Capacitance

Capacitance is the ratio of charge stored on a conductor pair to the voltage between them:

$C=\frac{Q}{V}\text{(F = C/V)}.$

Two conductors carrying $+Q$ and $-Q$ at potential difference $V$ form a capacitor. $C$ depends only on geometry (sizes, shapes, spacing) and the Permittivity of the dielectric between, not on $Q$ or $V$ separately. Pushing more charge raises the voltage in proportion; the ratio stays fixed.

Parallel-plate capacitor

Image: Parallel plate capacitor, CC BY-SA 4.0 — parallel-plate capacitor with uniform field $E$ between the plates.

Two parallel conducting plates of area $A$ separated by distance $d$, with dielectric of permittivity $ϵ$:

$C=\frac{ϵA}{d}.$

Derivation: charges $\pm Q$ distribute quasi-uniformly on the plates with surface density ${\rho }_s=Q/A$. Inside the dielectric, $\mathbf{E}$ is approximately uniform from the conductor boundary condition $E_n={\rho }_s/ϵ$, giving $E=Q/(ϵA)$. Voltage between plates: $V=Ed=Qd/(ϵA)$. Capacitance:

$C=\frac{Q}{V}=\frac{ϵA}{d}.$

The approximation neglects fringing fields at the edges; valid when plate dimensions are much greater than separation $d$.

Coaxial capacitor

Two concentric cylindrical conductors, inner radius $a$, outer radius $b$, length $l$, dielectric $ϵ$ between. Four steps, each an instance of a method that recurs throughout electrostatics.

Step 1 — Charge on the conductors. Put $+Q$ on the inner conductor; by induction $-Q$ accumulates on the inner surface of the outer conductor. Treat both as uniform line charge densities ${\rho }_l=Q/l$ on the inner conductor (using $l$ much greater than $b$, so end effects are negligible).

Step 2 — Field by Gauss’s law. Cylindrical symmetry → $\mathbf{E}=E_r(r)\hat{\mathbf{r}}$ inside the dielectric. Use a cylindrical Gaussian surface of radius $r$ ($a<r<b$) and length $\ell$ coaxial with the conductor pair. Flux through curved side: $E_r\cdot 2\pi r\ell$. Caps contribute zero. Enclosed charge: ${\rho }_l\ell$. So

$E_r\cdot 2\pi r\ell =\frac{{\rho }_l\ell }{ϵ}⟹E_r=\frac{{\rho }_l}{2\pi ϵr}=\frac{Q}{2\pi ϵrl}.$

The $1/r$ fall-off is characteristic of cylindrical symmetry.

Step 3 — Voltage from $\mathbf{E}$. Integrate $\mathbf{E}$ along a radial path from inner conductor ($r=a$) to outer ($r=b$):

$V_{ab}=-{\int }_b^a{E}_{r}dr={\int }_a^b\frac{Q}{2\pi ϵrl}dr=\frac{Q}{2\pi ϵl}\ln \frac{b}{a}.$

Sign convention: $V_{ab}>0$ when the inner conductor is at higher potential. Direction of integration is from low to high potential, picking up the negative sign on $E_r$.

Step 4 — Capacitance. Divide:

$C=\frac{Q}{V_{ab}}=\frac{2\pi ϵl}{\ln (b/a)}\text{(F)}.$

Capacitance per unit length: $C^{\prime }=2\pi ϵ/\ln (b/a)$ (F/m). This is one of the parameters that defines a coaxial Transmission line.

Sanity checks. Capacitance scales linearly with length, as expected for a uniform structure. The $\ln (b/a)$ factor diverges as $a\to 0$ (a thin inner wire concentrates field near the axis) and as $b\to \infty$ (the outer conductor recedes, V grows logarithmically). Doubling $ϵ$ doubles $C$: more dielectric polarization → more stored charge per volt.

Concentric sphere capacitor

Inner radius $a$, outer radius $b$, dielectric $ϵ$. By Gauss with spherical symmetry:

$E_r=\frac{Q}{4\pi ϵr^2},V=\frac{Q}{4\pi ϵ}\left(\frac{1}{a}-\frac{1}{b}\right).$

Capacitance:

$C=\frac{4\pi ϵ}{\frac{1}{a}-\frac{1}{b}}.$

Taking $b\to \infty$ gives the capacitance of an isolated sphere: $C=4\pi ϵa$.

RC product relation

For any two-conductor geometry with the same dielectric filling, the resistance $R$ of the dielectric (treating $\sigma$ as the leakage conductivity) and the capacitance $C$ satisfy

$RC=\frac{ϵ}{\sigma }.$

This holds independent of geometry. Proof sketch: both $R$ and $C$ depend on the same $\mathbf{E}$ pattern via integrals that share the geometry, so their product collapses to a material-only constant. Lets you predict one from the other (find $C$ from electrostatics, then $R=ϵ/(\sigma C)$ for the leakage path).

The ratio $ϵ/\sigma$ is the dielectric relaxation time, the time scale on which free charges placed inside a conductor relax to the surface. For copper, this is sub-femtosecond. For mica, ~15 hours.

General formula

For arbitrary geometry, capacitance can be computed from the field:

$C=\frac{{\oint }_Sϵ\mathbf{E}\cdot d\mathbf{s}}{-{\int }_l\mathbf{E}\cdot d\mathbf{l}}\text{(F)},$

where $S$ is any surface enclosing the $+Q$ conductor and $l$ is any path from conductor 2 to conductor 1.

Energy stored

Energy stored in a capacitor charged to voltage $V$ with charge $Q$:

$W_e=\frac{1}{2}C{V}^2=\frac{1}{2}\frac{Q^2}{C}=\frac{1}{2}QV.$

Derivation: bring charge in increments $dq$ at instantaneous voltage $v=q/C$, total work $={\int }_0^Q(q/C)dq=Q^2/(2C)$.

This energy lives in the electric field inside the dielectric, with energy density

$w_e=\frac{1}{2}ϵE^2{\text{(J/m}}^3\text{)}.$