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Put the positive plate at $z=d/2$ and the negative plate at $z=-d/2$. Reflection through $z=0$, combined with reversal of the two plate charges, makes the tangential fields cancel at every point of the mid-plane. Thus $E_x=E_y=0$ at the corner as well as inside the capacitor.

For the uniform-charge estimate at a capacitor corner, assemble four identical square pairs around their common corner. Their normal fields there are equal by rotation symmetry and add to the central field $\sigma /{ϵ}_0$ of the larger plate pair. With the bulk parallel-plate relation $\sigma \simeq {ϵ}_{0}V/d$, the conventional corner estimate isThis quarter-field result uses prescribed uniform surface charge, not symmetry alone for finite equipotential conductors. Real conducting plates redistribute charge near their edges; $d\ll a$ justifies the bulk relation away from the edges but does not remove the local corner fringe. For that exact conductor problem only the zero tangential components follow from the stated reflection symmetry; its normal coefficient requires a fringing-field boundary-value solution. The standard parallel-plate formula explicitly neglects edge effects.^[ref]

In vacuum, displacement current contributes the additional term ${ϵ}_0\partial \mathbf{E}/\partial t$:Taking the divergence and using Gauss law gives $\nabla \cdot \mathbf{J}+\partial \rho /\partial t=0$, the local conservation of charge. Without displacement current the curl equation would incorrectly require a divergence-free current even during charge accumulation.

For a charging capacitor, a surface cutting the wire carries conduction current while a surface passing through the gap carries changing electric flux. The displacement current term makes the circulation independent of the spanning surface. It is not transport of free charge across the vacuum gap. In matter the macroscopic equation is $\nabla \times \mathbf{H}={\mathbf{J}}_f+\partial \mathbf{D}/\partial t$, with electric polarization included in $\mathbf{D}$. Together with Faraday's law of induction, the vacuum equations give electromagnetic waves of speed $1/\sqrt{{\mu }_0{ϵ}_0}$.

In electrostatics, the Maxwell equations in matter areHere ${\rho }_f$ is free charge density and $\mathbf{P}$ is polarization density. For a homogeneous isotropic linear dielectric, the constitutive equation is $\mathbf{D}={ϵ}_0{ϵ}_r\mathbf{E}$. Integrating the first equation over a thin pillbox straddling an interface, with unit normal $\mathbf{n}$ from medium 1 to medium 2, gives $\mathbf{n}\cdot ({\mathbf{D}}_2-{\mathbf{D}}_1)={\sigma }_f$. Integrating the second around a vanishingly narrow rectangular loop gives $\mathbf{n}\times ({\mathbf{E}}_2-{\mathbf{E}}_1)=0$. These are the electrostatic interface boundary conditions. When there is no free surface electric charge, the normal electric displacement field and tangential electric field are continuous; the normal electric field generally is not.

Neglect end fringing field of a capacitor. In either axial segment of the coaxial cylindrical capacitor, a cylindrical Gaussian surface gives $D_r={\lambda }_f/(2\pi r)$, where ${\lambda }_f$ is free electric charge per length. Integrating the radial electric field between the conductors givesThe liquid and air segments share the same conductor electric potential difference, so their capacitances add in parallel. Thus the partially filled coaxial capacitor hasHere the question's $ϵ$ is relative permittivity, and air's relative permittivity is approximated by one.

Write $x$ for the rise above the external level. The wide reservoir keeps that level approximately fixed. At constant applied voltage, one must include work done by the source: the appropriate electrical contribution to the mechanical potential is $-C{V}^2/2$. The upward dielectric insertion force at fixed voltage is consequentlyThe excess hydrostatic pressure $\rho gx$ acts over the annular area $A=\pi (b^2-a^2)$. Alternatively its gravitational potential energy is $\rho gA{x}^2/2$, since liquid initially at the reservoir level must be raised through all heights from zero to $x$. Balancing the derivatives givesThis assumes $h+x<L$, a communicating liquid reservoir, negligible capillarity, and a sufficiently long capacitor that end effects are small. Differentiating gives the dielectric-rise voltmeter's sensitivity at a bias $V$:To obtain its numerical value in $\mu \mathrm{m}{\mathrm{V}}^{-1}$, multiply the right-hand side in SI units by $10^6$. A convenient parametric scale, with $\beta =b/a$, isFor a concrete estimate, choose $a=1\mathrm{mm}$, $b=2\mathrm{mm}$, mass density $\rho =10^3\mathrm{kg}{\mathrm{m}}^{-3}$, relative permittivity $ϵ=2$ and operating voltage $V=100\mathrm{V}$. These are illustrative choices, not supplied apparatus data. They giveThus this millimetre-scale example requires fine optical displacement measurement for volt-level changes. The parametric formula states exactly how the estimate changes with the chosen geometry, liquid and bias.

The displacement is quadratic in voltage, cannot distinguish its sign, and has zero first-order sensitivity at zero bias. Smaller gaps and larger relative permittivity increase sensitivity, but capillarity, leakage, dielectric breakdown, temperature-dependent liquid properties, and finite available rise then limit performance. A nonconducting liquid avoids a sustained conduction load, although charging still draws electric charge and leakage is possible. A high voltage can give a readable displacement; a low-voltage instrument would require fine displacement measurement and careful calibration.

A normal mode is a motion in which all coordinates have a fixed relative pattern and oscillate at one frequency. Near a stable equilibrium, let the quadratic kinetic and potential energies be $T={\dot{\mathbf{q}}}^{\mathsf{T}}\mathbf{M}\dot{\mathbf{q}}/2$ and $V={\mathbf{q}}^{\mathsf{T}}\mathbf{Kq}/2$. With positive-definite mass matrix $\mathbf{M}$, the equations and harmonic ansatz giveThe stiffness and mass matrices are symmetric. For distinct frequencies, their symmetry gives $({\omega }_s^2-{\omega }_r^2){\mathbf{a}}_r^{\mathsf{T}}\mathbf{M}{\mathbf{a}}_s=0$, so the mode vectors are orthogonal in the mass inner product. A basis can also be chosen in a degenerate eigenspace. Expanding $\mathbf{q}={\sum }_s{Q}_s{\mathbf{a}}_s$ decouples the equations into ${\ddot{Q}}_s+{\omega }_s^2{Q}_s=0$; initial displacements and velocities determine the coefficients. Positive stiffness gives real frequencies, a zero eigenvalue can describe a free symmetry motion, and a negative one signals instability rather than an oscillatory mode.^[ref]

For two equal masses between fixed walls with three equal springs, the stiffness matrix is $k\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$ and the mass matrix is $m1$. The patterns $(1,1)$ and $(1,-1)$ have frequencies $\sqrt{k/m}$ and $\sqrt{3k/m}$. An initial displacement of one mass excites both modes and can produce beats.

For a uniform taut string fixed at both ends, the wave equation ${\partial }_t^{2}y=c_s^2{\partial }_x^{2}y$, with $c_s=\sqrt{\mathcal{T}/{\mu }_{\ell }}$, has patterns and angular frequenciesThe boundary conditions select the spectrum. Fourier expansion of arbitrary admissible initial data provides the mode amplitudes, illustrating the extension from a finite set of coordinates to a field.

For the normal modes of inductively coupled LC circuits, take two identical circuits with self-inductance $L$, capacitance $C$ and mutual inductance $\mathcal{M}$, where $|\mathcal{M}|<L$. The capacitor charges obey $L{\ddot{q}}_1+\mathcal{M}{\ddot{q}}_2+q_1/C=0$ and the exchanged equation. In-phase and opposite-phase charge patterns haveThe positive inductance matrix for two identical coils is the physical counterpart of positive mass. Here inductance plays the role of the mass matrix. Mechanical, wave and electrical examples therefore share the same eigenvalue structure. Weak forcing excites modes according to its spatial overlap with each pattern; damping broadens the resonances, while strong nonlinearity can invalidate a fixed linear-mode decomposition.

Write the plate area as $S$ and separation as $d$, so the original capacitance is $C={\epsilon }_{0}S/d$. Neglect edge fields. The greatest capacitance occurs when each material occupies half the plate area and spans the full separation. These two regions are capacitors in parallel, sharing the same electric potential difference:The least capacitance occurs when a dielectric layer of thickness $d/2$ covers the whole area, leaving a vacuum layer of the same thickness. The two layers are capacitors in series: the normal electric displacement field is continuous and the voltage drops add. Their individual capacitances are $2ϵC$ and $2C$, givingThe arithmetic-harmonic bounds for dielectric capacitance show that these are genuine extrema over other arrangements of the same two volumes, within the ideal parallel-plate model. A uniform trial electric field in the fixed-voltage energy principle bounds $C$ above by the arithmetic mean of the two permittivities; a uniform trial electric displacement field in the fixed-charge principle bounds it below by their harmonic mean. The two layer arrangements attain those bounds. Both reduce to $C$ when $ϵ=1$.

Maxwell's displacement current is the electric-field contribution needed to make Ampère's law consistent with charge conservation. In macroscopic form,The displacement current density ${\mathbf{J}}_d={\partial }_t\mathbf{D}$ has units $\mathrm{A}{\mathrm{m}}^{-2}$. Since $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$, it contains the vacuum field term and, in matter, the polarization-current contribution. It need not represent free charge moving through a material.

Taking the divergence gives $0=\nabla \cdot {\mathbf{J}}_f+{\partial }_t(\nabla \cdot \mathbf{D})$. Gauss's law, $\nabla \cdot \mathbf{D}={\rho }_f$, then yields the continuity equation ${\partial }_t{\rho }_f+\nabla \cdot {\mathbf{J}}_f=0$. Without the displacement term, Ampère's original law would incorrectly require every current distribution to be divergence-free even while charge accumulates.

A charging parallel-plate capacitor makes this concrete. Ignoring fringing, $D=Q/A$, so the displacement current across its dielectric gap is $I_d=A{\partial }_{t}D=dQ/dt$, equal to the conduction electric current in its wire. The integral lawtherefore gives the same answer whether the spanning surface cuts the wire or passes through the gap. Together with Faraday's law, this field term also permits self-consistent electromagnetic waves in vacuum, with speed $c=1/\sqrt{{\mu }_0{ϵ}_0}$.

Integral conservation. Electric charge cannot be created by ordinary transport. For a fixed volume $V$, outward current reduces the enclosed charge:Here ${\rho }_e$ has units $\mathrm{C}{\mathrm{m}}^{-3}$ and $\mathbf{J}$ has units $\mathrm{A}{\mathrm{m}}^{-2}$. Applying the divergence theorem to an arbitrary volume gives the local continuity equationIt is local charge conservation: a net outflow must be accompanied by depletion. In a steady state, $\nabla \cdot \mathbf{J}=0$. Integrating over a small circuit junction then gives Kirchhoff's current law. During a transient the junction can accumulate charge and the incoming and outgoing conduction currents need not yet match.

Consistency with Maxwell's equations. In vacuum, take the divergence of the Ampère-Maxwell equation:Gauss's law, $\nabla \cdot \mathbf{E}={\rho }_e/{ϵ}_0$, reproduces the continuity equation. The displacement current term makes these equations consistent for time-dependent charge distributions, such as a charging capacitor. It is not a stream of free charges across the insulating gap. In matter, $\nabla \cdot \mathbf{D}={\rho }_f$ and $\nabla \times \mathbf{H}={\mathbf{J}}_f+{\partial }_t\mathbf{D}$ give the analogous conservation law for free charge; total charge also includes bound charge.

From molecular dipoles to macroscopic fields. A dielectric contains charges which remain bound to its atoms or molecules. An applied electric field displaces positive and negative charge centres and can orient permanent electric dipoles. The electric polarization $\mathbf{P}$ is electric dipole moment per unit volume, in $\mathrm{C}{\mathrm{m}}^{-2}$. Its associated bound charge isat the boundary of the polarized material, where $\hat{\mathbf{n}}$ points outward. A uniform electric polarization has no volume bound charge, but generally has surface bound charge. The electric field remains the force per unit test charge; it includes the fields of both free and bound charges.

Introducing the electric displacement field $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$ packages the free-charge source separately:For a homogeneous linear isotropic dielectric,The electric susceptibility ${\chi }_e$ is dimensionless; the permittivity $ϵ$ has units $\mathrm{F}{\mathrm{m}}^{-1}$. This is a constitutive approximation. An anisotropic dielectric requires a permittivity tensor, while a nonlinear dielectric need not have electric polarization proportional to electric field. The field in these equations is the macroscopic field, not necessarily the local field at an individual molecule.

Interfaces. Let $\hat{\mathbf{n}}$ point from medium 1 into medium 2. Integrating Gauss's law over a thin pillbox, and the electrostatic curl equation around a thin loop, gives the electrostatic interface boundary conditionsThus normal electric displacement is continuous when there is no free surface charge, whereas tangential electric field is continuous even if the permittivity changes. Normal electric field generally jumps because the interface carries bound charge. The electric potential is continuous for an ordinary interface without an ideal dipole sheet. These conditions, together with Laplace's equation in each homogeneous charge-free region, determine the field.

Energy and electrical sources. Reversibly assembling free charge requires incremental work $\delta W=\int \mathbf{E}\cdot \delta \mathbf{D}dV$. For a linear dielectric whose response can be reversibly integrated from zero field, the electrostatic energy isThis includes the material polarization response, rather than simply using ${ϵ}_0{E}^2/2$ everywhere. In a capacitor, $W=Q^2/(2C)=C{V}^2/2$. Inserting a dielectric increases capacitance. At fixed charge the stored energy decreases. At fixed voltage the stored field energy increases, but the voltage source supplies $VdQ$; the appropriate mechanical energy including that source changes by $-V^{2}dC/2$. Both cases predict attraction of dielectric into the capacitor. A force calculation must state which electrical quantity is fixed. A lossy or hysteretic dielectric also dissipates energy, so the reversible quadratic expression does not describe its full history.

Shape and depolarization. Suppose the dielectric is surrounded by vacuum and the distant imposed electric field is ${\mathbf{E}}_0$. Bound surface charge produces a depolarization field. For a uniformly polarized ellipsoid, along a principal axis,Here $N_d$ is the depolarization factor of a dielectric ellipsoid. This self-consistent field explains why the same material responds differently in different shapes.

An infinite slab with electric field normal to its faces has $N_d=1$, giving $E_{\mathrm{in}}=E_0/{ϵ}_r$; its opposite surface bound charges reduce the internal field. For electric field parallel to the faces, $N_d=0$, giving $E_{\mathrm{in}}=E_0$. End effects are neglected.

A very long circular rod in an axial electric field has negligible depolarization away from its ends, so $E_{\mathrm{in}}\simeq E_0$. In a transverse electric field the infinite-cylinder solution has $N_d=1/2$, giving $E_{\mathrm{in}}=2E_0/({ϵ}_r+1)$. To see the latter directly, take the applied field along $x$. The interior and exterior electric potentials can be written ${\varphi }_{\mathrm{in}}=-Br\cos \theta$ and ${\varphi }_{\mathrm{out}}=-E_{0}r\cos \theta +A\cos \theta /r$. Continuity of potential and normal electric displacement at $r=a$ gives $B=2E_0/({ϵ}_r+1)$.

For a sphere of radius $a$, $N_d=1/3$ andFor example, write ${\varphi }_{\mathrm{in}}=-Br\cos \theta$ and ${\varphi }_{\mathrm{out}}=-E_{0}r\cos \theta +C\cos \theta /r^2$. Continuity of potential gives $B=E_0-C/a^3$, and continuity of normal electric displacement gives $ϵB={ϵ}_0(E_0+2C/a^3)$. Solving yields $B=3E_0/({ϵ}_r+2)$ and $C=a^3{E}_0({ϵ}_r-1)/({ϵ}_r+2)$. The $1/r^2$ potential is that of the electric dipole moment above. For ${ϵ}_r=1$ all these induced effects vanish; as ${ϵ}_r$ becomes very large the normal-field sphere and slab approach the corresponding conductor field exclusion. The ideal infinitely long rod in an exactly axial imposed field has a different, singular end-free limit.

A dielectric responds to an electric field mainly by displacing or orienting bound charges, rather than by sustaining a steady conduction current. The polarization density $\mathbf{P}$ is electric dipole moment per unit volume. It creates bound chargeSeparating these charges from free charge motivatesFor an isotropic linear dielectric, $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$, soThe electric susceptibility ${\chi }_e$ describes material polarization and is not the magnetic susceptibility. More generally it is a tensor and can depend on frequency, field strength and temperature.

Electronic polarization displaces an electron cloud relative to its nucleus; ionic polarization displaces positive and negative ions; orientational polarization partially aligns permanent molecular dipoles against thermal disorder. For independent permanent moments of magnitude $p$, weak-field orientational response gives ${\chi }_e\simeq n{p}^2/(3{ϵ}_0{k}_{B}T)$ when local-field corrections and interactions are negligible. This explains a common inverse-temperature contribution without asserting that every dielectric follows that law.

At a static interface, $\mathbf{n}\cdot ({\mathbf{D}}_2-{\mathbf{D}}_1)={\sigma }_f$ and the tangential electric field is continuous. Bound surface charge is already included through $\mathbf{P}$. For a homogeneous filled capacitor, $C=ϵA/d$, and a reversible linear dielectric stores energy density $u=\mathbf{E}\cdot \mathbf{D}/2$. At fixed voltage, insertion raises the stored field energy but the voltage source also supplies energy; force arguments must include that source. At fixed charge the capacitor energy instead decreases on insertion.

At frequencies too high for a polarization mechanism to follow the field, its contribution weakens. A complex permittivity represents dissipative dielectric loss as well as phase lag; the sign of its imaginary part depends on the time convention. Strong fields can cause nonlinear response or dielectric breakdown, beyond the reversible linear model.

For the series-fed parallel LC voltage response, use phasors with time dependence $e^{i\omega t}$. The parallel branch has impedance $iX$, where $X=\omega L/(1-{\omega }^{2}LC)$. ThereforeAt the ideal parallel resonance ${\omega }_0=1/\sqrt{LC}$ its input impedance diverges. The input-current phase changes sign on crossing this pole; expressions there are interpreted by limits.

The capacitor voltage equals the parallel-branch voltage. Define $q=R(1-{\omega }^{2}LC)/(\omega L)$. ThenAt low frequency $H\sim i\omega L/R$; at high frequency $H\sim -i/(\omega RC)$; at resonance $H=1$. Thus it is a band-pass voltage filter or resonant frequency selector. Its phase decreases continuously from $+\pi /2$ through zero to $-\pi /2$. Half-power points satisfy $|q|=1$ and

For the supplied component values and $\omega =2\pi (50){\mathrm{s}}^{-1}$, $X=421.745\Omega$ and $V_{C,\mathrm{pk}}=99.3046\mathrm{V}$. The element peak currents areInductor and capacitor currents are opposite phasors; their magnitudes subtract here rather than add. The resistor carries the input current, withThese results assume ideal components and sinusoidal steady state.

In sinusoidal steady state, complex impedance is the voltage phasor divided by the current phasor. Its modulus is the amplitude ratio and its argument is the voltage lead. With the $e^{i\omega t}$ convention, $I=dq/dt=i\omega CV$, so $Z_C=1/(i\omega C)$.

The PDF labels $V_1$ across $R_1$, not across the complete driven network. If $I$ is the current through $R_1$, then $V_1=I{R}_1$ and $I=Y_p{V}_2$, whereThe resistor-referenced parallel-resonator response isWriting $b(\omega )=\omega C-1/(\omega L)$,The maximum occurs at ${\omega }_0=1/\sqrt{LC}$, where $|H|=R/R_1$ and the phase is zero. Below resonance the phase is positive, tending to $+\pi /2$; above it the phase is negative, tending to $-\pi /2$. The amplitude tends to zero at both extremes: the inductor shorts the parallel network at low frequency, and the capacitor does so at high frequency. At resonance their reactive currents cancel, leaving only the resistor current.

If instead $V_{\mathrm{in}}$ denotes the voltage across the complete circuit, its different transfer ratio is $V_2/V_{\mathrm{in}}=1/[1+R_1/R+i{R}_{1}b]$, with peak $R/(R+R_1)$. This distinction follows directly from the printed voltage endpoints.

In vacuum the electric and magnetic field energy densities areFor example, charging a capacitor requires $U={\int }_0^Q(q/C)dq=Q^2/(2C)=C{V}^2/2$. A parallel-plate capacitor with $C={ϵ}_{0}A/d$ and $E=V/d$ then gives $U={ϵ}_0{E}^{2}Ad/2$, identifying the electric field energy density. Similarly an inductor stores $L{I}^2/2$; applying this to a long solenoid yields $B^2/(2{\mu }_0)$ times its interior volume.

Electromagnetic energy is transported by the Poynting vector $\mathbf{S}=\mathbf{E}\times \mathbf{B}/{\mu }_0$. Poynting's theorem,expresses local energy conservation and transfer to matter. In a vacuum plane wave $E=cB$, so the two energy densities are equal and the energy flux has magnitude $c(u_E+u_B)$. For nondispersive linear media the corresponding reversible expressions are $\mathbf{E}\cdot \mathbf{D}/2$ and $\mathbf{B}\cdot \mathbf{H}/2$; they must not be applied unchanged to arbitrary dispersive or hysteretic materials.

For sinusoidal steady state with convention $e^{i\omega t}$, complex impedance is the voltage phasor divided by the current phasor. Its modulus is the amplitude ratio and its argument is the voltage-current phase difference. Since $V_L=LdI/dt$,The original circuit is a series RLC circuit with output across the inductor. Its transfer function isFor the original capacitance, the outputs are $10.15\mathrm{mV}$ at $9\mathrm{MHz}$ and $4.65\mathrm{mV}$ at $10\mathrm{MHz}$, a ratio $0.458$. The current resonance is $f_0=1/(2\pi \sqrt{LC})=8.967\mathrm{MHz}$, so the circuit favors the first transmitter.

The resistor voltage is in phase with the current; the inductor voltage leads it by $\pi /2$ and the capacitor voltage lags by $\pi /2$. Large reactive voltages therefore nearly cancel in their phasor sum. A large inductor voltage is consistent with a small supply voltage: it is inductor voltage magnification, not power gain in a passive circuit.

Reducing the capacitance raises $f_0$ to $9.452\mathrm{MHz}$, between the two transmitters. Direct evaluation gives $7.02\mathrm{mV}$ and $7.22\mathrm{mV}$, respectively. Their similar detunings give approximately equal amplitudes; they are not exactly equal, partly because the numerator of $H$ also depends on frequency.

The two-tone output of a linear resonant circuit, for coherent monochromatic sources of equal input amplitude and arbitrary fixed input phases, isWith $A\simeq B=V_o$, this is approximatelyThus a $9.5\mathrm{MHz}$ rapid oscillation has a beat envelope repeating in magnitude at $1\mathrm{MHz}$. The signed envelope cosine has frequency $0.5\mathrm{MHz}$. Unequal amplitudes leave a small nonzero envelope minimum; independently drifting transmitter phases make the beating drift. No new frequency is generated by this ideal linear circuit.

Use $e^{i\omega t}$ phasors. The relation $V=L\dot{I}$ gives $Z_L=i\omega L$. The three branches form a parallel RLC circuit and share the source voltage. The source-current phasor and its real time dependence areConsequently its amplitude is $V_0\sqrt{R^{-2}+(\omega C-1/(\omega L){)}^2}$ and its phase relative to the voltage is ${\tan }^{-1}[R(\omega C-1/(\omega L))]$. At low frequency it is inductive and lags toward $-\pi /2$; at high frequency it is capacitive and leads toward $+\pi /2$. The ideal amplitude diverges at either extreme and has a finite minimum between them; real component losses and source electrical impedance limit those asymptotes.

Zero phase requires cancellation of electrical susceptances:The inductor and capacitor currents are nonzero, equal in magnitude and opposite in phase, while the resistor supplies the in-phase source current. energy oscillates between the reactive elements; dissipation occurs in the shunt resistor. A source-current maximum would describe a different, series-resonant circuit.

The specified ratio isIt exceeds $1/\sqrt{2}$ when $|\omega C-1/(\omega L)|<1/R$. Define $b=1/(RC)$ and ${\omega }_0=1/\sqrt{LC}$. The two positive endpoints are ${\omega }_{\pm }=[\sqrt{b^2+4{\omega }_0^2}\pm b]/2$, soThe endpoints have geometric mean ${\omega }_0$. Increasing the parallel resistance reduces dissipative loading and narrows this band, in contrast with increasing the resistance of a series RLC circuit. Equivalently, $Q={\omega }_0{U}_{\max }/P_{\mathrm{loss}}$, with $U_{\max }=C{V}_0^2/2$ and $P_{\mathrm{loss}}=V_0^2/(2R)$. The parallel RLC bandwidth here belongs to the explicitly defined current ratio, not a half-maximum rule applied to a source-current peak that does not exist.

Use an RC low-pass filter: put the resistor in series with the input and the capacitor between the output node and the return conductor. Measure the output voltage across the capacitor, with negligible output loading. The voltage divider then has transfer functionAt the high frequency, the attenuation requirement isAt the low frequency, approximately unchanged amplitude means $2\pi f_{l}RC\ll 1$, soThe upper relation is an asymptotic design condition, because “approximately” specifies no numerical tolerance. For example, retaining at least $99\%$ of the low-frequency amplitude would sharpen the upper bound to approximately $68.7\mu \mathrm{F}$.

A practical illustrative choice is $C=22\mu \mathrm{F}$. Its voltage gain is approximately $0.99896$ at $1\mathrm{Hz}$ and $0.001096$ at $20\mathrm{kHz}$: the latter amplitude is reduced by a factor of about $912$. Its cutoff frequency is $1/(2\pi RC)\simeq 21.9\mathrm{Hz}$.

For an ideal air-gap parallel-plate capacitor, neglecting fringing, the electric field between oppositely charged plates is $E=\sigma /{ϵ}_0=Q/({ϵ}_{0}A)$. Dielectric breakdown therefore limits the magnitude of the electric charge on either plate toThe two plates carry $+Q$ and $-Q$; the quoted stored electric charge is the magnitude on one plate, rather than the net electric charge of the whole capacitor. The gap does not enter this ideal bound: $C={ϵ}_{0}A/d$ and $V_{\max }=E_{\mathrm{bd}}d$ make $C{V}_{\max }$ independent of $d$. Edge enhancement of the electric field can lower the practical limit.

Put ${\omega }_0^2=1/(LC)$. The initial conditions giveThe printed electric current arrow points toward the positively labelled capacitor plate, so its signed electric current is $I=\dot{q}$, not the positive discharge magnitude. HenceFor the stated electrical resistance,The voltage starts at $6\mathrm{V}$ with zero slope; the signed electric current starts at zero and initially becomes negative. Its initial slope is $\dot{I}(0)=-V_0/L=-6\mathrm{A}{\mathrm{s}}^{-1}$. Both oscillate with an $e^{-25t}$ envelope and a $40\mathrm{ms}$ amplitude time constant. The phase difference is close to a quarter period, with the small damping correction displayed explicitly in $V_C$.

The capacitor stores electrostatic energy, the inductor stores magnetic energy, and the resistor converts energy into heat. Their instantaneous values areThe sign convention for electric current does not change these expressions. With the initial voltage, $E_C(0)=C{V}_0^2/2=90\mu \mathrm{J}$ and $E_I(0)=P_R(0)=0$.

The two stored energies follow directly by squaring the expressions for electric charge and electric current:Both are nonnegative, with alternating maxima. Initially all stored energy is in the capacitor; at a zero of $q$ it is entirely in the inductor. The total isThe total energy decreases monotonically, with a horizontal tangent whenever the electric current is zero. Its envelope decays on the $1/(2\gamma )=20\mathrm{ms}$ scale, but the exact total is not a featureless exponential: the resistor's instantaneous loss varies within each oscillation. The figure in part (a) shows this gradual modulation of the downward slope as well as the component energies.

Complex impedance. In sinusoidal steady state write physical voltage and electric current as real parts of $\tilde{V}{e}^{i\omega t}$ and $\tilde{I}{e}^{i\omega t}$. The electrical impedance $Z=\tilde{V}/\tilde{I}$ is a ratio of phasors; its magnitude gives the voltage-to-current amplitude ratio and its argument gives their phase difference. It is measured in ohms.

For an ideal capacitor, $q=Cv$ and $i=dq/dt=Cdv/dt$. Consequently $\tilde{I}=i\omega C\tilde{V}$, givingWith this time convention the voltage lags the electric current by $\pi /2$.

Bridge output. The original diagram has $R_1$ and $C_1$ in parallel in the upper-left arm, and $R_2$ and $C_2$ in series in the lower-left arm. The point labelled $C$ is connected to the upper supply node. Set the lower supply node to zero, so $V_C=V_i$. Assuming the measured output draws negligible electric current, the two voltage dividers giveTheir difference is thereforeThe reactive-arm electrical impedances are

Balance. Setting $V_{AB}=0$ and cross-multiplying gives $R_3{Z}_2=R_4{Z}_1$. For this Wien bridge,The right-hand side must be the real number $R_4/R_3$. Equating imaginary and real parts givesPositive capacitances require $R_4/R_3>R_2/R_1$. In the equal-component case $R_1=R_2=R$ and $C_1=C_2=C$, this reduces to

Band-pass output and phase. Preserve the requested voltage polarity: $V_{AC}=V_A-V_C$. Since $V_C=V_i$,Put $u=\omega RC>0$. The equal-component ratio is $Z_2/Z_1=2+i(u-u^{-1})$, so the Wien bridge band-pass response isAt finite positive frequency, its imaginary part vanishes exactly when $u-u^{-1}=0$. ThusThe magnitude $|H|=[9+(u-u^{-1}{)}^2{]}^{-1/2}$ is maximal there and tends to zero at both frequency extremes, as expected for a band-pass filter.

A convenient continuous, unwrapped wave phase isIt decreases from $-\pi /2$ as $\omega \to 0^{+}$, passes through $-\pi$ at ${\omega }_r$, and tends to $-3\pi /2$ as $\omega \to \infty$. A principal-value plot instead jumps from $-\pi$ to $+\pi$ at resonance; both represent the same phasor. Reversing the output to $V_{CA}$ would shift the phase by $\pi$ and make the resonant ratio $+1/3$, but that is a different output polarity from the one printed.