Parallel Plate Capacitor

Equivalent forms

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

C = \frac{Q}{V}

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

Three geometric parameters fully determine the ability to store electric energy — a doorway from fields to circuits.

Symbol	Name	SI unit	Dimension	Range
$C$	Capacitanceoutput Charge stored per unit voltage	F	I²·T⁴·M⁻¹·L⁻²	1e-12 – 0.000001
$A$	Plate area Area of one capacitor plate	m²	L²	0.0001 – 1
$d$	Plate separation Distance between the two parallel plates	m	L	0.00001 – 0.1

Unit systems

Symbol	SI	CGS	Imperial
$C$	F	cm (esu)	F
$A$	m²	cm²	ft²
$d$	m	cm	in

Where it holds

Valid for ideal parallel plates where A >>

d^{2}

(uniform field approximation). Edge fringing effects become significant when plate separation approaches plate dimensions. Add dielectric constant

\kappa for

filled capacitors.

Dimensional analysis

$[C] = [\varepsilon _{0}]\cdot [$ A $]/[d] \to (F/m)(m^{2})(m^{-1}) = F \checkmark$

Discovery

Ewald Georg von Kleist / Pieter van Musschenbroek · 1745

The Leyden jar — the first capacitor — was independently discovered by von Kleist and van Musschenbroek. The parallel-plate formula was derived later from Gauss's law and the uniform field between plates.

Try this

How does your phone store charge without any chemical reaction?

A parallel-plate capacitor has plates of area 0.01 m² separated by 0.001 m in vacuum. Find the capacitance.

Research status: stable

Real-world applications

Touchscreen sensors in smartphones and tablets
DRAM memory cells in computers (capacitor stores 1 bit)
Power factor correction in AC circuits
Energy storage in camera flash units

Common misconceptions

Capacitance depends only on geometry and the dielectric, not on charge or voltage
The formula assumes a uniform field — it fails near the plate edges
Real capacitors have parasitic inductance and resistance not captured by this formula

Experimental verification

Measured using bridge circuits since the 19th century. Modern MEMS capacitors confirm the formula down to micrometer-scale separations. Deviations at nanometer scales reveal quantum capacitance effects.

Derivation

Apply Gauss's law to one plate:

E = \sigma /\varepsilon _{0} = Q/(\varepsilon _{0}A)

.

Voltage across the gap:

V = Ed = Qd/(\varepsilon _{0}A)

.

Therefore

C = Q/V = \varepsilon _{0}A/d

.

Limiting cases

d \to 0

⟶

C \to \infty

Infinitesimally close plates store infinite charge per volt — in practice, dielectric breakdown limits this.

A \to 0

⟶

C \to 0

Zero plate area means no charge storage.

d \to \infty

⟶

C \to 0

Widely separated plates cannot maintain a field to store energy.

What if…

What if you insert a glass dielectric

(\kappa = 5)

?

Capacitance increases $5\times$ : $C = 5 \times 88.54\,\mathrm{pF} = 442.7\,\mathrm{pF}$ . The dielectric polarizes and partially cancels the internal field, allowing more charge per volt.

What if the plate separation doubles?

Capacitance halves: $C = 44.27\,\mathrm{pF}$ . Wider gap means weaker field for the same charge, so less charge per volt.

What if you connect two identical capacitors in parallel?

Total capacitance doubles: C_total $= 2 \times 88.54 = 177.08\,\mathrm{pF}$ . Parallel connection effectively doubles the plate area.

1

Vacuum capacitor

Given ·

A:: 0.01
d:: 0.001

Find · C

Steps

Identify: $A = 0.01\,\mathrm{m}^{2}$ , $d = 0.001\,\mathrm{m}$ , vacuum (no dielectric)
Apply $C = \varepsilon _{0}A/d$
Compute: $8.8541878128\times 10^{-12} \times 0.01 = 8.854\times 10^{-14}$
Divide by d: $8.854\times 10^{-14} / 0.001 = 8.854\times 10^{-11} F$
Convert: 88.54 pF — typical of a small lab capacitor

Result ·

C = \varepsilon _{0} \times A / d = 8.854\times 10^{-12} \times 0.01 / 0.001 = 8.854\times 10^{-11} F = 88.54\,\mathrm{pF}

2

Energy stored in a charged capacitor

Given ·

C:: 8.854e-11
V applied:: 100

Find · Energy U

Steps

Capacitance from previous example: $C = 88.54\,\mathrm{pF}$
Applied voltage: $V = 100\,\mathrm{V}$
Apply $U = \tfrac{1}{2}CV^{2}$
$U = 0.5 \times 88.54\times 10^{-12} \times 10000 = 4.43\times 10^{-7} J$
This energy is stored in the electric field between the plates

Result ·

U = \tfrac{1}{2}CV^{2} = 0.5 \times 88.54\times 10^{-12} \times 100^{2} = 4.43\times 10^{-7} J = 0.443 \mu J

Gauss's Law→Electric Potential (Point Charge)→energy stored capacitor