A capacitor stores charge: two conductors, usually flat plates, separated by a gap. Connect it to a battery and charge $+Q$ builds on one plate and $-Q$ on the other, with the uniform field you met in Equipotentials and Field Strength between them.
Why it works
The field between closely spaced plates of area $A$ is $E = \dfrac{Q}{\varepsilon_0 A}$, where $\varepsilon_0 = 8.85 \times 10^{-12}\,\text{F/m}$. The voltage across the gap is then $V = Ed = \dfrac{Qd}{\varepsilon_0 A}$. So $Q$ is proportional to $V$, and the ratio $C = Q/V = \varepsilon_0 A/d$ depends only on the shape. Fill the gap with an insulator and its molecules polarize, weakening the field for the same charge by a factor $\kappa$, the dielectric constant, so $C$ grows by $\kappa$.
Key idea
$$C = \frac{Q}{V}, \qquad C = \frac{\kappa\varepsilon_0 A}{d}$$ Capacitance, in farads ($1\,\text{F} = 1\,\text{C/V}$), depends on the plates' area, their gap and the material between them, not on the charge or the voltage.
Worked examples
Worked example: Charge stored
How much charge does a $4.7\,\mu\text{F}$ capacitor hold at $9.0\,\text{V}$?
- $Q = CV = (4.7 \times 10^{-6}\,\text{F})(9.0\,\text{V}) = 4.2 \times 10^{-5}\,\text{C} = 42\,\mu\text{C}$.
Answer: $42\,\mu\text{C}$
Worked example: From the shape
Plates of area $0.020\,\text{m}^2$ sit $1.0\,\text{mm}$ apart in air. Find $C$.
- $$C = \dfrac{\varepsilon_0 A}{d} = \dfrac{(8.85 \times 10^{-12})(0.020)}{1.0 \times 10^{-3}} = 1.77 \times 10^{-10}\,\text{F} = 177\,\text{pF}.$$
- A farad is huge; everyday capacitors are measured in microfarads, nanofarads and picofarads.
Answer: $177\,\text{pF}$
Worked example: Changing the shape
That capacitor's gap is halved and filled with a material of $\kappa = 3$. Find the new $C$.
- Halving $d$ doubles $C$; the dielectric triples it.
- $C = 177\,\text{pF} \times 2 \times 3 = 1.06 \times 10^{3}\,\text{pF} \approx 1.1\,\text{nF}$.
Answer: About $1.06\,\text{nF}$
Common mistake
Thinking a capacitor's capacitance depends on how much charge it holds. Double the voltage and the charge doubles, so their ratio $C$ stays the same. Capacitance is like the width of a bucket: it tells how much the bucket holds per centimeter of depth, whether it is full or empty. Check: $\kappa\varepsilon_0 A/d$ contains no $Q$ or $V$.
When you change a capacitor (its gap, its area or its dielectric), ask what is held fixed. Still connected to a battery, its voltage stays the same and its charge changes. Disconnected, its charge has nowhere to go, so the charge stays the same and its voltage changes.