Capacitance — Physica

Equivalent forms

Q = CV

U = \tfrac{1}{2}CV^2

A single ratio Q/V that depends only on geometry and material, never on how much charge you happen to put on.

Symbol	Name	SI unit	Dimension	Range
$C$	Capacitanceoutput Charge stored per unit voltage	F	M⁻¹·L⁻²·T⁴·I²	1e-12 – 0.001
$Q$	Charge Charge on each plate	C	I·T	0 – 0.001
$V$	Voltage Potential difference across plates	V	M·L²·T⁻³·I⁻¹	0.1 – 100

Unit systems

Symbol	SI
$C$	F
$Q$	C
$V$	V

Where it holds

Defines capacitance for any linear conductor pair. For nonlinear dielectrics (ferroelectrics) or near breakdown, C becomes voltage-dependent and the simple ratio no longer holds.

Dimensional analysis

$[Q]/[V] = C/V = F \checkmark$

Discovery

Ewald Georg von Kleist / Pieter van Musschenbroek · 1745

The Leyden jar — the first capacitor — stored such a shock that Musschenbroek wrote he 'would not take a second for the kingdom of France.' Volta later named the property capacity.

Try this

Two metal plates that don't even touch can store charge for hours. How much charge fits per volt?

A capacitor holds 6 μC when charged to 3 V. What is its capacitance, and how much charge would 12 V store?

Research status: stable

Real-world applications

Energy storage and smoothing
Touch sensing
Timing circuits
Decoupling on every circuit board

Common misconceptions

Capacitance depends on how much charge you store — it doesn't; C is a fixed property, Q just follows V
A capacitor 'holds voltage' — it holds charge; the voltage is what that charge produces across the gap

Experimental verification

Charge a capacitor to several known voltages and measure stored charge (e.g. by integrating discharge current); Q vs V is a straight line whose slope is C.

Derivation

For a linear system the potential of a charged conductor is proportional to its charge,

V \propto Q

.

The constant of proportionality (inverse) is the capacitance,

C \equiv Q/V

, fixed by geometry and the dielectric.

Energy is the work to assemble the charge:

U = \int V dq = \int (q/C) dq = \tfrac{1}{2}CV^{2}

.

Limiting cases

V \to 0

⟶

Q \to 0

at fixed CNo voltage, no stored charge — but C itself stays the same, set by geometry.

plate

gap \to 0

⟶

C \to \infty

Closer plates hold more charge per volt; capacitance diverges as the gap closes (until breakdown).

What if…

What if you double the voltage?

Charge doubles but capacitance is unchanged — C is a property of the device, not the state.

What if you slide a dielectric between the plates?

C rises by the dielectric constant $\kappa$ , so the same voltage now stores $\kappa$ times more charge.

1

6 μC at 3 V

Given ·

Q:: 0.000006
V:: 3

Find · C and Q at 12 V

Steps

$C = Q/V = 6 \mu C / 3\,\mathrm{V} = 2 \mu F$
C is fixed, so at 12 V: $Q = CV$
$Q = (2\times 10^{-6})(12) = 24 \mu C (4\times the$ voltage $\to 4\times the$ charge)

Result ·

C = 2 \mu F

; quadrupling the voltage stores

24 \mu C

.

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