UY1: Dielectrics In Capacitors

A dielectric polarizes in an electric field, reducing the net field inside the material and changing capacitor behavior.

1) At a glance

• Prerequisites: C = Q/Δ V (Capacitance) and energy formulas (Energy Stored In Capacitors)
• Outcomes: compare isolated vs battery-connected insertion cases
• Key results: C = K C₀; isolated (Q fixed): V,E,U drop by K; battery-connected (V fixed): Q,U rise by K
• Key material factor: relative permittivity (dielectric constant) K = ε/ε₀
• Common trap: using the isolated-capacitor formulas when voltage is actually fixed by a battery

Motivation / intuition

A dielectric polarizes: its bound charges partially cancel the field produced by the free charges on the plates. That is why (for the same free charge) the field and voltage drop, and why capacitance increases.

2) Setup

For a fully filled linear dielectric in an ideal parallel-plate capacitor: C₀ = ε₀A/d, C = K C₀ = εA/d

We distinguish two common setups:

• Isolated capacitor: free charge Q fixed
• Battery-connected capacitor: voltage Δ V fixed

3) Core derivation/explanation

A) Isolated capacitor (Q fixed)

Since C → KC₀, Δ V = Q/C = (1/K)Q/C₀ = (Δ V₀)/K Field also drops: E = (Δ V)/d = E₀/K Energy: U = Q²/2C = U₀/K

B) Battery-connected capacitor (Δ V fixed)

Capacitance increases, so charge and energy increase: Q = CΔ V = KQ₀ U = 1/2 C(Δ V)² = K U₀

C) Induced charge and displacement form

Polarization induces bound charge opposing the free-charge field. In the ideal plate model, |σ_ind| = (1-1/K)σ_free with induced bound charge sign opposite to the adjacent free plate charge. A general linear-medium Gauss-law form is ∮ vector D · d vector A = Q_(free,encl), vector D = ε vector E

D) Dielectric breakdown

If electric field exceeds dielectric strength, the insulator conducts (breakdown). This limits capacitor voltage ratings.

4) Worked example(s)

A capacitor has C₀ = 100 pF and is charged to V₀ = 300 V, then disconnected from battery. A dielectric with K = 4 fully fills the gap.

• New capacitance: C = 400 pF
• Charge unchanged: Q = Q₀ = C₀V₀ = 30 nC
• New voltage: V = V₀/K = 75 V
• New energy: U = U₀/K

If instead battery stayed connected at 300 V, then Q and U would both become 4 times larger.

5) Practice set (with hints + answers)

1. Isolated capacitor: K = 2.5. How does voltage change?
2. Battery-connected capacitor: K = 3. How does stored energy change?
3. What physical quantity sets maximum safe operating field before failure?

Hints

1. Use fixed-Q result.
2. Use fixed-V result.
3. Think material limit in breakdown discussion.

Answers

1. Voltage becomes V₀/2.5.
2. Energy becomes 3U₀.
3. Dielectric strength.

6) Summary + next steps

• Dielectrics increase capacitance by factor K.
• What stays fixed (Q or Δ V) determines whether energy decreases or increases.
• Polarization viewpoint and vector D-form Gauss law unify the material description.

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