Dielectrics: Study Guide (Physics C: E&M 10.4)

Course note. This guide is for the calculus-based Physics C: Electricity and Magnetism course, Topic 10.4. The algebra-based Physics 2 course mentions dielectrics inside its capacitor topic; here they have a topic of their own, built on the Gauss’s-law results of Topic 10.3.

Constant used throughout: ε₀ = 8.85 × 10⁻¹² C²/(N·m²).

What happens inside a dielectric

A dielectric is an insulating material, such as plastic, glass, paper or ceramic. In a conductor, some electrons are free to move through the whole material. In a dielectric, the charges are bound to their atoms or molecules. An electric field can only shift them slightly. This is called polarization.

There are two ways it happens:

• Non-polar molecules have no dipole of their own. The applied field pulls the electron cloud one way and the nucleus the other, making a small induced dipole.
• Polar molecules (water is an example) already have a dipole. The field turns them partly into line, against the jostling of thermal motion.

Either way, the dipoles point along the field. Inside the material, the + end of one dipole sits next to the − end of the next, so their charges cancel. At the two faces they do not. The face next to the positive plate is left with a thin layer of negative bound charge; the face next to the negative plate gets positive bound charge. These are induced (bound) surface charges. They cannot leave the dielectric.

The induced field and the dielectric constant

The induced surface charges form a pair of oppositely charged sheets, just like a small capacitor inside the big one. Their field points from their + face to their − face. That is opposite to the applied field E₀ from the plates. By superposition, the net field in the material is smaller:

E = E₀ − E_induced

For a given material, E is a fixed fraction of E₀. We write that fraction as 1/κ:

E = E₀/κ

where κ (kappa) is the dielectric constant of the material. It has no unit. It also links the material’s permittivity ε to the permittivity of free space:

κ = ε/ε₀

Some values: κ = 1 for a vacuum (by definition), about 1.0006 for air, about 2.1 for PTFE and about 80 for water at 20 °C. Air is so close to 1 that the course treats air-filled capacitors as having κ = 1.

Size of the induced charge. If the plates carry free charge density σ, then E₀ = σ/ε₀. The induced sheets carry density σᵢ, so E_induced = σᵢ/ε₀. Setting E₀ − σᵢ/ε₀ = E₀/κ gives:

σᵢ = σ(1 − 1/κ)

σᵢ is always smaller than σ, so the field is reduced but never reversed. A conductor would be the limit κ → ∞: σᵢ = σ and the field inside drops to zero.

Capacitance with a dielectric

Fill the gap of an isolated parallel-plate capacitor with a dielectric. The free charge Q on the plates does not change. The field falls to E₀/κ everywhere in the gap, so the potential difference falls too:

ΔV = Ed = (E₀/κ)d = ΔV₀/κ

The capacitance therefore rises:

C = Q/ΔV = κQ/ΔV₀ = κC₀ = κε₀A/d

In words: the bound charges partly cancel the field of the free charges, so the plates can hold the same charge at a lower potential difference. That is exactly what “more capacitance” means. The formula C = κε₀A/d can also be written εA/d, with ε = κε₀.

This works for the other shapes too. Filling a spherical or coaxial capacitor completely with the material multiplies its capacitance by κ, because every field in the gap is divided by κ.

Isolated or connected: two different stories

As in Topic 10.3, the first question is what stays fixed. Insert a slab that fills the gap:

Quantity	Isolated (Q fixed)	Connected to battery (ΔV fixed)
C	×κ	×κ
Q	same	×κ (the battery supplies more charge)
ΔV	÷κ	same
E between plates	÷κ	same (E = ΔV/d)
U	÷κ, from U = Q²/(2C)	×κ, from U = ½C(ΔV)²

In the connected case, E is unchanged because ΔV and d are unchanged. The induced charge still cuts the field, but the battery sends extra free charge to the plates until E is back to ΔV/d.

Measuring κ in the lab

Figure 2 suggests a method. With a capacitance meter you can measure C directly.

1. Question: what is κ for a given sheet material?
2. Independent variable: something you can change in a controlled way, such as the plate area A (use different-sized plates) or the gap d (stack 1, 2, 3, … identical sheets).
3. Dependent variable: C, read from the meter.
4. Controls: the same material, plates pressed flat with no air gaps, the same leads, and the meter zeroed with the leads open to remove their own small capacitance.
5. Analysis: plot a graph that should be a straight line. C against A gives gradient κε₀/d; C against 1/d gives gradient κε₀A. Divide the gradient by ε₀ (and by A or 1/d) to find κ. A gradient uses all the data, so it is better than one reading.

Practice Question 6 gives a data set to analyse this way. Reading and simulations are useful preparation, but they do not replace hands-on laboratory work.

Worked example 1: inserting a slab into an isolated capacitor

Question. A parallel-plate capacitor has plate area 0.030 m² and gap 0.50 mm. It is charged to 20 V and disconnected. A slab with κ = 2.1 is slid in to fill the gap. Find C, Q, ΔV, E and U before and after, and the induced surface charge density.

Before.

1. C₀ = ε₀A/d = (8.85 × 10⁻¹²)(0.030) ÷ (0.50 × 10⁻³) = 5.31 × 10⁻¹⁰ F (531 pF).
2. Q = C₀ΔV₀ = 1.06 × 10⁻⁸ C.
3. E₀ = ΔV₀/d = 20 ÷ 0.50 × 10⁻³ = 4.0 × 10⁴ V/m.
4. U₀ = ½QΔV₀ = 1.06 × 10⁻⁷ J.

After (Q fixed at 1.06 × 10⁻⁸ C).

1. C = κC₀ = 1.12 × 10⁻⁹ F.
2. ΔV = Q/C = 20 ÷ 2.1 = 9.5 V.
3. E = E₀/κ = 1.9 × 10⁴ V/m.
4. U = Q²/(2C) = U₀/κ = 5.1 × 10⁻⁸ J.

Induced charge. σ = Q/A = 3.54 × 10⁻⁷ C/m², so σᵢ = σ(1 − 1/2.1) = 1.85 × 10⁻⁷ C/m².

Check. σᵢ/ε₀ = 2.1 × 10⁴ V/m, and E₀ − E = 4.0 × 10⁴ − 1.9 × 10⁴ = 2.1 × 10⁴ V/m. The induced field accounts exactly for the drop.

Interpretation. The stored energy fell by 5.6 × 10⁻⁸ J. That energy went into work done on the slab: the field pulls a dielectric into an isolated capacitor, and you would have to hold the slab back to stop it speeding up.

Worked example 2: inserting a slab with the battery connected

Question. A 120 pF air capacitor stays connected to a 9.0 V battery. A slab with κ = 3.0 is inserted to fill the gap. Find the change in charge, the change in stored energy and the work done by the battery.

1. Q₀ = C₀ΔV = (120 × 10⁻¹²)(9.0) = 1.08 × 10⁻⁹ C.
2. C = 3.0 × 120 pF = 360 pF, so Q = 3.24 × 10⁻⁹ C. The battery pushes ΔQ = 2.16 × 10⁻⁹ C onto the plates.
3. U₀ = ½C₀(ΔV)² = 4.86 × 10⁻⁹ J; U = ½C(ΔV)² = 1.46 × 10⁻⁸ J. ΔU = +9.72 × 10⁻⁹ J.
4. The battery moves ΔQ across a fixed 9.0 V, so it does work ΔQ × ΔV = 1.94 × 10⁻⁸ J.

Interpretation. The battery supplies twice the gain in stored energy. The other 9.72 × 10⁻⁹ J is the work done by the field pulling the slab in, just as in Worked example 1. E between the plates is the same before and after, 9.0 V ÷ d.

Common misconceptions

• “A dielectric adds charge to the plates.” With the capacitor isolated, the free charge is unchanged; only the bound charge appears, and it stays in the dielectric.
• “The dielectric cancels the field, like a conductor.” It only reduces it, by the factor κ. Only a conductor makes the field inside zero.
• “The induced field points the same way as the applied field.” It points the opposite way.
• “E always falls when a dielectric is inserted.” Only if Q is fixed. With a battery connected, E = ΔV/d stays the same.
• “Energy always increases when C increases.” With Q fixed, U = Q²/(2C) falls. With ΔV fixed, U = ½C(ΔV)² rises.
• Dividing C by κ. Capacitance goes up by κ; field and ΔV (for an isolated capacitor) go down.
• “κ can be less than 1.” For the materials in this course, κ ≥ 1, with κ = 1 only for a vacuum.

Where this leads

Capacitors, with or without dielectrics, are about to become circuit components. In Topic 11.1, Electric Current, charge starts to flow, and later in Unit 11 you will combine capacitors and watch them charge through resistors. Practise now with the practice questions, then use the revision notes and the topic checklist.