Inside a Capacitor: Area, Gap and Energy

A capacitor is two plates holding equal and opposite charge. Bigger plates hold more charge at the same voltage, a smaller gap pulls the charges closer together so they hold more too, and a dielectric between the plates lets the same voltage store even more. The energy sits in the electric field between the plates.

How it works

Capacitance and stored energy

In one sentence

For parallel plates of area $A$ a distance $d$ apart with a dielectric of constant $\kappa$ , $C = \frac{\kappa\varepsilon_0 A}{d}$ . A capacitor at voltage $V$ holds $Q = CV$ and stores $U = \frac{1}{2}CV^2$ , which equals the field energy density $\frac{1}{2}\kappa\varepsilon_0 E^2$ times the volume $Ad$ .

General formula

{\htmlData{sym=0}{C}} = \frac{{\htmlData{sym=1}{\kappa}}\,{\htmlData{sym=2}{\varepsilon_0}} {\htmlData{sym=3}{A}}}{\htmlData{sym=4}{{d}}}, \qquad {\htmlData{sym=5}{Q}} = {\htmlData{sym=0}{C}}{\htmlData{sym=6}{V}}, \qquad {\htmlData{sym=7}{U}} = \tfrac12 {\htmlData{sym=0}{C}}{\htmlData{sym=6}{V}}^2, \qquad {\htmlData{sym=8}{E}} = \frac{\htmlData{sym=6}{{V}}}{\htmlData{sym=4}{{d}}}

Capacitance measures how much charge a pair of conductors can hold per volt. Pulling the plates closer or making them bigger lets more charge crowd onto them for the same push.

What the symbols mean

$C$
Capacitance
Charge stored per volt, in farads (here picofarads).
$\kappa$
Dielectric constant
How much an insulator between the plates multiplies the capacitance: 1 for air.
$\varepsilon_0$
Permittivity of free space
8.85 × 10⁻¹² F/m, the constant that ties charge to field in a vacuum.
$A$
Plate area
The area of each plate.
$d$
Gap
The distance between the plates.
$Q$
Charge
The charge on each plate: +Q on one, −Q on the other.
$V$
Voltage
The potential difference between the plates.
$U$
Stored energy
The energy held in the field between the plates.
$E$
Field
The uniform electric field between the plates.

Charge $\pm Q$ spreads over plates of area $A$ , so the field between them is uniform: $E = \frac{Q}{\varepsilon_0 A}$ (Gauss's law for a sheet, doubled by the two plates).
The voltage is the field times the gap, $V = Ed = \frac{Qd}{\varepsilon_0 A}$ , so the charge per volt is $C = \frac{Q}{V} = \frac{\varepsilon_0 A}{d}$ ; a dielectric multiplies it by $\kappa$ .
Charging takes work: moving charge $dq$ across voltage $q/C$ costs $\frac{q}{C}\,dq$ , and integrating from $0$ to $Q$ gives $U = \frac{Q^2}{2C} = \frac12 CV^2$ .
Here $A$ is in cm², $d$ in mm and $V$ in volts, so $C$ comes out in picofarads, $Q$ in picocoulombs and $U$ in picojoules.

Learn the math

Capacitance and Stored EnergyE&M · Lesson 11, Electric Potential

Inside a Capacitor: Area, Gap and Energy