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AP Physics C: Electricity and Magnetism · Unit 10 · Topic 5 of 5 · 5 min lesson

Dielectrics

A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.

The big idea

Filling the gap with an insulator (a dielectric) multiplies the capacitance by its dielectric constant: C = κC0. The dielectric’s molecules polarize and partly cancel the field. What else changes depends on whether the battery is still connected.

1. Why it works

+ + + + + + − − − − − − − − − − − + + + + + E = E0/κ dielectric (κ) C = κC0
Induced surface charge on the dielectric opposes the plates’ field, so less voltage is needed for the same charge.
C = κC0κ ≥ 1 (air ≈ 1)
E = E0/κfor the same plate charge
Isolated (Q fixed)ΔV → ΔV/κ, U → U/κ
Battery (ΔV fixed)Q → κQ, U → κU

2. Partly filled capacitors

Side by side (dielectric fills part of the area, full gap): the two parts share the same ΔV, so they’re in parallel: add capacitances.

Layered (dielectric fills part of the gap, full area): the same charge passes through each layer, so they’re in series: add reciprocals.

Isolated capacitor: sliding in a dielectric lowers the energy, so the capacitor actually pulls the slab in.

With a battery connected, the battery supplies extra charge and the stored energy goes up by κ.

3. Example, then try it

Example

A 10 pF air capacitor is filled with a dielectric of κ = 3.0. Find its new capacitance. If it was isolated with 50 V across it, what is the new voltage?

C = (3.0)(10) = 30 pF.

Isolated, Q is fixed, so ΔV = 50 ÷ 3.0 ≈ 17 V.

Try it · work it on paper first

An air-filled parallel-plate capacitor has C0 = 40 pF. A slab with κ = 4.0 is inserted in two ways: (a) filling the left half of the area, full gap; (b) filling the bottom half of the gap, full area. Find C each way.

1

(a) Side by side → parallel: each half-area part is C0/2 = 20 pF of air; the filled half becomes κ(20) = 80 pF.

2

(a) Total: 20 + 80 = 100 pF (2.5C0).

3

(b) Layered → series: each half-gap layer alone is 2C0 = 80 pF of air; the dielectric layer becomes κ(80) = 320 pF.

4

(b) Total: 1/C = 1/80 + 1/320 = 5/320, so C = 64 pF (1.6C0).

Answer: (a) 100 pF; (b) 64 pF.

4. Common mistakes

Not checking the battery.Isolated: Q fixed. Connected: ΔV fixed.
Mixing up the partial-fill combinations.Side by side is parallel; layered is series.
Thinking a dielectric conducts.It’s an insulator; its charges shift but don’t flow.
Check yourself

A dielectric with constant κ is slid into an isolated, charged capacitor. What happens to the electric field between the plates?

Q is fixed, and the polarized dielectric partly cancels the plates’ field: E = E₀/κ, and so ΔV = Ed drops by κ too.

Ready to practice?

5 questions on Dielectrics, each with a hint and an explanation for every wrong answer.

Unit 10 topics

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