Capacitance & calculating it

Two conductors that store equal and opposite charge make a capacitor. Capacitance is how much charge it holds per volt.

Source: Serway & Jewett, Physics for Scientists and Engineers, 7th ed., §26.1–26.2.

Before you start

What you need first

Topic 9 — voltage $V$, and $V = Ed$ between parallel plates.
Topic 7 — the uniform field of a charged sheet/plate.
Powers of ten — µF, nF, pF.

What you'll be able to do

Say what a capacitor is and what it stores.
Use $C = Q/V$ (and $Q = CV$), and the farad.
Use the parallel-plate formula $C = \varepsilon_0 A/d$.
Say what makes a capacitance bigger.

The idea

What is a capacitor?

A capacitor is two conductors (often two plates) with a gap between them. Connect a battery and one plate gains $+Q$, the other $-Q$ — the capacitor stores that charge and the energy with it.

Think of it as a tiny tank for charge: a bigger tank holds more charge at the same voltage.

Two plates of area $A$, a gap $d$ apart, holding $+Q$ and $-Q$.

Capacitance — the definition

Capacitance $C$ measures how much charge a capacitor holds per volt across it:

$$ C = \frac{Q}{V} \qquad\Longleftrightarrow\qquad Q = CV $$

where:
Symbol	Meaning	SI unit
C	capacitance (charge stored per volt)	F = C/V
Q	charge on each plate (one $+Q$, one $-Q$)	C
V	voltage across the plates	V

The farad is huge. Real capacitors are far smaller, so you'll see µF $(10^{-6})$, nF $(10^{-9})$, and pF $(10^{-12})$. Convert to farads before using any formula.

📐 Worked example 1

Capacitance and charge stored

(a) A capacitor stores $Q = 30\ \mu\text{C}$ at $V = 10$ V — find $C$. (b) A $C = 5\ \mu\text{F}$ capacitor is connected to $V = 12$ V — find the charge stored.

1Capacitance from $C = Q/V$:

$$ C = \frac{30\times10^{-6}}{10} = 3\times10^{-6}\ \text{F} = 3\ \mu\text{F} $$

2Charge from $Q = CV$:

$$ Q = (5\times10^{-6})(12) = 6\times10^{-5}\ \text{C} = 60\ \mu\text{C} $$

Capacitance of parallel plates

For two flat plates, the capacitance is set entirely by the geometry — area and gap (with air or vacuum between):

$$ C = \frac{\varepsilon_0 A}{d} $$

where:
Symbol	Meaning	SI unit
A	area of each plate	m²
d	gap between the plates	m
ε₀	permittivity of free space $=8.85\times10^{-12}$	C²/(N·m²)

Bigger plates (large $A$) → more room for charge → bigger $C$. Smaller gap (small $d$) → bigger $C$. Real capacitors use large plates very close together, often rolled into a tiny cylinder.

📐 Worked example 2

A parallel-plate capacitor

Two plates of area $A = 0.020$ m² sit a gap $d = 0.0010$ m (1 mm) apart, with air between. Find the capacitance.

1Use $C = \varepsilon_0 A/d$:

$$ C = \frac{(8.85\times10^{-12})(0.020)}{0.0010} $$

2Work it out:

$$ C \approx 1.8\times10^{-10}\ \text{F} = 180\ \text{pF} $$

✏️ Try it yourself

A parallel-plate capacitor has plates of area $A = 0.050$ m² with an air gap $d = 0.0020$ m.

(a) Find its capacitance.
(b) If you halve the gap, what happens to the capacitance?

(a) $C = \dfrac{\varepsilon_0 A}{d} = \dfrac{(8.85\times10^{-12})(0.050)}{0.0020} \approx 2.2\times10^{-10}\ \text{F} = 220\ \text{pF}.$ (b) $C$ is inversely proportional to $d$, so halving the gap doubles $C$ to about $440$ pF.

Common mistakes

Mistake	Fix
Thinking $C$ depends on $Q$ or $V$.	$C$ is fixed by the geometry ($A$, $d$); $C=Q/V$ just measures it. More $V$ → proportionally more $Q$, same $C$.
Leaving capacitance in µF/pF in a formula.	Convert to farads first.
Putting $d$ on top of the plate formula.	$C = \varepsilon_0 A/d$: the gap is on the bottom — smaller gap, bigger $C$.
Forgetting both plates carry the same size charge.	One plate $+Q$, the other $-Q$; "the charge" $Q$ is that size.

Recap — the whole topic on one screen

Idea	What you own now
Capacitor	Two conductors storing $+Q$ and $-Q$; a charge "tank."
Capacitance	$C = Q/V$, $Q = CV$; unit farad (C/V), usually µF–pF.
Parallel plates	$C = \dfrac{\varepsilon_0 A}{d}$ — set by geometry.
Bigger C	Large area, small gap.

Next topic

Capacitors in series & parallel

Wire several capacitors together and they behave like one. Next we find the rules — and meet the circuit symbol for a capacitor.

→ Topic 13 · Capacitors in series & parallel

Symbol	Meaning	SI unit
C	capacitance (charge stored per volt)	F = C/V
Q	charge on each plate (one \(+Q\), one \(-Q\))	C
V	voltage across the plates	V

Mistake	Fix
Thinking \(C\) depends on \(Q\) or \(V\).	\(C\) is fixed by the geometry (\(A\), \(d\)); \(C=Q/V\) just measures it. More \(V\) → proportionally more \(Q\), same \(C\).
Leaving capacitance in µF/pF in a formula.	Convert to farads first.
Putting \(d\) on top of the plate formula.	\(C = \varepsilon_0 A/d\): the gap is on the bottom — smaller gap, bigger \(C\).
Forgetting both plates carry the same size charge.	One plate \(+Q\), the other \(-Q\); "the charge" \(Q\) is that size.

Idea	What you own now
Capacitor	Two conductors storing \(+Q\) and \(-Q\); a charge "tank."
Capacitance	\(C = Q/V\), \(Q = CV\); unit farad (C/V), usually µF–pF.
Parallel plates	\(C = \dfrac{\varepsilon_0 A}{d}\) — set by geometry.
Bigger C	Large area, small gap.