How a Camera Flash Stores Energy for an Instant Burst of Light

A camera flash charges up with a faint whine and then releases its energy in a split second. The part doing the storing is a capacitor — two conductors separated by an insulator. Understanding it needs the behaviour of conductors and dielectrics in electric fields.

This part covers conductors, dielectrics and polarisation, the parallel plate capacitor, and capacitor combinations with stored energy.

**Free electrons in a conductor move until they cancel any field inside, so in electrostatic equilibrium the field inside is zero, excess charge sits on the outer surface, the conductor is an equipotential, and the field just outside is perpendicular to the surface with magnitude $\frac{\sigma }{{\epsilon }_0}$.

Free and bound charges:

- Conductors contain free charges, such as electrons in metals, that move through the material
- Insulators contain bound charges held within their atoms and molecules

Results for a conductor in electrostatics:**

- $E=0$ inside the material
- The field at the surface is normal to the surface
- Excess charge resides only on the surface
- The whole conductor is at one potential
- The field just outside is $E=\frac{\sigma }{{\epsilon }_0}$
- Electrostatic shielding — a charge-free cavity inside has zero field

Worked example. A conductor has surface charge density $\sigma =8.85\times 10^{-7}$ C m${}^{-2}$. The field just outside is

$$E=\frac{8.85\times 10^{-7}}{8.85\times 10^{-12}}=10^5{\text{ N C}}^{-1}$$

An everyday example. Sensitive instruments are housed in metal cases, which shield them from outside electric fields.

The substance. Charge crowds at the sharp points of a conductor, where the field is strongest — the idea behind lightning conductors.

**A dielectric is an insulator whose molecules become polarised in an electric field, producing an internal field that opposes the applied one; this reduces the net field by a factor $K$, the dielectric constant, and increases the capacitance by the same factor.

Two kinds of molecules:

- Non-polar** molecules, such as O${}_2$ and N${}_2$ — the field shifts their charge centres apart, inducing dipoles
- Polar molecules, such as H${}_2$O and HCl — permanent dipoles that the field partly aligns

Polarisation. The dipole moment per unit volume is the polarisation $\vec{P}$. Bound surface charges appear on the dielectric's faces and reduce the field inside:

$$E=\frac{E_0}{K}$$

Effect on capacitance. For the same charge, the potential difference falls by $K$, so

$$C=K{C}_0$$

Worked example. A $10$ pF capacitor filled with mica of $K=6$ has capacitance $C=6\times 10=60$ pF.

An everyday example. The capacitor in a ceiling fan uses a thin dielectric layer so that a small component can hold a large charge.

The substance. A dielectric does not add charge — it weakens the field, so the same voltage can store more charge.

**Plates of area $A$ separated by $d$ with charges $\pm Q$ produce a uniform field $E=\frac{\sigma }{{\epsilon }_0}$ between them, so $V=\frac{Qd}{{\epsilon }_{0}A}$ and $C_0=\frac{{\epsilon }_{0}A}{d}$; filling the gap with a dielectric of constant $K$ gives $C=\frac{K{\epsilon }_{0}A}{d}$.

Derivation in vacuum:**

- Each plate acts as a charged sheet with field $\frac{\sigma }{2{\epsilon }_0}$; between the plates the fields add to $E=\frac{\sigma }{{\epsilon }_0}$, and outside they cancel
- $V=Ed=\frac{\sigma d}{{\epsilon }_0}=\frac{Qd}{{\epsilon }_{0}A}$
- $C_0=\frac{Q}{V}=\frac{{\epsilon }_{0}A}{d}$

With a dielectric. The field falls to $\frac{E}{K}$, so $V$ falls to $\frac{V}{K}$ and

$$C=\frac{K{\epsilon }_{0}A}{d}$$

Worked example. Plates of area $0.02$ m${}^2$ are $1$ mm apart:

$$C_0=\frac{8.85\times 10^{-12}\times 0.02}{10^{-3}}=1.77\times 10^{-10}\text{ F}=177\text{ pF}$$

Filling the gap with a dielectric of $K=5$ raises this to $885$ pF.

An everyday example. A phone touchscreen senses your finger because the finger changes the capacitance between tiny transparent electrodes in the screen.

The substance. Capacitance depends only on geometry and the medium, not on the charge or voltage applied.

**Capacitors in series carry the same charge and combine as $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\cdots$; in parallel they share the same voltage and combine as $C=C_1+C_2+\cdots$; and a capacitor stores energy $U=\frac{1}{2}C{V}^2=\frac{Q^2}{2C}$.

Series:**

- The same charge $Q$ on each capacitor, with voltages adding
- $\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\cdots$, so the total is less than the smallest

Parallel:

- The same voltage $V$ across each, with charges adding
- $C_{eq}=C_1+C_2+\cdots$

Worked example. Capacitors of $2\mu$F and $3\mu$F in parallel are joined in series with a $5\mu$F capacitor across $100$ V.

$$C_p=2+3=5\mu \text{F},\frac{1}{C_{eq}}=\frac{1}{5}+\frac{1}{5}\Rightarrow C_{eq}=2.5\mu \text{F}$$

$$Q=C_{eq}V=2.5\times 10^{-4}\text{ C},U=\frac{1}{2}{C}_{eq}{V}^2=\frac{1}{2}\times 2.5\times 10^{-6}\times 100^2=0.0125\text{ J}$$

An everyday example. A camera flash charges a capacitor slowly from a small battery, then releases the stored $\frac{1}{2}C{V}^2$ energy into the lamp in a fraction of a second.

The substance. The rules are the reverse of those for resistors — capacitors in parallel add directly, while resistors in series do.