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A capacitor is one of the simplest — and yet most important — components in electronics. In its purest form it is nothing more than two metal plates held a small distance apart, separated by an insulator. Despite this simplicity, capacitors appear in every radio, every power supply, every camera flash, every computer motherboard and almost every sensor you are ever likely to meet. The quantity that characterises a capacitor is its capacitance, and that is the subject of this lesson.
This lesson begins OCR A-Level Physics A Module 6.1 — Capacitors. Module 6 as a whole is about storing and manipulating charge, and it leans heavily on ideas from Module 4 (electricity) and Module 5 (fields). Capacitance is the bridge between them: it tells us how much charge a conductor can hold for a given potential, and the formulae we derive here will unlock every later calculation in the topic.
Connect a resistor to a battery and current flows through it continuously, dissipating energy as heat. Connect a capacitor to the same battery, and something very different happens. A brief current flows — milliseconds or microseconds — and then stops. The capacitor has stored charge on its plates, and now holds a potential difference equal to the battery e.m.f. It is, in effect, a tiny reservoir of electrical energy.
Key idea: A capacitor does not allow steady current through it. It stores charge on its plates and a corresponding amount of energy in the electric field between them.
Disconnect the battery and connect the charged capacitor to a small lamp. The lamp lights briefly as charge leaves the plates and flows through the lamp, then goes out again. The capacitor has discharged.
This "charge up quickly, release quickly" behaviour is exploited in camera flash units (tens of joules released in a millisecond), defibrillators (hundreds of joules released in a few milliseconds), and the timing circuits that make computers tick.
Suppose we put a charge Q on a capacitor and measure the potential difference V that develops across its terminals. Experiment shows that the two quantities are proportional:
Q ∝ V
The constant of proportionality is called the capacitance, symbol C:
C = Q / V
The SI unit of capacitance is the farad (F), named after Michael Faraday. From the defining equation,
1 F = 1 C V⁻¹
One farad is an enormous amount of capacitance. Most real capacitors in circuits are measured in microfarads (μF, 10⁻⁶ F), nanofarads (nF, 10⁻⁹ F) or picofarads (pF, 10⁻¹² F). A 1 F capacitor is roughly the size of a fist.
| Symbol | Meaning | Unit |
|---|---|---|
C | Capacitance | farad (F) |
Q | Charge stored on one plate | coulomb (C) |
V | Potential difference between plates | volt (V) |
Watch the notation! The letter C appears twice in capacitor calculations — as the symbol for capacitance (italicised: C) and as the symbol for the unit coulomb (upright: C). Context always makes it clear, but exam answers must use the correct unit.
A 47 μF capacitor is connected to a 12 V battery. How much charge is stored on each plate?
Q = CV
= 47 × 10⁻⁶ × 12
= 5.64 × 10⁻⁴ C
= 564 μC
The plates carry equal and opposite charges: +564 μC on the plate connected to the positive terminal, −564 μC on the other. The capacitor as a whole is electrically neutral; what it "stores" is the separation of charge, not any net charge.
The simplest capacitor geometry is two parallel metal plates of area A, separated by a small distance d. For a vacuum (or, to an excellent approximation, air) between the plates, the capacitance is
C = ε₀A / d
where ε₀ = 8.85 × 10⁻¹² F m⁻¹ is the permittivity of free space.
This formula makes intuitive sense:
graph LR
A[Plate + Q] ---|separation d| B[Plate − Q]
C[Area A each] --> A
C --> B
Two square metal plates of side 15 cm are held 2.0 mm apart in air. Calculate the capacitance.
A = 0.15 × 0.15 = 0.0225 m²
d = 2.0 × 10⁻³ m
C = ε₀A / d
= (8.85 × 10⁻¹²)(0.0225) / (2.0 × 10⁻³)
= 9.96 × 10⁻¹¹ F
≈ 100 pF
Even a reasonably large pair of plates in air gives only about 100 picofarads — a tiny capacitance. This is why real capacitors are built with very thin insulators and often with rolled or interleaved layers of metal foil to get large areas into small volumes.
If the space between the plates is filled with an insulator — called a dielectric — the capacitance increases. The formula becomes
C = ε₀ε_r A / d
where ε_r is the dimensionless relative permittivity (sometimes called the dielectric constant) of the material.
| Material | Relative permittivity ε_r |
|---|---|
| Vacuum | 1 (exactly) |
| Air | ~1.0006 |
| Paper | ~2.0 – 4.0 |
| Polythene | ~2.3 |
| Mica | ~7 |
| Ceramic (low-K) | ~10 |
| Ceramic (high-K) | up to 10 000 |
| Water (at 20°C) | ~80 |
Why does a dielectric raise the capacitance? When the dielectric is placed in the field between the plates, its molecules become polarised: their positive and negative charge centres shift slightly in opposite directions. The surfaces of the dielectric develop layers of induced charge that partially cancel the field due to the plates. For the same charge on the plates, the field — and hence the voltage — is reduced, so C = Q / V is larger.
The plates from the previous example are now separated by a 2.0 mm sheet of mica with ε_r = 7.0. What is the new capacitance?
C = ε₀ε_r A / d
= 7.0 × (9.96 × 10⁻¹¹ F)
≈ 7.0 × 10⁻¹⁰ F
≈ 700 pF
Inserting a dielectric has multiplied the capacitance by ε_r, exactly as the formula predicts.
OCR H556 is unusual among A-Level specifications in requiring students to know the capacitance of an isolated conducting sphere. A conducting sphere of radius R held at potential V relative to a notional "zero" at infinity carries charge
Q = 4πε₀RV
Dividing by V, we obtain the capacitance:
C = 4πε₀R
This formula tells you how much charge a sphere can hold per volt of potential — and the answer is surprisingly small. Even a sphere the size of the Earth has a capacitance of only a few hundred microfarads. Lightning is a reminder that a small capacitance at a very high voltage can still store a lot of energy.
A metal sphere of radius 5.0 cm is isolated in air. Calculate its capacitance.
C = 4πε₀R
= 4π × (8.85 × 10⁻¹²) × 0.050
= 5.56 × 10⁻¹² F
≈ 5.6 pF
A 5 cm sphere is therefore essentially a 5.6 pF capacitor. Contrast this with the 100 pF we calculated for two 15 cm plates: the parallel-plate geometry wins easily, which is why every practical capacitor is built that way.
Exam Tip: OCR loves the isolated sphere formula because it connects capacitance back to the potential of a point charge (Module 6.3). If you can derive
C = 4πε₀RfromV = Q/(4πε₀R)in the exam, you will pick up marks other candidates miss.
Real capacitors span about 12 orders of magnitude:
| Device | Typical capacitance |
|---|---|
| RF tuning capacitor | 1 – 100 pF |
| Ceramic disc (bypass) | 1 – 100 nF |
| Film capacitor | 10 nF – 10 μF |
| Electrolytic (power supply) | 1 μF – 10 000 μF |
| Supercapacitor (memory backup) | 0.1 – 100 F |
| Earth as an isolated sphere | ~700 μF |
C = ε₀ε_r A/d, not C = ε₀A/d with a note saying "with dielectric".C = 4πε₀R.Q in C = Q/V is the charge on one plate — either the positive or (in magnitude) the negative.4.70 × 10⁻⁴ F, not 4.70 × 10⁻⁷ F. Practise converting μF, nF and pF until it is automatic.C = Q/V, unit farad.C = ε₀A/d.C = ε₀ε_r A/d.C = 4πε₀R.C = Q/V; unit the farad.C = ε₀A/d.ε_r multiplies the capacitance by ε_r.R has C = 4πε₀R — an OCR-specific result.