absorb that 2.13 µJ; release it instead and the same energy becomes kinetic, the slab overshoots, and it settles only once friction and circuit resistance have dissipated it. Fixed charge, full stroke: with no source, the 0.625 µJ the capacitor loses is the whole of the work done on the slab. Run either case backwards and every sign inverts — withdrawing the slab at a fixed 100 V costs you 2.13 µJ of mechanical work, and the source recovers 4.25 µJ as the extra 42.5 nC flows back into it.