Dielectrics in Parallel Plate Capacitors Revision Notes | A Level AQA Physics

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]

1. $A = 0.080 \text{ m} \times 0.080 \text{ m} = 6.4 \times 10^{-3} \text{ m}^2$ $d = 2.0 \text{ mm} = 2.0 \times 10^{-3} \text{ m}$ — 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]
2. $C = \frac{A \varepsilon_0 \varepsilon_r}{d}$ $A = 6.4 \times 10^{-3} \text{ m}^2,\quad \varepsilon_0 = 8.85 \times 10^{-12} \text{ F m}^{-1},\quad \varepsilon_r = 4.5,\quad d = 2.0 \times 10^{-3} \text{ m}$ — State the formula first — this secures the method mark even if arithmetic slips later. [M1]
3. 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. $10^{-3} \times 10^{-12} = 10^{-15}$.
4. $C = \frac{2.5488 \times 10^{-13}}{2.0 \times 10^{-3}}$ $= \frac{2.5488}{2.0} \times 10^{-13-(-3)}$ $= 1.2744 \times 10^{-10} \text{ F}$ — $10^{-13} \div 10^{-3} = 10^{-13+3} = 10^{-10}$. Dividing the coefficients: $2.5488 \div 2.0 = 1.2744$.
5. $C \approx 1.3 \times 10^{-10} \text{ F} = 130 \text{ pF}$ — 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.

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.