Section 02
The Idea, Step by Step
Start: a tiny charge bucket
A capacitor is two metal plates facing each other across a small gap. Hook them to a battery and electrons pile onto one plate while the other is left short — like water filling a tank. Unplug the battery and the charge just sits there, ready to dump out in a quick burst. That is exactly how a camera flash works: it sips charge slowly from the battery, then releases it all in a few thousandths of a second.
Build: capacitance is charge stored per volt
How much charge a capacitor holds for each volt you push onto it is its capacitance $C$, measured in farads. The rule is just $C = Q/V$: a bigger $C$ means more charge $Q$ packed in for the same voltage $V$. For two flat plates the geometry sets it: $C = \kappa\varepsilon_0 A/d$, so wider plates (area $A$) and a thinner gap ($d$) both store more, and a dielectric filling ($\kappa$) multiplies it. Worked number: plates of area $A=0.01\,\text{m}^2$ a gap $d=1\,\text{mm}$ apart in air ($\kappa=1$) give $C=(1)(8.85\times10^{-12})(0.01)/(0.001)\approx 89\,\text{pF}$. Charge that to 12 V and it holds $Q=CV\approx 1.1\,\text{nC}$.
Capacitance
$$C=\frac{Q}{V}=\kappa\varepsilon_0\frac{A}{d}$$
Deepen: charging takes time, energy grows as V²
Charging is not instant — a resistor $R$ in the wire sets the pace. The voltage climbs as $V_C(t)=V_0\left(1-e^{-t/RC}\right)$, reaching about 63% of full after one time constant $\tau=RC$ and essentially full after $5\tau$. The energy banked in the field between the plates is $U=\tfrac12 CV^2$ — notice it grows with the square of voltage, so doubling $V$ quadruples the stored energy (that quadratic is why high-voltage capacitors pack such a punch). The sliders map straight onto these symbols: $C$ and $V$ fix how much charge and energy, $R$ fixes how fast (watch $\tau$ change), and $d$ and $\kappa$ reshape the plate field $E=V/d$ and the capacitance itself.
RC charging & stored energy
$$V_C(t)=V_0\left(1-e^{-t/RC}\right),\qquad U=\tfrac12 CV^2$$
Try this in the sim above
Slide the resistance $R$ up and watch the $V_C(t)$ curve stretch out — the time constant $\tau=RC$ grows, so charging slows. Switch to Dielectric mode and drag $\kappa$ from 1 toward 80 (water) to see the capacitance balloon for the very same gap. Then in Energy mode, double the voltage and confirm the stored energy roughly quadruples rather than doubles.
Section 03
Equations & Derivation
Capacitance & Charge
$$C = \frac{Q}{V},\quad C_{\text{parallel plate}} = \kappa\varepsilon_0\frac{A}{d}$$
Energy Stored
$$U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV$$
Capacitors in Series & Parallel
$$\frac{1}{C_s} = \frac{1}{C_1}+\frac{1}{C_2}+\cdots,\quad C_p = C_1+C_2+\cdots$$
RC Charging
$$V_C(t) = V_0\!\left(1-e^{-t/RC}\right),\quad Q(t) = CV_0\!\left(1-e^{-t/RC}\right)$$
Symbol Definitions
| Symbol | Quantity | SI Unit |
|---|
| $C$ | Capacitance | F |
| $Q$ | Charge on capacitor | C |
| $V_C$ | Voltage across capacitor | V |
| $U$ | Energy stored | J |
| $\kappa$ | Dielectric constant (relative permittivity) | dimensionless |
| $d$ | Plate separation | m |
| $A$ | Plate area | m² |
| $\varepsilon_0 = 8.85\times10^{-12}$ | Permittivity of vacuum | F m⁻¹ |
1
Capacitance. $C=Q/V$ — charge stored per volt. For parallel plates: $C=\kappa\varepsilon_0 A/d$. Larger area → more charge storage. Smaller gap → stronger field, more charge for same voltage. Dielectric increases $C$ by reducing effective field.
2
Energy storage. $U=\frac{1}{2}CV^2$. Energy is stored in the electric field between plates: $u=\frac{1}{2}\varepsilon_0 E^2$ J/m³. Capacitors can release energy very quickly (unlike batteries) — used in camera flashes, defibrillators, particle accelerators.
3
Series capacitors. Same charge on each (charge conservation). Total capacitance is smaller than smallest: $1/C_s=\sum 1/C_i$. Good for high-voltage applications — each capacitor sees only part of the total voltage.
4
Dielectrics. Inserting a dielectric (polarisable material) increases capacitance by $\kappa$ (dielectric constant). Water: $\kappa=80$; mica: $\kappa=7$. The polarised dielectric reduces the effective electric field, allowing more charge to be stored for the same voltage.
Ref: Halliday, Resnick & Walker 10th Ed., Ch. 25; Griffiths — Introduction to Electrodynamics (4th Ed.), §2.5; Horowitz & Hill — The Art of Electronics.
Section 05
Common Misconceptions
❌ Capacitors in series always reduce the total capacitance.
✅ Yes — series capacitors always give smaller total capacitance than the smallest individual capacitor: $1/C_s=\sum1/C_i$. This is the opposite of resistors: capacitors in series behave like resistors in parallel, and vice versa. This is because $C=Q/V$ and adding series capacitors means the same charge $Q$ distributes over higher total voltage $V$.
📖 HRW 10th Ed., §25-5.
❌ A fully charged capacitor allows current to flow freely.
✅ A fully charged capacitor blocks DC current. Once $V_C=V_0$, no more current flows ($I=CdV_C/dt=0$). Capacitors "block DC" but pass AC — because $V_C$ continuously changes for AC, so $I=CdV/dt\neq0$. Capacitors are therefore used as AC coupling and DC blocking elements.
📖 HRW 10th Ed., §25-4.
❌ Doubling voltage doubles energy stored in a capacitor.
✅ Energy $U=\frac{1}{2}CV^2\propto V^2$. Doubling voltage quadruples energy stored. Doubling capacitance doubles energy. This quadratic dependence on voltage means high-voltage capacitors store enormous energy — a reason defibrillators use ~1000V across relatively small capacitors (~200μF).
📖 HRW 10th Ed., §25-3.
Misconception research: McDermott — Tutorials in Introductory Physics; Arons — A Guide to Introductory Physics Teaching.