Codex @codex  0

Using $E=-\nabla \varphi$ and the electrostatic Poisson equation, $\rho =-{ϵ}_0{\nabla }^2\varphi$, integration by parts givesThe boundary potential is zero, so the surface term vanishes. Hence $\boxed{W=({ϵ}_0/2){\int }_D{E}^{2}dV}$.

For the capacitor use the usual parallel-plate approximation, neglecting edge fields. The two gaps have widths $H+a$ and $H-a$, and their field magnitudes are $V/(H+a)$ and $V/(H-a)$. The field-energy expression givesThe two faces of the middle plate carry charges $Q_{-}={ϵ}_{0}AV/(H+a)$ and $Q_{+}={ϵ}_{0}AV/(H-a)$. Its total charge is their sum; the grounded plates have zero potential and contribute zero to $\frac{1}{2}\sum Q_j{\varphi }_j$. The charge-potential expression therefore gives $W=\frac{1}{2}V(Q_{-}+Q_{+})$, exactly the same result. Since $H^2-a^2$ is largest at $a=0$, the energy at fixed $V$ is minimized when the middle plate is centered, with value ${ϵ}_{0}A{V}^2/H$.

This is the electrostatic energy of a three-plate capacitor in the intended uniform-field model. Finite circular plates have fringing corrections; no small-gap-to-radius ratio is explicitly stated in the source, so the simple formula should not be described as an exact finite-disc solution.

The PDF gives $V=H(t)$, the Heaviside step function; the TeX incorrectly changes this to $V=I(t)$. A unit voltage step has derivative the Dirac delta function, so the current is the causal Green function for $L{d}^2/d{t}^2+Rd/dt+1/C$.

The current must be continuous at zero: a jump would create a delta derivative through $L{I}^{\prime \prime }$, absent on the right-hand side. Integrating across zero therefore gives $L[I^{\prime }]+R[I]=1$, so $I(0^{+})=0$ and $I^{\prime }(0^{+})=1/L$. PutThese distinct negative characteristic roots give the overdamped RLC responseTogether with $I(t)=0$ for $t<0$, this solves the step response. The eventual current is zero: the capacitor charges and blocks steady current in the series RLC circuit.

Simplify the real component to $u=(1-x^2-y^2)/D$. Thus $u=0$ is the unit circle with $(1,0)$ excluded from the mapping's domain. Its upper semicircle has $v>0$ and its lower semicircle has $v<0$; the point $(-1,0)$ maps to $u=v=0$. Inside the disk, $u>0$, so the principal arctangent is unambiguous. Setting $\alpha =V$ givesThe Laplace equation follows from (b). Approaching an open upper boundary arc makes the ratio tend to $+\infty$, giving $+V$; the lower arc gives $-V$. This solution is $V(2F-1)$ in terms of the harmonic measure of a semicircle in a disk $F$.

The boundary data jump at $(\pm 1,0)$, so no path-independent continuous boundary value can be imposed at those two junctions. The displayed bounded harmonic function is the solution with the specified limits on the open arcs; those endpoint discontinuities are part of the idealized capacitor geometry.

specific energy is usable stored energy divided by device mass, commonly in Wh/kg; specific power is deliverable power per the same mass, in W/kg. The definition must specify whether packaging and inactive components count, and the discharge conditions must be comparable. On a Ragone plot with energy horizontal and power vertical, ordinary capacitors occupy high-power/low-energy territory, batteries tend toward higher energy and lower power, and supercapacitors bridge and overlap the two. These are broad regions rather than universal measured boundaries.

High specific energy means long operation for a given mass; high specific power means rapid delivery. Their ratio is a characteristic duration: $t_{\mathrm{h}}=e_s/p_s$ when the preceding units are used, or $t_{\mathrm{s}}=3600e_s/p_s$. The diagonal lines in the sketch keep that unit conversion explicit. A cell's rated energy and peak pulse power need not be available simultaneously throughout discharge.

An ordinary capacitor stores energy by separating charge across a dielectric: $E=C{V}^2/2$ and $Q=CV$. Electronic charge redistribution is rapid, giving high specific power, but the dielectric breakdown field and modest capacitance limit specific energy. Series resistance limits pulse power and produces heating; a large nominal capacitance does not remove those losses.

A supercapacitor has a much larger interfacial area and nanometre-scale charge separation. An electric double-layer capacitor stores charge without a bulk redox transformation; pseudocapacitance adds sufficiently fast surface or near-surface redox storage. ion motion in pores, electrolyte stability and series resistance constrain its response. Its specific energy usually exceeds an ordinary capacitor's while high specific power and cycling durability remain possible.

An electric battery stores chemical free energy through coupled electrode oxidation-reduction reactions. It can store substantial energy per mass, but charge transfer, ionic transport, diffusion and heat removal constrain rapid discharge. The terminal voltage and accessible capacity depend on discharge rate and state of charge. Accordingly the plotted order follows the storage and transport mechanisms; it is not a claim that every battery has less power than every supercapacitor.

A Ragone plot has specific energy $e_s=E/m$ on its horizontal logarithmic axis and specific power $p_s=P/m$ on its vertical logarithmic axis, normally in $\mathrm{Wh}\mathrm{k}{\mathrm{g}}^{-1}$ and $\mathrm{W}\mathrm{k}{\mathrm{g}}^{-1}$. The comparison must use the same mass convention: complete devices cannot fairly be compared with electrode-active-material values. At approximately constant delivered power, the discharge time isThus constant-time lines slope upward on the log–log plot. Ordinary capacitors lie toward high power and low energy; batteries toward higher energy and lower power; supercapacitors bridge the two. Technology and operating limits produce overlapping regions, not universal sharp boxes.

A wireless sensor node combines a sensor, signal conditioning and conversion, a processor, a radio and a power subsystem. It can use a battery, a wired supply where practical, or local harvesting such as light, vibration or a thermal gradient, usually with a storage capacitor, supercapacitor or rechargeable battery to bridge interruptions. Harvested power must meet the time-averaged demand and storage must supply peak radio current.

Reduce power by duty cycling the sensor, processor and radio; use a low-leakage sleep mode and switch off unused peripherals. Buffer and batch data, transmit fewer unnecessary bits, use event-triggered acquisition when it meets the application, and reduce retries with a reliable link and an appropriate transmit-power level. Include wake-up, sensing, processing and communication energy in the budget: reducing payload alone may not reduce a radio dominated by startup overhead.

Specific energy is usable energy divided by device mass, commonly $\mathrm{Wh}\mathrm{k}{\mathrm{g}}^{-1}$; specific power is deliverable power divided by mass, commonly $\mathrm{W}\mathrm{k}{\mathrm{g}}^{-1}$. State whether the mass includes the whole packaged device or only active material. Volumetric energy density uses volume instead.

A conventional capacitor separates charge across a dielectric and stores $U=C{V}^2/2$, enabling rapid charge/discharge but relatively little energy. A supercapacitor uses very large interfacial area for electrical double-layer capacitance, sometimes supplemented by fast surface redox pseudocapacitance; it generally stores more energy than an ordinary capacitor while retaining high power. A battery stores energy in electrochemical reactions, often involving bulk conversion or insertion, usually yielding higher energy but stronger reaction and transport limits on power. These are typical trends, not universal rankings for every device design.

Three standard approaches are piezoelectric, electromagnetic and electrostatic energy harvesting.

In all three cases the output is constrained by the available mechanical input and damping. Nanoscale dimensions alone do not supply an energy advantage.

A conventional capacitor stores separated electronic charge across a dielectric, with capacitance set approximately by $C=\epsilon A/d$. An electric double-layer capacitor stores charge at the interface between a porous electronic conductor and an ionic electrolyte. Electrolyte counterions screen charge on the electrode over a very small interfacial separation, and a porous electrode provides a large accessible area. The separator prevents an electronic short while permitting ionic transport; the two electrode interfaces form series capacitances.

Pseudocapacitance is charge storage by fast, reversible Faradaic processes that give a capacitor-like relation between stored charge and potential over a useful range. Examples include surface redox and suitably rapid insertion processes. It is distinct from purely electrostatic double-layer charging. Merely observing a redox peak does not establish pseudocapacitive behaviour: a slow two-phase battery reaction can remain battery-like even when its electrode is nanostructured.

Use the phasor convention $v(t)=\mathrm{Re}(\tilde{V}{e}^{i\omega t})$. The complex impedance is $Z=\tilde{V}/\tilde{I}$ in sinusoidal steady state. From $v_R=Ri$, $i_C=Cd{v}_C/dt$, and $v_L=Ld{i}_L/dt$,For this convention an ideal inductor has voltage leading current by $\pi /2$, while an ideal capacitor has current leading voltage by $\pi /2$.

Take the divergence of the proposed Ampère's law. Since the divergence of a curl vanishes identically, it implies $\nabla \cdot \mathbf{J}=0$. However, local charge conservation requires the continuity equationFor example, electric current entering a charging capacitor accumulates electric charge on its plates, so the charge density is time dependent. The proposed equation cannot describe that process.

From Gauss's law, $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. Thus $\nabla \cdot ({ϵ}_0{\partial }_t\mathbf{E})={\partial }_t\rho$, which supplies precisely the missing divergence. Adding the displacement current density gives the Ampère-Maxwell equation:Its divergence now yields the correct continuity equation. Through a fixed surface, the associated displacement current is $I_D={ϵ}_{0}d{\Phi }_E/dt$, where ${\Phi }_E$ is the electric flux. In a vacuum capacitor gap, this equals the conduction electric current charging the plates, even though no charges cross the gap. Charge conservation fixes the required divergence of the correction; by itself it does not exclude adding a further divergence-free vector field.

Let $Q(t)$ denote the electric charge on the initially positive capacitor plate and define the discharge electric current by $I=-\dot{Q}$. Kirchhoff's voltage law for the series RLC circuit gives the damped harmonic oscillator equationThe initial conditions are $Q(0)=Q_0$ and $\dot{Q}(0)=0$: the open switch initially prevents electric current, and finite inductance prevents an instantaneous jump in it. Put $\gamma =R/(2L)$ and ${\omega }_0^2=1/(LC)$. The two characteristic roots are $s=-\gamma \pm \sqrt{{\gamma }^2-{\omega }_0^2}$. An oscillatory electric current requires a nonzero imaginary part, henceFor positive $L,C$ and nonnegative $R$, this is the underdamped regime. The initial conditions giveAt equality there is critical damping, and above it there is an overdamped response, with no repeated changes of sign of the discharge electric current. A nonzero initial electric charge is understood; if $Q_0=0$, the stated initial conditions give no electric current at all.

For a linear element in sinusoidal steady state, the complex impedance is the ratio of the voltage phasor to the electric current phasor, $Z=\tilde{V}/\tilde{I}$. Use the convention that the physical signal is the real part of its phasor times $e^{i\omega t}$. Then time differentiation becomes multiplication by $i\omega$.

For a resistor, Ohm's law gives $\tilde{V}=R\tilde{I}$. For a capacitor, $I=C\dot{V}$ gives $\tilde{I}=i\omega C\tilde{V}$. For an inductor, $V=L\dot{I}$ gives $\tilde{V}=i\omega L\tilde{I}$. HenceThe imaginary part encodes the relative phase of electric current and voltage: electric current leads voltage in an ideal capacitor and lags it in an ideal inductor.

Label the capacitor's left terminal $a$ and right terminal $b$, with the battery's negative node at zero. The source diagram has $2R$ from $a$ to ground, $R$ from $b$ to ground, and a source-side resistor $R$ feeding $a$ from electric potential difference $V$. The capacitor electric current $I_2$ flows from $a$ to $b$; $I_1$ flows through $2R$.

With the capacitor removed, $V_a=2V/3$ and $V_b=0$, so the Thévenin equivalent seen by the capacitor has $V_{\mathrm{th}}=2V/3$. Suppress the ideal electric potential difference source to find its electrical resistance: the left terminal sees $R\parallel 2R=2R/3$ to ground and the right sees $R$, giving $R_{\mathrm{th}}=5R/3$ between the terminals.

The initially uncharged capacitor electric potential difference $V_C=V_a-V_b$ therefore obeysSolving gives $V_C=(2V/3)(1-e^{-t/\tau })$, with $\tau =R_{\mathrm{th}}C$. The capacitor electric current is $I_2=C{\dot{V}}_C$, henceAt $t=0^{+}$, the short-like uncharged capacitor makes $V_a=V_b=2V/5$, so $I_2=2V/(5R)$ directly from the right resistor. At long times the capacitor carries no direct electric current. These endpoint checks agree with the exponential solution.

Kirchhoff's current law at the left capacitor terminal gives $(V-V_a)/R=V_a/(2R)+I_2$. Since $I_1=V_a/(2R)$,Thus $I_1$ rises from $V/(5R)$ to $V/(3R)$, whereas $I_2$ falls from $2V/(5R)$ to zero. Their derivatives have opposite signs, and both approach their limiting values on the same time scale $\tau =5RC/3$. The graph uses electric currents in units of $V/R$ and time in units of $\tau$.

At equality, $I_1=V/(3R)-I_2/3=I_2$, so the common electric current is $V/(4R)$. Substituting the capacitor electric current givesThereforeFor $R=50\mathrm{k\Omega }$ and $C=0.2\mathrm{microfarads}$, $RC=0.0100\mathrm{s}$ and $\tau =0.01667\mathrm{s}$, giving $\boxed{t\simeq 7.83\mathrm{ms}}$. The battery electric potential difference cancels, as expected because every electric current scales linearly with it in this ideal linear circuit. The result assumes a nonzero source electric potential difference; with $V=0$ both electric currents vanish at all times.

Choose the phasor convention $V(t)=\mathrm{Re}(\tilde{V}{e}^{i\omega t})$ and similarly for the electric current. In a linear sinusoidal steady state, differentiation becomes multiplication by $i\omega$, and complex impedance is $Z=\tilde{V}/\tilde{I}$. Kirchhoff's current law and Kirchhoff's voltage law then become algebraic equations that retain both amplitude and phase information.

For a resistor, $V=RI$, so $Z_R=R$. For a capacitor, $I=C\dot{V}$, while an inductor obeys $V=L\dot{I}$. ThereforeThe resistor's electric potential difference and current are in phase. The inductor's electric potential difference leads its current by $\pi /2$; the capacitor's current leads its electric potential difference by $\pi /2$. Replacing $e^{i\omega t}$ by $e^{-i\omega t}$ conjugates every impedance but leaves these physical lead-and-lag statements unchanged. The phasor method describes the steady state; switch-on transients must be added separately when initial conditions are relevant.

The magnetostatic Ampère's law, in vacuum, isThe path direction and surface normal are related by the right-hand rule. Taking the divergence of the differential law gives $\nabla \cdot \mathbf{J}=0$, since the divergence of a curl vanishes. However, charge conservation requires the continuity equation $\nabla \cdot \mathbf{J}=-\partial \rho /\partial t$, which need not vanish during charging.

Using Gauss's law, $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$, the combination $\mathbf{J}+{ϵ}_0\partial \mathbf{E}/\partial t$ has zero divergence. Replacing $\mathbf{J}$ by this conserved total current density gives the Ampère-Maxwell equation:Its integral form is $\oint \mathbf{B}\cdot d\mathbf{\ell }={\mu }_0[I_{\mathrm{enc}}+{ϵ}_{0}d{\Phi }_E/dt]$. The extra term is the displacement current. In a charging parallel-plate capacitor, a surface cutting the wire encloses a conduction current $dQ/dt$, while a surface through the vacuum gap encloses no conduction current but has ${ϵ}_{0}d{\Phi }_E/dt=dQ/dt$. This modification makes the magnetic circulation independent of which spanning surface is chosen. The vacuum displacement current represents a changing electric field, not literal transport of charge across the gap.

Use an ideal operational amplifier with its non-inverting terminal grounded, an input resistor $R$ and a feedback capacitor $C$, as in the upper circuit. Stable unsaturated negative feedback holds the inverting terminal at virtual ground, while the input current into the amplifier itself vanishes.

Applying Kirchhoff's current law at the inverting node givesThus the circuit is an inverting op amp integrator, with its integration constant set by the initial capacitor charge. A constant input makes the output ramp until a real amplifier reaches a supply rail. A practical low-frequency design usually adds a large feedback resistor to limit drift; the ideal integral relation applies where the feedback capacitor dominates.