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Dielectrics

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Polarization weakens the field inside

A dielectric is an insulator whose molecules polarize in an applied field: dipoles stretch or rotate slightly, leaving thin layers of induced bound charge on the slab’s faces — negative facing the positive plate, positive facing the negative plate. This bound charge creates an internal field opposing the applied one, so the net field inside the material drops to E = E₀/κ, where κ ≥ 1 is the dielectric constant (κ ≈ 1.0006 for air, ≈ 80 for water). Unlike a conductor, a dielectric only partially cancels the field — no charge is free to travel across it.

Capacitance with a dielectric
C = κC₀ = κε₀A/d, effective permittivity ε = κε₀
Filling the gap always *raises* capacitance: for the same Q, the weakened field means a smaller V = Ed, and C = Q/V grows by κ.

Scenario 1 — battery disconnected: Q is fixed

Charge a capacitor, disconnect the battery, then slide in a slab with constant κ. With no conducting path, the plate charge Q cannot change. Everything else adjusts: C → κC₀, so V = Q/C → V₀/κ, the field E = V/d → E₀/κ, and the energy U = Q²/2C → U₀/κ. The stored energy decreases — where did it go? The fringing field at the gap’s edge pulls the slab inward, doing positive work on it. You would have to do negative work (hold the slab back) to insert it slowly.

Scenario 2 — battery connected: V is fixed

Keep the battery attached instead and the voltage is pinned at V₀; now the charge adjusts. C → κC₀ forces Q = CV → κQ₀ — the battery pumps extra charge onto the plates. The field E = V/d is unchanged, and the stored energy U = ½CV² → κU₀ increases. The battery supplies energy ΔQ·V₀ = (κ − 1)Q₀V₀, which is exactly twice the capacitor’s gain — the other half goes into work done pulling the slab in. The two scenarios differ in every quantity except C, so identify the circuit condition before computing anything.

Worked example

A 2.0 μF capacitor is charged to 120 V and disconnected from the battery. A κ = 4.0 slab then fills the gap. Find the new charge, capacitance, voltage, and stored energy.

  1. 1.Disconnected ⇒ Q is fixed: Q = C₀V₀ = (2.0 × 10⁻⁶)(120) = 2.4 × 10⁻⁴ C, before and after.
  2. 2.Capacitance: C = κC₀ = 4.0 × (2.0 μF) = 8.0 μF.
  3. 3.Voltage: V = Q/C = (2.4 × 10⁻⁴)/(8.0 × 10⁻⁶) = 30 V — exactly V₀/κ ✓.
  4. 4.Energy before: U₀ = ½C₀V₀² = ½(2.0 × 10⁻⁶)(1.44 × 10⁴) = 1.44 × 10⁻² J.
  5. 5.Energy after: U = Q²/2C = (5.76 × 10⁻⁸)/(1.6 × 10⁻⁵) = 3.6 × 10⁻³ J = U₀/4 ✓. The missing 1.08 × 10⁻² J went into work done by the field drawing the slab in.
Answer: Q = 2.4 × 10⁻⁴ C (unchanged); C = 8.0 μF; V = 30 V; U = 3.6 × 10⁻³ J (down by κ)
Worked example

Now a 2.0 μF capacitor stays connected to a 100 V battery while a κ = 3.0 slab fills it. Find the new charge and energy, and the energy the battery supplies.

  1. 1.Connected ⇒ V is fixed at 100 V. Capacitance: C = κC₀ = 6.0 μF.
  2. 2.Charge: Q₀ = C₀V = 2.0 × 10⁻⁴ C before; Q = CV = 6.0 × 10⁻⁴ C after — the battery delivered ΔQ = 4.0 × 10⁻⁴ C.
  3. 3.Energy: U₀ = ½C₀V² = ½(2.0 × 10⁻⁶)(1.0 × 10⁴) = 1.0 × 10⁻² J; U = ½CV² = 3.0 × 10⁻² J = κU₀ ✓.
  4. 4.Battery output: W_bat = ΔQ·V = (4.0 × 10⁻⁴)(100) = 4.0 × 10⁻² J.
  5. 5.Bookkeeping: the capacitor gained 2.0 × 10⁻² J — half of W_bat. The other 2.0 × 10⁻² J is the work done pulling the slab into the gap.
Answer: Q = 6.0 × 10⁻⁴ C; U = 3.0 × 10⁻² J; battery supplies 4.0 × 10⁻² J, half stored and half spent drawing the slab in
Watch out

The classic dielectric trap: with the battery disconnected, it is Q that stays fixed — not V. Students reflexively hold V constant and conclude the energy rises; in fact V drops to V₀/κ and U falls to U₀/κ. Ask one question before anything else: is there still a conducting path to a battery? No path ⇒ Q fixed. Path ⇒ V fixed.

On the exam

Dielectric quick table — battery disconnected (Q fixed): C ↑ ×κ, V ↓ ÷κ, E ↓ ÷κ, U ↓ ÷κ. Battery connected (V fixed): C ↑ ×κ, Q ↑ ×κ, E unchanged, U ↑ ×κ. In both cases the slab is pulled inward. Reproduce this table from C = κC₀ plus “what’s held fixed” rather than memorizing blindly.

Checkpoint

A charged capacitor is disconnected from its battery, and a κ = 2.0 dielectric is inserted. Which quantity is unchanged?

Checkpoint

A capacitor remains connected to a 12 V battery while a κ = 3.0 dielectric fills its gap. The charge on the plates:

Answer the 2 checkpoints as you read.

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