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An electric machine uses electromagnetism to transfer energy between electrical and mechanical forms, or between electrical circuits. An electric motor converts electrical input into mechanical output; an electrical generator converts mechanical input into electrical output. The wider machine category also includes the stationary electrical transformer. A rotating electric machine combines force on current-carrying conductors with electromagnetic induction; the same rotating conductors that develop torque can develop an opposing induced electric potential difference.^[ref]

For an original idealized direct current motor example, let $e=k_e\omega$, $\tau =k_{t}I$, and $V=e+IR$, where $e$ is the induced opposing electric potential difference, $\omega$ the angular velocity, $\tau$ the torque, and $R$ the winding electrical resistance. In coherent SI units, assume $k_e=k_t=0.1$, $\omega =100\mathrm{rad}{\mathrm{s}}^{-1}$, $I=2\mathrm{A}$, and $R=1\Omega$. Thenand the electromechanical conversion power balance isThe $20/24$ conversion fraction is about $83.3\%$ before friction and other losses. At rest the induced term vanishes, so applying the same $12\mathrm{V}$ gives $12\mathrm{A}$ in this simplified model. This makes the opposing induced electric potential difference experimentally visible: at fixed applied electric potential difference, a shaft-speed measurement and an electric current measurement distinguish a turning motor from its stalled state. Actual starting control and additional losses change the idealized numbers, not the need to balance electrical input, mechanical output, and heating.

An electric motor converts electrical energy into mechanical work, usually rotation, through electromagnetic forces. For a coil with $N$ turns, area $A$ and electric current $I$ in a uniform magnetic field $B$, the torque magnitude iswhere $\theta$ is the angle between the coil's normal and the magnetic field. Repeated rotation requires the current or effective field arrangement to keep supplying torque in the intended direction. Electrical input power must cover useful output plus losses.

A coil with $N=100$, $I=0.5\mathrm{A}$, $A=0.01{\mathrm{m}}^2$ and $B=0.2\mathrm{T}$ has maximum torque $0.1\mathrm{N}\mathrm{m}$. At angular speed $\omega$, instantaneous mechanical power is $P=\tau \omega$; the maximum-torque value alone does not specify a motor's sustained output.

Michael Faraday's 1821 rotation experiment made the conversion visible: a current-carrying wire moved around a fixed magnet while a pool of Mercury maintained electrical contact. The Royal Institution preserves his 1822 apparatus, consisting of a wire, glass vessel and fixed bar magnet connected through the conducting liquid. Continuous rotation, rather than a single deflection, established a route from steady electric current to repeated mechanical motion.^[ref]

In electrostatics, the electric potential $\Phi$ tells us the interaction potential energy per unit test electric charge in the electric field of prescribed sources. Its most directly useful quantity is an electric potential difference: how much work per unit electric charge an external agent must supply to move that test electric charge slowly between two points. This replaces the bookkeeping of a vector force along a path with the difference of two scalar values. The test electric charge must be small enough not to rearrange the sources appreciably.

For motion from $A$ to $B$, write $\Delta \Phi =\Phi (B)-\Phi (A)$. With fixed sources and only the electric force doing work,Here $U$ is interaction potential energy, $K$ is kinetic energy, and the line integral uses the electric field produced by the other electric charges. For slow transport with unchanged kinetic energy, the external work is $W_{\mathrm{ext}}=\Delta U$, so $\Delta \Phi =W_{\mathrm{ext}}/q$ for nonzero $q$. This external work has the opposite sign to the work done by the electric field. The SI unit is the volt,A volt measures energy per unit electric charge, not energy itself. For example, crossing an electric potential difference of $+5\mathrm{V}$ changes the potential energy of $q=+2\mathrm{nC}$ by $+10\mathrm{nJ}$, but changes that of $q=-2\mathrm{nC}$ by $-10\mathrm{nJ}$. Both test electric charges encounter the same electric potential difference; their opposite signs reverse the energy change.^[ref]

The electrostatic potential is a scalar function of position. The electric field is a conservative vector field, and all three components follow from the negative gradient of the electric potential:The electric field points toward decreasing electric potential. A positive test electric charge feels electric force in that direction; a negative test electric charge feels electric force in the opposite direction, since $\mathbf{F}=q\mathbf{E}$. A freely accelerated negative test electric charge therefore moves toward higher electric potential while lowering its potential energy $U=q\Phi$.^[ref]

Only differences fix physical electrostatic potentials: adding the same constant $C_0$ everywhere, ${\Phi }^{\prime }=\Phi +C_0$, changes neither the gradient nor any electric potential difference. Choosing an electrical reference potential is this elementary gauge transformation. Setting zero at infinity is convenient for finitely many localized point charges; it is not available for an ideal uniform electric field extending to infinity. There we can choose a finite reference point. Assigning zero at ground is a convention, whereas physically grounding an electrical conductor also allows exchange of electric charge to impose a boundary value. Zero electric potential at a point never by itself implies zero electric field.^[ref]

For a fixed point charge $Q$ at the origin in vacuum, choose zero at infinity. Coulomb's law giveswhere ${\epsilon }_0$ is the vacuum permittivity. Use the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ and $Q=+2\mathrm{nC}$. At $r=0.20\mathrm{m}$,A test electric charge $q=-3\mathrm{nC}$ has $U=-2.697\times 10^{-7}\mathrm{J}$ and radial electric force $F_r=-1.3485\times 10^{-6}\mathrm{N}$, directed inward. Moving it inward from $r_A=0.20\mathrm{m}$ to $r_B=0.10\mathrm{m}$ raises the electric potential from $89.9$ to $179.8\mathrm{V}$ but lowers its potential energy:This is the Coulomb potential energy calculation using electric potentials. The source's singular self-field at $r=0$ is excluded; when finding the force on a source point charge, use only the electric field of the other sources.^[ref]

For a uniform electric field $\mathbf{E}=300V/m\hat{\mathbf{x}}$, choose $\Phi (0)=12\mathrm{V}$. ThenMoving from $x_A=0$ to $x_B=0.04\mathrm{m}$ gives $\Delta \Phi =-12\mathrm{V}$. For $q=+2\mathrm{nC}$ the change in potential energy is $-24\mathrm{nJ}$ and the electric field does $+24\mathrm{nJ}$ of work. For $q=-2\mathrm{nC}$, the same prescribed displacement gives $\Delta U=+24\mathrm{nJ}$: the external agent must supply that work for slow transport. This uniform electric field approximates the interior of a parallel-plate capacitor sufficiently far from the edges; the fringing field of a capacitor limits the approximation.^[ref]

An equipotential is a level set $\Phi =\text{constant}$. On a smooth equipotential with nonzero gradient, every tangent displacement has $d\Phi =\nabla \Phi \cdot d\mathbf{l}=0$, so the electric field is normal to the equipotential. Thus moving a test electric charge along an equipotential requires no net electrical work. Around a point charge, the equipotentials are spheres; the circles in the diagram are their sections, not an assumption of two-dimensional Coulomb's law. A uniform electric field has parallel plane equipotentials. For contours separated by the same small electric potential difference, smaller normal separation means a stronger electric field, approximately $|\mathbf{E}|\approx |\Delta \Phi |/\Delta n$. At a point where $\nabla \Phi =0$, the electric field is zero and has no direction to call normal.^[ref]

The superposition principle adds electric potentials as signed scalars. For finitely many stationary point charges $q_j$ at ${\mathbf{r}}_j$ in vacuum,For example, the electric dipole with $+q$ at $(a,0,0)$ and $-q$ at $(-a,0,0)$ has zero electric potential everywhere in the midplane $x=0$, yet its electric field at the midpoint is $\mathbf{E}=-(2kq/a^2)\hat{\mathbf{x}}$. With $q=2\mathrm{nC}$ and $a=0.10\mathrm{m}$, this is $-3596N/C\hat{\mathbf{x}}$. At $(2a,0,0)$, the same model gives $\Phi =2kq/(3a)\approx 119.867\mathrm{V}$ and $E_x=8kq/(9a^2)\approx 1598.222N/C$. Cancellation of electric potentials is quite different from cancellation of vector electric fields.

A motivating measurement is conductive-paper electric potential mapping. A documented teaching experiment uses slightly conductive paper on an insulating board, two circular electrodes in highly conducting ink, a steady supply, and a digital voltmeter. Connect the supply to the two electrodes, keep the voltmeter common lead on the negative electrode, and use its other lead as a movable probe. Adjust the supply using the measured electric potential difference between electrodes; the published procedure uses $10\mathrm{V}$. The observable is the voltmeter reading $\Phi (P)-\Phi (\mathrm{COM})$ at the probe position $P$. Locate several points reading $2\mathrm{V}$, record their positions on a separate map, and join them into an equipotential; repeat at other readings. Lift the probe between contacts so it does not damage the paper. This procedure makes the electrical reference potential concrete: every reading compares two contacts.^[ref]

From neighboring readings infer an electric field component by a finite difference, $E_x\approx -[\Phi (x+h)-\Phi (x)]/h$. Reconstruct two in-plane components with displacements along two independent directions. Alternatively, draw the electric field normal to mapped equipotentials, toward lower electric potential. The second diagram illustrates the same measurement principle with bar electrodes spanning a uniform sheet, insulating top and bottom edges, and a synthetic model $\Phi (x)=(50V/m)x$. The model readings $4.0\mathrm{V}$ at $x=0.08\mathrm{m}$ and $5.0\mathrm{V}$ at $x=0.10\mathrm{m}$ imply $E_x=-50V/m$. These values are calculated illustrations, not reported experimental data. The documented circular-electrode experiment similarly estimates the electric field from readings at successive centimetre marks.^[ref]

Useful controls include checking the voltmeter zero, checking that the electric potential difference along each painted electrode is negligible, and repeating reference readings to detect supply drift. Reversing the supply should reverse the signs relative to the same reference contact. Contact resistance, leakage electric currents, and touching the sheet can distort the map. A voltmeter has finite input electrical resistance and draws a small electric current; voltmeter loading changes the quantity being measured if that electrical resistance is insufficiently large relative to the sheet's effective source electrical resistance.^[ref][ref]

The paper carries a steady electric current, so this is a planar analogue rather than a direct measurement of isolated point charges in three-dimensional vacuum. For constant electrical conductivity, Ohm's law and steady charge conservation give $\mathbf{j}=-\sigma \nabla \Phi$ and $\nabla \cdot \mathbf{j}=0$, hence the two-dimensional Laplace equation away from electrodes. Insulating sheet edges impose zero normal electric current, while finite electrodes fix boundary values. This derivation explains both the analogy and its limits: inhomogeneous paper, finite boundaries, and finite electrodes can change the contours. One should not fit the three-dimensional point charge formula $kQ/r$ to a planar map without testing that model. Small probe spacing resolves spatial variation better, but amplifies uncertainty in the inferred finite difference. The published laboratory instructions explicitly ask students to assess whether their proposed field fit is appropriate.^[ref][ref]

The scalar description has a crucial limit. Faraday's law of induction for a stationary closed path $\Gamma$ sayswhere ${\Phi }_B$ is magnetic flux. Nonzero changing magnetic flux gives nonzero circulation, incompatible with a single-valued electric potential alone. More generally, a scalar electric potential and a magnetic vector potential together giveConsequently, in the presence of induction, two voltmeter lead routes between the same contacts can produce different readings if the enclosed magnetic flux changes. This follows by subtracting the two path line integrals; it is a deduction from Faraday's law of induction, not a claimed outcome of the paper experiment. The steady setup and negligible induction justify interpreting its readings as electric potential differences. Merely introducing a scalar electric potential that depends on time cannot supply a nonzero curl of the electric field.^[ref]

A short mathematical justification connects the statements above. Along a smooth path, the chain rule gives $d\Phi =\nabla \Phi \cdot d\mathbf{l}$. Integrating and using $\mathbf{E}=-\nabla \Phi$ yields the endpoint formula by the fundamental theorem for line integrals, proving path independence. Conversely, vanishing circulation around every closed path lets us define $\Phi (P)=-{\int }_{P_0}^P\mathbf{E}\cdot d\mathbf{l}$ independently of the route, with $\Phi (P_0)=0$. Zero curl is a local test; for a continuously differentiable curl-free vector field, a simply connected open region suffices for a global potential. The explicit finite-source formula already gives a global electric potential away from its source points.^[ref]

An electric potential difference compares electric potentials at two points. For a move from $a$ to $b$, define $\Delta \Phi =\Phi (b)-\Phi (a)$; the terminal convention $V_{AB}=\Phi (A)-\Phi (B)$ instead means the potential at $A$ relative to $B$, so the order must always be stated. In electrostatics, the line integral of the electric field givesHere $\Delta U$ is the change in interaction potential energy of the test electric charge $q$ with fixed sources, while $W_{\mathrm{field}}$ is work done by the electric field. For an external agent moving the charge slowly without a change in kinetic energy, its work is $\Delta U$. A positive electric charge gains kinetic energy when accelerated toward lower electric potential; a negative electric charge gains kinetic energy when accelerated toward higher electric potential. Both motions lower $q\Phi$.^[ref]

For example, fix $Q=+2\mathrm{nC}$ at the origin and compare distances $r_a=0.20\mathrm{m}$ and $r_b=0.10\mathrm{m}$. With the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ and zero at infinity,Moving $q=-3\mathrm{nC}$ inward therefore givesIf only the electric force does work, conservation of energy makes the gain in kinetic energy $2.697\times 10^{-7}\mathrm{J}$. Direct integration confirms the signs: ${\int }_{r_a}^{r_b}kQdr/r^2=kQ(1/r_a-1/r_b)=-89.9\mathrm{V}$, which multiplied by the negative test electric charge gives positive work. The electric potential difference is unchanged by choosing a different electrical reference potential.

In an electrical circuit, an electric potential difference is the voltage across a component's terminals. In an electrical network, a vertex voltage is specified relative to a reference vertex; its differences across edges drive current flow. For a uniform electric field, the integral reduces to $\Delta \Phi =-\mathbf{E}\cdot ({\mathbf{r}}_b-{\mathbf{r}}_a)$. A field $\mathbf{E}=300V/m\hat{\mathbf{x}}$ across a displacement $0.04\mathrm{m}\hat{\mathbf{x}}$ therefore gives $\Delta \Phi =-12\mathrm{V}$, as in the interior approximation for a parallel-plate capacitor. The volt, measured in joules per coulomb, expresses energy transfer per unit electric charge, which makes it useful for comparing components without specifying the amount of charge transported.

The instantaneous rate of energy transfer through a circuit port is $P(t)=V(t)I(t)$, with positive electric current entering the positive voltage terminal for the passive sign convention. Peak phasors in sinusoidal steady state give $⟨P⟩=\frac{1}{2}\mathrm{Re}(\tilde{V}{\tilde{I}}^{\ast })$. The factor $1/2$ is absent for root mean square phasors.

Electric power balance during a demand step tracks where the additional energy comes from when an electrical load suddenly requests more power. At an explicitly chosen system boundary, conservation of energy giveswhere $E_s$ includes the energy storage inside that boundary. A negative right-hand side means that stored energy is decreasing. An external electric power system, an electric battery and a rotating electrical generator supply different physical buffers; they should not all be counted as independent incoming supplies if some are already included in $E_s$.

For an original lumped rotating-machine example, take moment of inertia $J=10\mathrm{kg}{\mathrm{m}}^2$ and initial angular velocity ${\omega }_0=100\mathrm{rad}{\mathrm{s}}^{-1}$. Suppose mechanical input initially remains unchanged while electrical output increases by $2\mathrm{kW}$, with losses already accounted for. The rotational kinetic energy then obeysAfter $0.5\mathrm{s}$, $\omega \approx 98.995\mathrm{rad}{\mathrm{s}}^{-1}$: the extra $1\mathrm{kJ}$ came from rotation. This stipulated model assumes the extra delivered power can be maintained during that interval; it does not predict the load's own response to falling electric potential difference or frequency.

For a rotating electrical generator with $p$ magnetic pole pairs, the generated electrical frequency is $f=p\omega /(2\pi )$. A speed change therefore links the energy balance to a measurable waveform change. The DOE derivation relates generated frequency to pole count and rotational speed.^[ref] This rotating-machine argument does not describe every electronically controlled supply.

A proposed demand-step test records the imposed electrical load, source and storage power, terminal electric potential difference, frequency and timing of control action. The important distinction is between an initial buffer and a sustainable new operating point. If available continuous power remains too small, increasing stored energy alone does not remove that limit; electrical load shedding can instead reduce the requested output.

Electric power distribution is the part of an electric power system that delivers electrical energy from supply points to connected electrical loads. It uses conductors, switching and protective equipment, and often electrical transformers to provide suitable local electric potential differences. Electric power transmission mainly transfers bulk power between supply regions and major receiving points; the distinction concerns network function rather than one universal voltage threshold.

A one-line diagram suppresses individual phase conductors to show the supply-to-load structure. The DOE handbook's apparatus example connects utility and standby supplies, buses, switching and protection to the consuming circuits.^[ref] Following the actual connection points is necessary before assigning losses or assuming that two loads share a return path.

For an original two-wire direct current example, a source holds $V_s=240\mathrm{V}$ and supplies a resistive electrical load $R_L=12\Omega$ through an outgoing-plus-return conductor electrical resistance $R_w=0.4\Omega$. Ohm's law givesThe conductor heating is $I^2{R}_w\approx 149.84\mathrm{W}$, while the load receives $I^2{R}_L\approx 4.4953\mathrm{kW}$. Their sum equals the source output $V_{s}I\approx 4.6452\mathrm{kW}$. Assigning the full source electric potential difference to the load would miss both its reduced terminal value and the location of the dissipated energy.

A proposed bench demonstration uses a current-limited low-voltage supply, a known series resistor representing the wires and a separate load resistor. Simultaneously measuring source and load electric potential differences and branch electric current tests the loss calculation. Real alternating current networks add electrical reactance, phase imbalance and protection constraints; the simple example isolates conductor heating rather than representing every distribution effect.

An electric power system is an interconnected arrangement that produces, transfers, controls, and consumes electrical energy. Electricity generation supplies electrical power, electric power transmission carries it between regions, and electric power distribution delivers it to electrical loads. Electrical generators, electrical transformers, and transmission lines perform different parts of this task. The one-line apparatus diagram in the DOE handbook makes these connections concrete: it follows supplies through switching and protection equipment to consuming circuits.^[ref]

A useful calculation separates delivered power from conductor heating. Consider an idealized two-conductor direct current link delivering $P=10\mathrm{MW}$ at load electric potential difference $V=10\mathrm{kV}$, with total outgoing-plus-return electrical resistance $R=0.1\Omega$. Its electric current and resistive loss areDelivering the same power at $100\mathrm{kV}$ reduces the electric current to $100\mathrm{A}$ and the loss to $1\mathrm{kW}$. This is a fixed-load, fixed-resistance comparison; it omits converter loss, insulation constraints, and changes in conductor temperature. In an alternating current network, an electrical transformer changes the operating electric potential difference without itself generating the delivered energy.^[ref]

At each instant, generation, imports, and storage discharge must supply electrical load, exports, losses, and storage charging. A sudden imbalance can initially draw on stored energy; it cannot persist indefinitely without changing the operating state. This electric power balance during a demand step connects network calculations to control and to the smaller, deliberately bounded microgrid.

Electroluminescence is nonthermal light emission driven by electrical excitation. In a light-emitting diode, injected electrons and holes can undergo radiative electron-hole recombination, converting part of the supplied electrical energy into photons.

The emitted optical spectrum depends on the available transitions rather than solely on temperature, unlike an ideal blackbody. Measuring both the electric current and emitted radiant power tests conversion efficiency. Increasing electric current need not increase useful emission proportionally: nonradiative recombination, heating and extraction losses can all matter.

Electrolysis uses an external source of electrical work to drive redox chemical reactions at the electrodes of an electrolytic cell. Its characteristic application is forcing a chemical reaction in a direction that is nonspontaneous under the specified conditions: a discharging galvanic cell supplies electrical work, whereas electrolysis consumes electrical work. This makes electrolysis useful for producing otherwise difficult-to-obtain chemical substances, charging a rechargeable battery, and depositing a metal coating on a conducting object.^[ref]

Always identify the anode by oxidation and the cathode by reduction. In ordinary driven two-terminal operation, the source withdraws electrons from the positive anode and delivers electrons to the negative cathode. In a discharging galvanic cell, the anode is negative and the cathode positive. Thus terminal signs change with operating mode, while the reaction definitions remain valid. Through the electrolyte, ions transport electric charge; they need not all be consumed at an electrode. The electrolyte and electrolytic cell articles explain this transport and its conventional current direction.

For a net chemical equation transferring $z$ electrons per reaction event, the molar reaction Gibbs free energy obeyswhere $F$ is the Faraday constant and $E_{\mathrm{chem}}$ is the reversible cell electromotive force for the reaction in the written direction. A nonspontaneous direction has ${\Delta }_{r}G>0$ and $E_{\mathrm{chem}}<0$ at fixed temperature and pressure. The ideal opposing voltage magnitude is ${\Delta }_{r}G/(zF)$; sustaining a finite electric current generally needs additional drive because of electrical resistance and finite electrode reaction rates. This thermodynamic voltage concerns the actual composition, rather than necessarily standard conditions.^[ref] Not every use of electrolysis increases bulk chemical Gibbs free energy: transferring copper between equivalent surfaces can have zero bulk ${\Delta }_{r}G$ while the source still supplies dissipated electrical work. See the existing electrolytic cell example.

A particularly revealing experiment measures copper transfer instead of merely looking for a coating. Two separated copper electrodes stand in aqueous $\mathrm{CuS}{\mathrm{O}}_4$ electrolyte in a vessel, with insulated leads to a regulated direct-current power supply, a series ammeter and a switch. A balance reads each electrode's mass before and after a timed run. The Institute of Physics describes copper cells with ammeters, adjustable current and removable cathodes; its comparison uses three similar cells at $1\mathrm{A}$ for ten minutes, $0.5\mathrm{A}$ for ten minutes, and $0.5\mathrm{A}$ for twenty minutes. The specified observation to test is that deposited copper mass is proportional to electric current times time. These are documented experimental settings, not measurements generated here.^[ref]

The ideal half-reactions areEach deposited copper chemical atom requires two electrons, so $z=2$. The anode loses copper and the cathode gains it; ideally their mass changes have equal magnitudes, and the bulk copper cation inventory stays constant. Local concentrations near the electrodes can nevertheless differ. The sulfate anions participate in ionic transport without appearing in these net half-reactions; electronic electric current in the wires becomes ionic electric current in the electrolyte through the two interfaces.^[ref]

Clean, rinse, dry and identify the electrodes before their initial weighing. After interrupting the electric current, rinse away adhering electrolyte, dry again and weigh without rubbing off the deposit. KIT's university practical explicitly calls for distilled-water rinsing, careful drying and weighing both electrodes; it warns that loose copper grains can be lost during handling.^[ref] Useful additional controls are repeated drying and weighing until the reading stabilizes, and a matched unpowered immersion followed by the same rinse/dry cycle. These controls test retained liquid, residual dissolved salts, handling loss and chemical changes unrelated to passed electric charge; they are proposed checks, not additional observations reported by KIT. Keep immersed electrode area, spacing, electrolyte composition and temperature comparable across runs, and keep the electrodes from touching.

Record the electric current as a function of time, together with the exact switch-on and switch-off times. Taking $I(t)\ge 0$ in the chosen deposition direction, the transported electric charge isFor constant electric current, $Q=I(t_1-t_0)$; with sampled readings, use numerical integration rather than assuming that one ammeter reading remained constant. The two equal-charge Institute of Physics settings each give $600\mathrm{C}$, while its shorter $0.5\mathrm{A}$ setting gives $300\mathrm{C}$. With $M_{\mathrm{Cu}}=63.546\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, the mass per amount of substance tabulated by KIT, and $F\approx 96485.33212\mathrm{C}\mathrm{mo}{\mathrm{l}}^{-1}$, ideal Faraday's laws of electrolysis predictThese are calculated predictions for complete conversion of the measured electric charge into retained copper, not experimental results. For a stable process, doubling the electric charge doubles the deposited mass; equal electric charge gives equal deposited mass even when the electric current and duration differ.

The balance reading is a test of the whole model. Competing reduction of water or other chemical species can divert electric current from copper deposition; transient electrode capacitance can also consume part of a short pulse without producing copper. A fragile deposit can detach even when its formation used the expected electric charge. Conversely, retained electrolyte or salts can increase an apparent mass gain. Comparing anode loss with cathode gain helps detect such problems but does not identify every cause. A slope below $M_{\mathrm{Cu}}/(2F)$ alone cannot distinguish competing chemical reactions from lost deposit. Current efficiency measures the fraction of transported electric charge assigned to a specified electrode reaction, rather than the fraction of electrical energy stored chemically.^[ref]

The beautiful connection is that an electrical measurement counts chemical events: integrating electric current and reading a balance provide two ways of following the same electron stoichiometry. This makes electrolysis useful for controlling coating quantity and for testing a charge-to-amount of substance relation. It does not determine coating uniformity, adhesion or the required voltage, which depend on transport and electrode chemical kinetics as well.

A conducting rod sliding on rails separated by $\ell$ in a transverse magnetic field has induced current magnitude $B\ell v/R$ when self-inductance is neglected and resistance is $R$. The Lorentz force opposes motion with magnitude $B^2{\ell }^{2}v/R$, so free motion decays exponentially on the displayed time constant. Joule heating integrates to the rod's lost kinetic energy. Current direction follows the Lenz's law.

A particle calorimeter designed to measure electron and photon energies through particle showers. Track reconstruction helps distinguish charged electrons from neutral photons.

Electromagnetic damping removes mechanical energy through induced electric currents and their associated magnetic forces. Motion changes the linked magnetic flux or moves a conductor through a field; the induced current then couples the motion to resistive heating.

A conducting rod of length $\ell$ sliding on rails through a perpendicular uniform magnetic field $B$ provides a simple model. With total circuit resistance $R$ and negligible inductance, $\mathcal{E}=B\ell v$, $I=B\ell v/R$ and $F=-B^2{\ell }^{2}v/R$. The mechanical power loss is $B^2{\ell }^2{v}^2/R=I^{2}R$, accounting for where the lost energy goes.

The instantaneous-current model requires the electrical relaxation time to be short compared with the motion's variation time. Significant inductance stores magnetic energy and changes the transient response. Replacing the resistor with an ideal lossless circuit therefore changes the energy balance rather than preserving the same damping law.

In vacuum the electromagnetic energy density isand the field energy in a region is $U_{\mathrm{EM}}=\int udV$.

In a linear isotropic nondispersive medium with time-independent permittivity $ϵ$ and magnetic permeability $\mu$, the stored electromagnetic energy per volume is $u=ϵ|\mathbf{E}{|}^2/2+|\mathbf{B}{|}^2/(2\mu )$. For a short derivation, the Maxwell equations in matter and the divergence identity for a cross product give $\nabla \cdot (\mathbf{E}\times \mathbf{H})=-\mathbf{H}\cdot {\partial }_t\mathbf{B}-\mathbf{E}\cdot {\partial }_t\mathbf{D}-{\mathbf{J}}_f\cdot \mathbf{E}$. The constitutive relations turn the first two terms into $-{\partial }_{t}u$, proving the Poynting theorem. Choose $u=0$ when the electric field and magnetic field both vanish. Dispersive or time-dependent material response requires a different accounting of stored energy.

An electromagnetic field carries momentum as well as energy. In vacuum, its SI momentum density iswhere $\mathbf{S}=\mathbf{E}\times \mathbf{B}/{\mu }_0$ is the Poynting vector. Thus the field momentum in a region is ${\mathbf{P}}_{\mathrm{field}}=\int \mathbf{g}dV$. ^[ref]

For a directed vacuum plane wave, $\mathbf{S}=cu\hat{\mathbf{n}}$, so a pulse with total energy $U$ carries momentum $U\hat{\mathbf{n}}/c$. Complete absorption transfers that momentum to matter. Reversing the pulse at a stationary perfect mirror transfers twice as much normal momentum, explaining radiation pressure.

Conservation joins field momentum to the mechanical momentum of matter and momentum crossing the chosen boundary. Changing a field can therefore accelerate matter even when no material particle carries the momentum across the intervening space. Static configurations can also contain nonzero field momentum; the vacuum formula does not require radiation. ^[ref]

The electromagnetic momentum density in vacuum is $\mathbf{g}={ϵ}_0\mathbf{E}\times \mathbf{B}$. It equals the Poynting vector divided by $c^2$, expressing the relativistic relation between energy flux and momentum density.

Electromagnetic radiation is a propagating disturbance of the electric field and magnetic field that carries energy and momentum away from its source. It can be a sustained electromagnetic wave or a finite wave packet: periodic oscillation is not required. In vacuum the disturbance travels at the speed of light. A receiver responds to fields arriving now from an earlier source event, as expressed by retarded time. This is how electrical activity at one place can deliver energy and information elsewhere without a connecting wire.

Stored electromagnetic energy alone does not establish outgoing electromagnetic radiation. A stationary electric charge in electrostatics, or a steady localized electric current in magnetostatics, can store electromagnetic energy with zero Poynting vector. Static combinations can have circulating energy flux without net outward power. Near a changing source, reactive energy can return later. The near and far field distinction separates these contributions from outgoing radiation zone fields.^[ref]

A single traveling vacuum plane electromagnetic wave obeying the Maxwell equations has $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$. Its Poynting vector gives the energy density and flux of a plane electromagnetic wave:Here $I$ is wave intensity; ${ϵ}_0$ and ${\mu }_0$ are vacuum permittivity and vacuum permeability. For a monochromatic wave with linear polarization and peak amplitude $E_0$, squared cosine averages to $1/2$. Pulses require the instantaneous Poynting vector; arbitrary incident-and-reflected superpositions need not obey this relation.^[ref]

The electromagnetic momentum density is $\mathbf{g}=\mathbf{S}/c^2$. Completely absorbing a unidirectional vacuum pulse of energy $U$ transfers momentum $U/c$; see radiation pressure on a perfectly absorbing surface. A calculated $1.0\mathrm{mJ}$ pulse transfers $3.34\times 10^{-12}\mathrm{kg}m/s$. Symmetric emission can lose energy without net recoil because opposite momenta cancel.^[ref]

The direction of emission matters as much as its total power. For an oscillating electric dipole moment along $z$, the angular distribution of electric dipole radiation is strongest across the axis and vanishes along it. Its origin is geometric: the electric dipole radiation field is the transverse projection of the second time derivative of the electric dipole moment, so its magnitude contains $\sin \theta$. Squaring this electric field to obtain wave intensity gives ${\sin }^2\theta$. The electric-dipole approximation for radiation requires a source much smaller than its wavelength and observation in the radiation zone. ^[ref]

For example, the power per solid angle at $\theta =30^{\circ }$ is one quarter of its equatorial value, while the electric field amplitude is one half. To normalize the angular distribution to total power, use spherical coordinates with $d\Omega =\sin \theta d\theta d\varphi$:The integral follows by substituting $x=\cos \theta$, giving ${\int }_{-1}^1(1-x^2)dx=4/3$. This is a probability-like normalization over direction, with units of inverse steradian; it does not make the dipole an isotropic radiator. The same angular integral underlies the Larmor formula for nonrelativistic accelerated electric charge.

Detection requires the incident electromagnetic field to do something observable. An antenna develops an electrical response; a thermopile converts absorbed energy into a thermal and then electrical signal; a photodetector can respond through photon absorption. Hertz spark transmission and detection experiment makes this conversion visible as a tiny spark. Detector orientation, sensitivity, and surrounding reflections all matter: failure to see a signal does not prove that no electromagnetic radiation reached the receiver.

A cascade driven primarily by electron and Positron bremsstrahlung and photon pair production in matter. An electromagnetic calorimeter estimates the incident energy from the deposited cascade. Escaping shower particles and incomplete instrument coverage require containment and response corrections.

Electromagnetism describes how electric charge and electric current interact through the electric field $\mathbf{E}$ and magnetic field $\mathbf{B}$. The Maxwell equations describe the fields and their sources; the Lorentz force describes the response of a particle of electric charge $q$ and velocity $\mathbf{v}$:The electric field can change the particle's energy. The magnetic part of the Lorentz force is perpendicular to velocity, so magnetic forces do no work; they can bend a trajectory. These are local interactions: a particle responds to the electromagnetic field at its position. ^[ref].

The remarkable unification is that the same Maxwell equations connect Coulomb's law, Faraday's law of induction, and freely propagating electromagnetic waves. A changing electric field contributes to magnetic circulation through displacement current; together with Faraday's law, this permits waves in vacuum without conducting matter. The Poynting theorem then explains how the fields store and transport energy. This connects the force on a charge to the energy carried by light.

In his 1865 paper, James Clerk Maxwell derived transverse disturbances and compared the propagation speed inferred from electrical measurements with independently measured light speeds. Part VI, especially §§95–97, argues that their agreement supports identifying light as an electromagnetic wave. This was a comparison with finite experimental accuracy, rather than a numerical identity between the historical measurements. ^[ref]; ^[ref].