Unit 10 · Topic 10.4

10.4 Dielectrics

A dielectric is an insulating material placed between a capacitor's plates. It becomes polarized, its own field partly cancels the plates' field, and the capacitance goes up by a factor κ, the dielectric constant. What happens to the charge, voltage and energy depends on whether the capacitor is isolated or still connected to a battery.

Key terms

dielectric
dielectric constant (κ)
polarization
permittivity
isolated vs. connected capacitor

What a dielectric does

Put an insulator into an electric field and its molecules polarize: each one's electrons shift slightly opposite to the field, and molecules that are already lopsided, like water, tend to line up with it. No charge flows through the material, but its surfaces end up with a thin layer of charge. The surface next to the + plate gets a negative layer, and the surface next to the − plate gets a positive layer.

Those induced layers make a field that points opposite to the plates' field. The net field inside the dielectric is weaker. For the same charge on the plates, a dielectric that fills the gap cuts the field by a factor κ: $E = \dfrac{E_0}{\kappa}$ . (If a battery stays connected, it adds charge to the plates and the field ends up unchanged; see the table below.)

The dielectric constant

The dielectric constant κ (kappa) is a number with no units that says how strongly a material weakens the field. It's 1 for a vacuum, barely more than 1 for air, a few for paper or plastic, and about 80 for water. A material's permittivity is $\varepsilon = \kappa\varepsilon_0$ .

For a parallel-plate capacitor completely filled with a dielectric, $C = \frac{\kappa\varepsilon_0 A}{d} = \kappa C_0$ where $C_0$ is the capacitance with nothing in the gap. A weaker field means less voltage for the same charge, and so more capacitance.

Isolated versus connected

What happens when you slide in a dielectric depends on what's held fixed:

Quantity	Isolated (Q fixed)	Connected to battery (ΔV fixed)
C	× κ	× κ
Q	same	× κ
ΔV	÷ κ	same
E between plates	÷ κ	same
U	÷ κ	× κ

Where the energy goes

Isolated: $U = Q^2/2C$ falls, because C grows while Q stays put. The lost energy goes into pulling the dielectric in: the plates' charges attract the induced surface charges, so the field does work on the slab as it enters.

Connected: $U = \tfrac{1}{2}C(\Delta V)^2$ rises. The battery supplies the extra energy by pushing more charge onto the plates.

Dielectric slab versus metal slab

Compare a slab of metal of thickness t slid into the gap, not touching either plate. The metal is a conductor, so its free electrons move until the field inside it is zero (10.1). The field now exists only across the remaining gap, d − t, so $C = \dfrac{\varepsilon_0 A}{d - t}$ . A conductor is like a dielectric with an enormous κ: it cancels the field inside it completely, while a dielectric only weakens it.

Either slab raises the capacitance, and neither changes the plates' charge if the capacitor is isolated. The difference is how much of the field survives inside the slab.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Inserting a dielectric, two ways
A 50 pF air-filled capacitor is charged by a 12 V battery. A dielectric with κ = 3.0 then fills the gap. Find the new Q, ΔV and U (a) if the battery was disconnected first and (b) if the battery stays connected.
Show the solutionHide the solution
1. Step 1: Before: Q = CΔV = (50 × 10⁻¹²)(12) = 600 pC, and $U = \tfrac{1}{2}Q\Delta V = 3.6\times10^{-9}$ J. With the dielectric, C = 3.0 × 50 = 150 pF.
2. Step 2: (a) Isolated: Q stays 600 pC. $\Delta V = Q/C = 600/150 = 4.0$ V. $U = \tfrac{1}{2}(600\times10^{-12})(4.0) = 1.2\times10^{-9}$ J, one-third of before.
3. Step 3: (b) Connected: ΔV stays 12 V. Q = (150 pF)(12 V) = 1800 pC. $U = \tfrac{1}{2}(1800\times10^{-12})(12) \approx 1.1\times10^{-8}$ J, three times before.
Answer: (a) Q = 600 pC, ΔV = 4.0 V, U = 1.2 × 10⁻⁹ J; (b) Q = 1800 pC, ΔV = 12 V, U ≈ 1.1 × 10⁻⁸ J
Example 2Calculator allowed
Finding κ from a voltage reading
An isolated charged capacitor reads 90 V. When a slab of unknown material fills the gap, the reading drops to 36 V. Find the dielectric constant.
Show the solutionHide the solution
1. Step 1: The capacitor is isolated, so Q is fixed and $\Delta V$ is inversely proportional to C.
2. Step 2: $\kappa = \dfrac{C}{C_0} = \dfrac{\Delta V_0}{\Delta V} = \dfrac{90}{36} = 2.5$ .
Answer: κ = 2.5
Example 3Calculator allowed
The field inside a dielectric (classic trap)
Parallel plates carry a free surface charge density of 4.0 × 10⁻⁶ C/m². The gap is filled with a dielectric of κ = 2.0. Find the field in the dielectric, and the charge density induced on each face of the dielectric.
Show the solutionHide the solution
1. Step 1: Without the dielectric: $E_0 = \dfrac{\sigma}{\varepsilon_0} = \dfrac{4.0\times10^{-6}}{8.85\times10^{-12}} \approx 4.5\times10^{5}$ N/C.
2. Step 2: With it: $E = \dfrac{E_0}{\kappa} \approx 2.3\times10^{5}$ N/C.
3. Step 3: The field acts as if the net surface charge were σ/κ = 2.0 × 10⁻⁶ C/m². So each face of the dielectric carries an induced charge density of 4.0 × 10⁻⁶ − 2.0 × 10⁻⁶ = 2.0 × 10⁻⁶ C/m², opposite in sign to the plate it touches.
4. Step 4: The trap is thinking the dielectric removes charge from the plates. The plates' charge is unchanged; the dielectric's induced surface charge partly cancels its field.
Answer: E ≈ 2.3 × 10⁵ N/C; induced charge density 2.0 × 10⁻⁶ C/m² on each face, opposite to the neighboring plate

Common mistakes

Dividing C by κ. A dielectric always increases capacitance.
Thinking charge flows through the dielectric. It's an insulator: its charges shift slightly but stay bound to their molecules.
Applying the isolated-capacitor results to a connected one (or the reverse). Decide first whether Q or ΔV is fixed.
Saying the stored energy always increases with a dielectric. It decreases if the capacitor is isolated.

On the exam

Expect tables or multiple-choice items asking how C, Q, ΔV, E and U change when a dielectric is inserted, with the battery connected or not.
In explanations, describe polarization: the induced surface charges make a field opposite to the original, so the net field and the voltage are smaller for the same plate charge.
Unless a problem says otherwise, the exam treats capacitors as air-filled, so κ ≈ 1. Use κ only when a dielectric is named.

Connected topics

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Check yourself

4 questions on 10.4 Dielectrics. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A charged capacitor is disconnected from its battery and stores energy U₀. A dielectric with κ = 3 is then slid in to fill the gap completely. What is the new stored energy?

Question 2 of 4Calculator allowed

A capacitor stays connected to a battery while a dielectric of constant κ is slid in to fill the gap. What happens to the charge on the plates and the stored energy?

Question 3 of 4Calculator allowed

A 50 pF air-filled capacitor is connected to a 20 V battery. While it stays connected, the gap is filled with a dielectric of κ = 2.5. How much more charge flows onto the positive plate?

Question 4 of 4Calculator allowed

Why does putting a dielectric between a capacitor's plates increase its capacitance?

0 of 4 answered