Dielectrics in Parallel Plate Capacitors Revision Notes | A Level AQA Physics
A dielectric between the plates increases capacitance, and C = Aε₀εr/d includes its relative permittivity εr. Free AQA notes and 11 questions answered.
Dielectrics in Parallel Plate Capacitors is part of Capacitance (A-level only) in A Level AQA Physics.
6 min
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31
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A Level AQA
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What this lesson covers
A dielectric between the plates increases capacitance, and C = Aε₀εr/d includes its relative permittivity εr. Free AQA notes and 11 questions answered.
Describe the action of a dielectric material in a parallel plate capacitor at the molecular level
Explain how polar molecules rotate in the presence of an electric field and why this reduces the field between the plates
Define relative permittivity and the dielectric constant, using correct AQA terminology
Apply the equation $C = \frac{A \varepsilon_0 \varepsilon_r}{d}$ to calculate capacitance with a dielectric present
Key ideas to keep in view
Use this lesson to stay inside Capacitance (A-level only) while connecting the detail back to the wider Fields and their consequences (A-level only) topic.
Dielectric: An electrically insulating material placed between the plates of a capacitor that increases capacitance by reducing the effective electric field between the plates through polarisation of its molecules.
Relative permittivity (dielectric constant): A dimensionless ratio, symbol $\varepsilon_r$, that describes how much more charge a capacitor can store with a given dielectric compared to a vacuum between its plates. Also called the dielectric constant.
Polarisation: The process by which molecules in a dielectric develop an induced dipole moment — or rotate if already polar — so that positive and negative charge centres separate and align with an applied electric field.
Permittivity of free space: The constant $\varepsilon_0 = 8.85 \times 10^{-12}$ F m$^{-1}$, representing the ability of a vacuum to permit an electric field. It appears in the capacitance equation for a parallel plate capacitor.
Revision notes
Read the core explanations from Dielectrics in Parallel Plate Capacitors before testing yourself from memory.
1. Why Dielectrics Matter
Here is a puzzle: two identical parallel plate capacitors sit on a bench. You slide a slab of material between the plates of one of them — without changing the plate area or separation. Its capacitance increases. No extra charge was added, no geometry was changed. So what is going on inside that material?
The answer lies in the behaviour of molecules under an electric field. Understanding this is not only conceptually satisfying — it is a direct AQA specification requirement, and examiners test it with both descriptive and calculation questions.
2. The Parallel Plate Capacitor — A Quick Anchor
Recall that a parallel plate capacitor stores charge by maintaining opposite charges on two conducting plates separated by a gap. The electric field between the plates points from the positive plate to the negative plate, and the capacitance $C$ tells you how much charge $Q$ you store per volt of potential difference: $C = Q/V$.
With only a vacuum (or air, which is a very good approximation) between the plates, the capacitance depends on three things: the plate area $A$, the separation $d$, and the permittivity of free space $\varepsilon_0$. The relationship is $C = \varepsilon_0 A / d$. Now watch what happens when you fill that gap with a dielectric.
3. What a Dielectric Does — The Molecular Picture
Picture the space between the plates filled with molecules. Some of these molecules are polar — they already carry a permanent dipole moment, meaning one end is slightly negative and the other slightly positive. Water is the classic example, but many polymers used in real capacitors behave similarly.
Before the capacitor is charged, these polar molecules point in random directions, so their effects cancel out. The moment the capacitor is connected to a supply and a field builds up between the plates, something changes: each polar molecule experiences a torque — a turning force — because the field exerts a torque on the molecule as opposite ends experience forces in opposite directions, causing rotation rather than net translation.
The molecules rotate and align with the field. This is the key process the AQA specification asks you to describe: a simple polar molecule rotates in the presence of an electric field so that its dipole aligns with that field.
Now zoom in on what those aligned dipoles do collectively. The negative ends of the molecules cluster near the positive plate; the positive ends cluster near the negative plate. This creates an internal electric field inside the dielectric that opposes the original field from the plates. The net field between the plates is therefore reduced.
A reduced field means a reduced potential difference across the plates for the same stored charge. Since $C = Q/V$, a smaller $V$ for the same $Q$ means a larger capacitance. The dielectric has effectively allowed the capacitor to store more charge per volt — which is precisely why dielectrics are used in real capacitors.
4. Relative Permittivity and the Dielectric Constant
Different materials polarise to different extents. A material that polarises strongly will oppose the field more, reduce $V$ more, and boost capacitance more. To quantify this, physicists define the relative permittivity $\varepsilon_r$, also called the dielectric constant.
For a vacuum, $\varepsilon_r = 1$ by definition. For air, $\varepsilon_r \approx 1.0006$, which is why air is treated as equivalent to a vacuum in most calculations. For materials used in real capacitors — such as mica ($\varepsilon_r \approx 5$), ceramic ($\varepsilon_r$ up to several thousand), or certain polymers — $\varepsilon_r$ is significantly greater than 1, which is why these materials allow compact, high-capacitance devices.
5. The Capacitance Equation with a Dielectric
Combining the geometry of the parallel plate capacitor with the material property $\varepsilon_r$ gives the full equation you need for AQA calculations.
Notice that $\varepsilon_r$ multiplies $\varepsilon_0$. The product $\varepsilon_0 \varepsilon_r$ is sometimes written as $\varepsilon$ (the absolute permittivity of the material). Doubling $\varepsilon_r$ doubles $C$; halving $d$ also doubles $C$. These proportionalities are worth internalising because examiners often ask you to compare two configurations rather than calculate a raw number.
6. Worked Calculation — Applying the Equation
Theory is only useful when you can deploy it under exam conditions. The following worked example walks through a standard AQA-style calculation, with every substitution shown explicitly so you can follow the mark-winning method.
Worked example: Calculating capacitance with a dielectric
A parallel plate capacitor has square plates of side 8.0 cm, separated by 2.0 mm. A dielectric of relative permittivity 4.5 fills the gap. Calculate the capacitance. ($\varepsilon_0 = 8.85 \times 10^{-12}$ F m$^{-1}$) [3 marks]
$A = 0.080 \text{ m} \times 0.080 \text{ m} = 6.4 \times 10^{-3} \text{ m}^2$
$d = 2.0 \text{ mm} = 2.0 \times 10^{-3} \text{ m}$ — Plate side is 8.0 cm = 0.080 m; area = side squared. Separation must be in metres. Converting before substituting avoids unit errors — a common source of dropped marks. [M1]
$C = \frac{A \varepsilon_0 \varepsilon_r}{d}$
$A = 6.4 \times 10^{-3} \text{ m}^2,\quad \varepsilon_0 = 8.85 \times 10^{-12} \text{ F m}^{-1},\quad \varepsilon_r = 4.5,\quad d = 2.0 \times 10^{-3} \text{ m}$ — State the formula first — this secures the method mark even if arithmetic slips later. [M1]
Numerator $= 6.4 \times 10^{-3} \times 8.85 \times 10^{-12} \times 4.5$
$= 6.4 \times 8.85 \times 4.5 \times 10^{-15}$
$6.4 \times 8.85 = 56.64$
$56.64 \times 4.5 = 254.88$
Numerator $= 254.88 \times 10^{-15} = 2.5488 \times 10^{-13}$ — Work through the powers of ten separately from the coefficients to avoid errors. $10^{-3} \times 10^{-12} = 10^{-15}$.
$C \approx 1.3 \times 10^{-10} \text{ F} = 130 \text{ pF}$ — The data is given to 2 significant figures, so round to 2 s.f. Always state the unit — F or pF are both acceptable here. [A1]
Answer: $C \approx 1.3 \times 10^{-10}$ F (130 pF)
7. Comparing Conditions — Constant Voltage vs Constant Charge
A subtlety that AQA examiners exploit is the difference between inserting a dielectric while the capacitor remains connected to a battery (constant voltage) versus inserting it after the battery has been disconnected (constant charge). The equation $C = A\varepsilon_0\varepsilon_r / d$ tells you the new capacitance in both cases, but the consequences for $Q$ and $V$ differ.
Condition
What stays fixed
Effect on $C$
Effect on $Q$
Effect on $V$
Battery connected (constant $V$)
$V$ fixed by battery
$C$ increases (more $\varepsilon_r$)
$Q = CV$ increases
$V$ unchanged
Battery disconnected (constant $Q$)
$Q$ fixed (nowhere to go)
$C$ increases (more $\varepsilon_r$)
$Q$ unchanged
$V = Q/C$ decreases
Both scenarios start from the same equation for $C$. The difference is which quantity the external circuit constrains. Keeping this distinction clear is the mark between a grade B and a grade A response on a 'explain' question.
8. Exam Technique and Common Mistakes
The dielectric topic rewards precision in both language and calculation. The following guidance is drawn from the patterns that cost students marks in AQA mark schemes.
One final point on exam language: the specification uses both 'relative permittivity' and 'dielectric constant' for $\varepsilon_r$. Either term is accepted in AQA answers, but 'relative permittivity' is the more precise term and is preferred in definitions.
Taken together, the three specification requirements for this topic form a coherent story: polar molecules rotate in a field (molecular description), this rotation increases capacitance by a factor $\varepsilon_r$ (material property), and the full equation $C = A\varepsilon_0\varepsilon_r / d$ lets you calculate the result quantitatively (application). Examiners can — and do — test all three levels in a single multi-part question.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course.
The relative permittivity of air is about 1.0006, so close to 1 that air is treated as equivalent to a vacuum in most calculations. A vacuum has a relative permittivity of 1 by definition, and materials used in real capacitors have values significantly greater than 1.
What is relative permittivity?
Relative permittivity εr, also called the dielectric constant, is the ratio of a capacitor's capacitance with the dielectric filling the gap to its capacitance with a vacuum between the plates. It is dimensionless and always at least 1: a vacuum is 1, air about 1.0006, mica about 5, and ceramic up to several thousand.
How do you calculate capacitance with a dielectric?
Use C = Aε₀εr/d. For square plates of side 8.0 cm (A = 6.4 × 10⁻³ m²), 2.0 mm apart, with a dielectric of εr = 4.5, C = 6.4 × 10⁻³ × 8.85 × 10⁻¹² × 4.5 ÷ 2.0 × 10⁻³ ≈ 1.3 × 10⁻¹⁰ F, or 130 pF. Doubling εr doubles C, as does halving d.
What is absolute permittivity?
The absolute permittivity of a material, written ε, is the product ε₀εᵣ: the permittivity of free space multiplied by the relative permittivity. It is the factor that appears with plate area and separation in C = Aε₀εᵣ/d.
What does a dielectric do in a capacitor?
A dielectric is a material placed between the plates of a capacitor that increases its capacitance. When a field is applied, its polar molecules rotate to align with the field, creating an internal field that opposes the field from the plates. The reduced net field means a smaller pd for the same charge, so C = Q/V increases.
How do polar molecules behave in an electric field?
A polar molecule has a permanent dipole, with one end slightly negative and the other slightly positive. In an electric field, opposite ends experience forces in opposite directions, producing a torque that rotates the molecule until its dipole aligns with the field, rather than causing net translation.
What happens when a dielectric is inserted with the battery connected or disconnected?
Either way, the capacitance increases. With the battery still connected, V is fixed by the battery, so the charge Q = CV increases. With the battery disconnected, Q is fixed because the charge has nowhere to go, so the pd V = Q/C decreases.
What are typical dielectric constants?
Mica has a relative permittivity of about 5, ceramics can reach several thousand, and certain polymers are also significantly greater than 1. These high values allow compact, high-capacitance devices, because the capacitance is multiplied by the relative permittivity of the material between the plates.
How do non-polar molecules behave in a dielectric?
Non-polar molecules in a dielectric develop induced dipoles, because the electric field displaces their electron clouds. As with rotated polar molecules, this leaves a layer of negative charge at the surface nearest the positive plate and a layer of positive charge nearest the negative plate.
How does plate separation affect capacitance?
Halving the plate separation doubles the capacitance, because C = Aε₀εᵣ/d, so C goes up as d goes down. Doubling the relative permittivity also doubles C, and increasing the plate area A increases the capacitance in the same proportion.
What is the capacitance of parallel plates in air?
The capacitance of parallel plates with a vacuum or air between them is C = ε₀A/d, depending on the plate area A, the separation d and the permittivity of free space, ε₀ = 8.85 × 10⁻¹² F m⁻¹. Filling the gap with a dielectric multiplies this by the relative permittivity.