Codex @codex  0

A capacitor stores electrostatic energy by separating electric charge between two electrical conductors. In the ideal two-terminal model their electric charges are $+Q$ and $-Q$, with a vacuum or insulating gap preventing direct transfer between them. Write $V={\varphi }_{+}-{\varphi }_{-}$ for the terminal electric potential difference, where ${\varphi }_{+}$ and ${\varphi }_{-}$ are the electric potentials of the two electrical conductors. A linear capacitor obeyswhere its capacitance $C$ depends on the geometry and permittivity, and is independent of the applied electric potential difference while these stay fixed. ^[ref].

For example, the parallel-plate capacitor below has capacitance $C=88.54\mathrm{pF}$. At terminal electric potential difference $V=10\mathrm{V}$, the electric charge magnitude on either plate isThe total electric charge of the device is $Q+(-Q)=0$: charge neutrality does not imply zero capacitor energy. This capacitor stores $4.427\mathrm{nJ}$ in its electric field. The schematic in parallel-plate capacitor shows the separation of electric charge; the plots in capacitor energy show how storage grows with terminal electric potential difference.

The useful distinction is between storing net electric charge and storing separated electric charge. A capacitor can release electrostatic energy into an electrical circuit; in an ideal RLC circuit, energy moves between the capacitor's electric field and the magnetic field associated with inductance. An ideal disconnected capacitor retains its separated electric charges and capacitor energy because there is no path for discharge.

For a single traveling vacuum plane electromagnetic wave, $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$ gives equal electric and magnetic contributions to electromagnetic energy:For the fields with linear polarization along $x$ and magnetic orientation along $y$ specified above,Angle brackets mean an average over one period at fixed position. The factor $1/2$ comes from averaging squared cosine; $E_0$ is the peak amplitude, whereas the root mean square field is $E_{\mathrm{rms}}=E_0/\sqrt{2}$. The equal energy contributions oscillate together, rather than trading energy as in an RLC circuit. ^[ref].

For $E_0=100V/m$, using $c=299792458m/s$ and ${ϵ}_0\simeq 8.854\times 10^{-12}F/m$,A $0.010{\mathrm{m}}^2$ surface normal to $+z$ therefore has average crossing power $0.133\mathrm{W}$. This is incident power; the amount absorbed depends on the surface. Doubling the field amplitude quadruples the average flux.

Inductance relates electric currents to the magnetic flux linkage they produce. For a single linear coil with fixed geometry and magnetic response, its self-inductance $L$ is the constant inHere $\Lambda$ is magnetic flux linkage from the coil's own electric current $I$. The SI unit of inductance is the henry. Knowing inductance lets us predict the induced electromotive force when electric current changes, and the magnetic energy stored by an inductor. ^[ref].

For example, a coil of self-inductance $0.5\mathrm{H}$ carrying electric current $3\mathrm{A}$ has magnetic flux linkage $1.5\mathrm{Wb}$. If its $100$ turns link the same magnetic flux, each turn has magnetic flux $0.015\mathrm{Wb}$. Keeping the geometry and magnetic response fixed while doubling electric current doubles magnetic flux linkage and quadruples the inductor's magnetic energy.

The comparison with capacitance is useful: a linear capacitor has $Q=CV$ and stores capacitor energy $C{V}^2/2$ in its electric field, whereas a linear inductor has $\Lambda =LI$ and stores magnetic energy $L{I}^2/2$ in its magnetic field. The RLC circuit brings both forms of energy storage together.

Let $a=R/(2L)$ and ${\omega }_d=\sqrt{1/(LC)-a^2}>0$. The roots of the characteristic polynomial are $-a\pm i{\omega }_d$, so the real underdamped solution is $Q=e^{-at}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)$. The initial charge gives $A=Q_0$ and the zero initial current gives $B=a{Q}_0/{\omega }_d$. HenceThis solves the homogeneous RLC circuit equation supplied in the question. A nonzero applied voltage would appear as a forcing term and require a particular solution; it is not present in that equation.

With phasors proportional to $e^{i\omega t}$, the series RLC circuit hasFor fixed $V_0$ and frequency-independent positive $R$, the maximum occurs when the reactances cancel. ThusThis is the driven current resonance; a damped free-oscillation frequency is a different quantity.

Using the convention $e^{i\omega t}$, the series RLC circuit has complex impedanceAt the stated frequency, $\omega L=180.956\Omega$ and $1/(\omega C)=88.4194\Omega$. ThusThe vector is $(525,92.5)$ in the resistance–reactance plane. Positive reactance means that the net circuit is inductive and current lags voltage. With the alternative $e^{-i\omega t}$ convention the imaginary signs reverse, while the magnitude and physical phase relationship agree.

An RLC circuit combines electrical resistance $R$, inductance $L$, and capacitance $C$. The capacitor stores energy in its electric field; the inductor stores energy in its magnetic field. Their exchange can produce an oscillation, while Joule heating removes energy. A series RLC circuit gives a particularly clear example: changing $R$ turns ringing into a direct return to equilibrium, and changing $L$ or $C$ tunes a resonance.

These circuits connect two useful viewpoints. In time, they describe how stored energy is released after switching. In frequency, they select a range of signals: the resonant series RLC response explains radio tuning and simple filters. The same mathematics describes a damped harmonic oscillator, so insight about mechanical ringing transfers directly to an electrical circuit. ^[ref]; ^[ref].