University Physics II · Capacitance and Dielectrics · 5.5
Capacitors with Dielectrics
A linear dielectric raises capacitance by the factor K. Whether the field, the charge and the stored energy rise or fall depends on whether the source is still connected — but the slab is pulled in either way.
Build the model
Connect the measurement to the mechanism.
Filling a capacitor's gap with a linear dielectric multiplies its capacitance by one number, the dielectric constant K = C/C₀. Everything else depends on which quantity the circuit holds fixed. Cut the source and the free charge is locked: bound charge weakens the field to E₀/K, ΔV falls by K, and the stored energy drops to U₀/K.
Leave the source connected and ΔV is locked instead: the field is unchanged, charge flows in until Q = KQ₀, the energy rises to KU₀, and the source delivers twice that increase. The two cases disagree about the sign of ΔU, which tempts you into thinking they disagree about the force. They do not.
A partly inserted slab is two capacitors whose combined C grows as it enters, and the electrical force F = ½ΔV² dC/dx points toward larger capacitance under either constraint. The slab is always pulled in; only the ledger of who paid for it changes.
- Simple definition
- The dielectric constant K is the factor by which filling a capacitor's gap with an insulating material multiplies its capacitance: C = KC₀, with K = 1 for vacuum and K > 1 for every real dielectric.
- Example
- A 177 pF air gap filled by a K = 3.40 polymer film becomes 602 pF. Held at fixed charge, its voltage falls from 100 V to 29.4 V; held at a fixed 100 V, it draws 42.5 nC more charge from the source.
Defined by measurement — the factor by which a material multiplies the capacitance of the same electrodes.
Dimensionless, K ≥ 1: vacuum 1, dry air 1.0006, paper ≈ 3.5, mica ≈ 5.4
A = 0.0200 m² at d = 1.00 mm gives C₀ = 177 pF; with K = 3.40 the same plates give 602 pF.
A in m², d in m, C in F; the slab must span the whole gap and the whole area
Bound charge weakens the field, so the same stranded charge now sits at a lower potential difference.
Source disconnected; use U = Q²/(2C), the form written in the constant
Charge flows in to offset the bound charge, and the source pays twice what the capacitor keeps.
Source connected; use U = ½CΔV², and E = ΔV/d is unchanged
Side by side the parts are in parallel; stacked they are in series. K alone works only for a full fill.
Slid in a depth x; a slab of thickness t < d across the area gives ε₀A/(d − t + t/K)
One expression for both constraints, and it always points toward larger capacitance: inward.
F in N; ΔV is the instantaneous value, so at fixed Q the force falls as the slab enters
K is a measured capacitance ratio
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 σb = σf(1 − 1/K) — here 0.706 σf — 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.
Cut the source and the charge is what stays
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.
Leave the source on and the voltage is what stays
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.
A slab part-way in is two capacitors
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.
The force points toward larger capacitance
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.
The energy ledger closes both ways
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.
Change one variable at a time
Make the relationship visible.
Drag the slab all the way in at K = 3.4: both branches start at U₀ = 0.885 µJ and end at the same 602 pF, yet the stored energy finishes at 3.010 µJ with the source connected and 0.260 µJ with the leads cut. Slide K back to 1 and both branches flatten onto the dashed line — a vacuum slab is no slab.
CAPACITANCE C(x)304.6 pF
U WITH ΔV HELD AT 100 V1.523 µJ
U WITH Q HELD AT 17.7 nC0.515 µJ
INWARD PULL AT 100 V10.62 µN
Live interpretationCAPACITANCE C(x): 304.6 pF. U WITH ΔV HELD AT 100 V: 1.523 µJ. U WITH Q HELD AT 17.7 nC: 0.515 µJ. INWARD PULL AT 100 V: 10.62 µN
Catch the common trap
Explain before calculating.
A capacitor is charged by a 12 V source, and the source is left connected while a slab of dielectric constant K = 3 is slid in until it fills the gap completely. Compared with the empty capacitor, what happens to the free charge Q, the stored energy U, and the work done by the source?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyPlates of area A = 0.0150 m² sit 0.500 mm apart. The capacitor is charged to 60.0 V and the leads are then cut. A slab of dielectric constant K = 2.60 is slid in until it fills the gap. Find the new capacitance, the charge, the potential difference and the field.
- Empty gap first: C₀ = ε₀A/d = (8.854 × 10⁻¹² F m⁻¹)(0.0150 m²) ÷ (5.00 × 10⁻⁴ m) = 2.66 × 10⁻¹⁰ F = 266 pF.
- Filling the gap multiplies that by K and changes nothing else: C = KC₀ = 2.60 × 266 pF = 691 pF. The geometry never moved; only the material between the plates did.
- The leads were cut, so the free charge is stranded: Q = C₀ΔV₀ = (2.66 × 10⁻¹⁰ F)(60.0 V) = 1.59 × 10⁻⁸ C = 15.9 nC, before the slab and after it.
- Same Q, larger C: ΔV = Q/C = ΔV₀/K = 60.0 V ÷ 2.60 = 23.1 V, and the field falls in step, E = ΔV/d = 23.1 V ÷ 5.00 × 10⁻⁴ m = 4.62 × 10⁴ V m⁻¹.
AnswerC = 691 pF; Q holds at 15.9 nC; ΔV falls to 23.1 V and E to 4.62 × 10⁴ V m⁻¹.
MediumRepeat the insertion on the same 266 pF capacitor with the 60.0 V source left connected. Find the free charge afterwards, the change in stored energy, and the energy the source delivers.
- ΔV is pinned at 60.0 V now, but C belongs to the plates and the filling, not to the circuit, so it is still C = KC₀ = 691 pF and E = ΔV/d is unchanged.
- Charge is the free variable: Q = CΔV = (6.91 × 10⁻¹⁰ F)(60.0 V) = 4.14 × 10⁻⁸ C = 41.4 nC, against Q₀ = 15.9 nC, so ΔQ = 25.5 nC flowed in from the source.
- Use the energy form written in the quantity being held, U = ½CΔV²: U₀ = ½(2.66 × 10⁻¹⁰ F)(60.0 V)² = 0.478 µJ and U = KU₀ = 1.243 µJ, a gain of ΔU = 0.765 µJ.
- The source pushed that charge through the full 60.0 V: W = ΔVΔQ = (60.0 V)(2.55 × 10⁻⁸ C) = 1.53 µJ — exactly twice ΔU, as ½ΔV²ΔC against ΔV²ΔC always is.
- The other half is not missing: 0.765 µJ is the work the field did on the slab while pulling it in.
AnswerQ rises to 41.4 nC and U to 1.243 µJ (ΔU = +0.765 µJ), while the source delivers 1.53 µJ — twice the gain.
HardPlates of length L = 0.120 m and width w = 0.0500 m are 0.800 mm apart and held at a fixed 250 V. A slab of dielectric constant K = 4.50 that fills the gap thickness is slid in lengthwise a depth x = 0.0450 m. Find C(x), the force on the slab, and the work needed to pull it back out.
- Part-way in is not a K problem. The filled length x and the empty length L − x hang between the same plates at the same ΔV, so they are two capacitors in parallel: C(x) = ε₀w[L + (K − 1)x]/d.
- Collect the constant: ε₀w/d = (8.854 × 10⁻¹² F m⁻¹)(0.0500 m) ÷ (8.00 × 10⁻⁴ m) = 5.534 × 10⁻¹⁰ F m⁻¹. With L + (K − 1)x = 0.120 m + 3.50(0.0450 m) = 0.2775 m, C = 1.54 × 10⁻¹⁰ F = 154 pF — between 66.4 pF empty and 299 pF full.
- The pull needs the slope, not the value: dC/dx = ε₀w(K − 1)/d = (5.534 × 10⁻¹⁰ F m⁻¹)(3.50) = 1.937 × 10⁻⁹ F m⁻¹, so F = ½ΔV²(dC/dx) = ½(250 V)²(1.937 × 10⁻⁹ F m⁻¹) = 6.05 × 10⁻⁵ N = 60.5 µN, pointing inward.
- At fixed ΔV that slope is constant, so the force is the same everywhere along the stroke: dragging the slab back out to x = 0 costs you W = FΔx = (6.05 × 10⁻⁵ N)(0.0450 m) = 2.72 × 10⁻⁶ J = 2.72 µJ.
- Audit it against the ledger: C falls by 87.2 pF, so the capacitor gives up ½ΔC ΔV² = 2.72 µJ while the source takes back ΔC ΔV² = 5.45 µJ. Your 2.72 µJ is the difference — the same number reached two ways.
AnswerC(0.0450 m) = 154 pF; the slab is pulled in with 60.5 µN; withdrawing it fully at a fixed 250 V costs 2.72 µJ of mechanical work.