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A projectile of mass $m_1$, electric charge $q_1$, and nonrelativistic laboratory kinetic energy $K_{\mathrm{lab}}$ approaches an initially stationary free target of mass $m_2$ and like-sign electric charge $q_2$. The relative-motion kinetic energy is $K_{\ast }=m_2{K}_{\mathrm{lab}}/(m_1+m_2)$. At a head-on turning point it is converted to the repulsive Coulomb potential $k_e{q}_1{q}_2/r_{\min }$, proving the displayed separation. The remaining laboratory kinetic energy is the translation of the centre of mass; it cannot be converted into the pair's separation energy. An externally fixed target is the infinite-target-mass limit.

The Coulomb potential energy is the interaction potential energy of two point charges in vacuum. With separation $r>0$, electric charges $q_1,q_2$, $k=1/(4\pi {\epsilon }_0)$ and zero potential energy at infinite separation,This is mutual interaction energy; it does not include the divergent self-energies of ideal point charges. Unlike electric potential, which has units of volts, $U$ has SI units of joules. With charge 1 fixed and positive radial direction away from it, the conservative force in Coulomb's law givesThus the $1/r$ potential energy and the inverse-square law for force are two descriptions of the same interaction. ^[ref].

For $q_1=+2\mathrm{nC}$, $q_2=-3\mathrm{nC}$ and $r=0.20\mathrm{m}$, using the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$,At $r=0.40\mathrm{m}$, $U=-1.3485\times 10^{-7}\mathrm{J}$. Separating the pair from $0.20$ to $0.40\mathrm{m}$ therefore requires external work $\Delta U=+1.3485\times 10^{-7}\mathrm{J}$ if the initial and final kinetic energies agree; the electric force does $-1.3485\times 10^{-7}\mathrm{J}$ of work. Separating them all the way to infinity requires $+2.697\times 10^{-7}\mathrm{J}$. The negative initial potential energy means that the pair lies below the chosen separated reference; it does not mean a negative kinetic energy.

Like-signed electric charges have positive $U$, which decreases as they separate, so their electric force is repulsive. Opposite-signed electric charges have negative $U$, which decreases as they approach, so their electric force is attractive. This attractive interaction supplies the Coulomb potential used in the hydrogen atom; determining its stationary quantum states also requires quantum mechanics. ^[ref]. Dividing $U$ by a nonzero test electric charge $q_2$ gives the source's electric potential ${\Phi }_1=k{q}_1/r$. More generally, the electrostatic potential energy of point charges sums each unordered charge pair once. This energy description is useful because work between two separations follows from endpoint values, without integrating a complicated trajectory.

Coulomb's law gives the mutual electric force between stationary point charges in vacuum. Put electric charges $q_1$ and $q_2$ at ${\mathbf{r}}_1$ and ${\mathbf{r}}_2$, and define the displacement from charge 1 to charge 2 by $\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1$, with separation $r=|\mathbf{r}|>0$ and unit vector $\hat{\mathbf{r}}=\mathbf{r}/r$. In SI units, the force on charge 2 due to charge 1 isHere ${\epsilon }_0$ is the vacuum permittivity. The magnitude is $k|q_1{q}_2|/r^2$. If $q_1{q}_2>0$, the force on charge 2 points away from charge 1; if $q_1{q}_2<0$, it points toward charge 1. Interchanging the charges gives ${\mathbf{F}}_{1\leftarrow 2}=-{\mathbf{F}}_{2\leftarrow 1}$, consistent with Newton's third law in this electrostatic setting. ^[ref].

For a fully computed example, stipulate $q_1=+2\mathrm{nC}$, $q_2=-3\mathrm{nC}$, ${\mathbf{r}}_1=0$, ${\mathbf{r}}_2=0.20\mathrm{m}\hat{\mathbf{x}}$ and the approximate value $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$. ThenThe force is attractive, of magnitude $1.3485\times 10^{-6}\mathrm{N}$. The source electric field at charge 2 is ${\mathbf{E}}_1({\mathbf{r}}_2)=449.5N/C\hat{\mathbf{x}}$, pointing away from the positive source. Multiplication by the negative $q_2$ reverses the direction to give the attractive force. This field excludes charge 2's own self-field.

Holding the electric charges fixed while changing the separation from $R$ to $2R$ or $3R$ gives the normalized inverse-square law valuesFor the stipulated example, separations $0.20$, $0.40$ and $0.60\mathrm{m}$ give magnitudes $1.3485\times 10^{-6}$, $3.37125\times 10^{-7}$ and $1.498333\dots \times 10^{-7}\mathrm{N}$. The diagram computes these ratios and uses one common arrow-length scale for both sign choices.

The inverse-square law has a geometric explanation through Gauss's law: a source point charge gives fixed total electric flux through a concentric sphere, whose surface area is $4\pi r^2$. Dividing that flux by the area gives a radial electric field proportional to $1/r^2$. This connects a pairwise force calculation to a global conservation-like accounting of field sources. For several stationary point charges, the superposition principle adds the individual electric fields as vectors, and ${\mathbf{F}}_i=q_i{\mathbf{E}}_{\mathrm{other}}({\mathbf{r}}_i)$ uses all sources except charge $i$ itself. The Coulomb potential energy gives an alternative route when the desired quantity is work rather than instantaneous force.

For an ideal two-plate capacitor with opposite charges, a stated inside-minus-outside charge difference is twice the inner plate charge. The voltage is inner plate charge divided by capacitance. Confusing a plate charge with the difference of plate charges creates a factor-two error in inferred membrane voltage.

A critically damped RLC response is the boundary between underdamped RLC response and overdamped RLC response. The free series RLC circuit has a repeated characteristic root $-{\omega }_0$, with general solutionFor initial conditions $q(0)=q_0$, $i(0)=0$,In the plotted normalization, $q(1)=2/e\approx 0.735759$ and $i(1)=-1/e\approx -0.367879$. The electric charge returns without ringing; the polynomial factor $1+{\omega }_{0}t$ matters, so a single exponential function does not give the whole response.

For fixed $L,C$ and these initial conditions, critical damping avoids ringing while giving a faster asymptotic return than any larger $R$: the overdamped RLC response has a root closer to zero. This makes critical damping a useful tuning target for settling after switching. An MIT laboratory exercise makes the distinction observable by asking students to compare all three cases with $L=47\mathrm{mH}$ and $C=47\mathrm{nF}$. Those values give $R_{\mathrm{crit}}=2\sqrt{L/C}=2000\Omega$ and ${\omega }_0\approx 21276.6{\mathrm{s}}^{-1}$. ^[ref].

A crystal X-ray spectrometer selects a narrow wavelength interval by diffraction at known interplanar spacing $d$ and Bragg angle $\theta$. For fixed diffraction order $n$, a known $d$ lets an angular measurement determine $\lambda =2d\sin \theta /n$. This reverses the crystallography problem of finding $d$ from a known wavelength.

Siegbahn's laboratory developed X-ray tubes and evacuated spectrographs for longer wavelengths. Rotating the crystal selected reflections; photographic plates recorded positions. Opposite-side exposures measured angles on both sides of the direct beam. Later tube instruments mounted a beam-defining slit opposite the plate holder. A separate double instrument compared two reflected beams with two ionization chambers. These were different instruments, not one universal assembly.^[ref]

The Nobel biography describes constant-temperature spectrographs and a precision angular scale read by diametrically opposed microscopes.^[ref] Temperature control matters because thermal expansion changes $d$; an accurately read angle cannot compensate for an unknown interplanar spacing.

The schematic separates preparation, selection and readout. An X-ray tube target supplies Characteristic X-rays with a possible bremsstrahlung background. A collimator limits incident directions; evacuation reduces air X-ray attenuation. Rotating the crystal brings different wavelengths into reflection. A photographic plate stores line positions over an exposure, whereas an ionization chamber can measure collected electric charge at successive angles. Neither readout alone establishes absolute wavelength: an angular zero, angular-scale calibration, diffraction order, $d$ and any required refraction correction enter the inference. Narrower slits reduce accepted angular spread but also reduce signal; a weak line can become harder to locate despite a narrower instrumental response.

A control reported in Siegbahn's lecture photographed a line system with copper-sulphate and calcium-sulphate crystals. Irregular relative shifts around copper's characteristic frequency supported anomalous dispersion; the arrangement excluded differing absorption across that frequency as the explanation.^[ref] A finer angular scale alone cannot remove this physical refraction correction.

Marie Curie's thesis describes powdered material on the lower plate of a capacitor, biased by a battery. Radiation made the intervening air conduct. The upper plate connected to an electrometer and a loaded piezoelectric crystal of quartz. Adjusting the load supplied a known compensating electric charge while keeping the electrometer at zero; charge divided by elapsed time determined the electric current. Voltage scans identified a collection plateau. Contact potentials affected weak-field readings; excessively strong fields could instead introduce additional ionization. Fixed geometry and plateau operation made comparisons meaningful.^[ref]

The null method turns a weak signal into a balance against a known quantity. In the diagram's chosen constant-current model, $I=10\mathrm{pA}$ for $\Delta t=30\mathrm{s}$ givesSupplying $-300\mathrm{pC}$ over the same interval leaves zero net electric charge at the sensing node. The electrometer detects imbalance; the compensating-charge calibration supplies the numerical measurement.

This electric current measures collected charge produced by ionization. Converting it to becquerels additionally requires the energy deposition and charge-collection response per transformation. The apparatus can compare preparations without already possessing that complete source-to-signal calibration.

The current density $\mathbf{J}(\mathbf{r},t)$ describes the amount and direction of local electric charge transport per area per time. Its SI unit is $A/{\mathrm{m}}^2$. Through a small area element with unit normal $\mathbf{n}$, the contribution to electric current is $dI=\mathbf{J}\cdot \mathbf{n}dA$: the dot product selects transport through the surface. A current density tangent to the surface contributes zero. If current density is uniform and perpendicular to a wire's transverse slice, its magnitude $j=|\mathbf{J}|$ satisfiesHere $A$ is the wire cross-sectional area. For nonuniform current density, $I/A$ is only the average normal component, and the surface integral is needed. ^[ref].

Consider two separate uniform-flow cases, each carrying electric current $I=2\mathrm{A}$. A wire cross-sectional area of $1\mathrm{m}{\mathrm{m}}^2=10^{-6}{\mathrm{m}}^2$ gives $j=2\times 10^{6}A/{\mathrm{m}}^2$; doubling the wire cross-sectional area to $2\mathrm{m}{\mathrm{m}}^2$ gives $j=10^{6}A/{\mathrm{m}}^2$. Equal electric current can therefore mean different local transport intensity. This distinction matters for Joule heating, whose local power per volume depends on current density and electric field.

For a toy electron transport model, stipulate mobile electron number density $n=8\times 10^{28}{\mathrm{m}}^{-3}$ and drift velocity ${\mathbf{v}}_d=-10^{-4}m/s\hat{\mathbf{x}}$, with the positive unit normal $\hat{\mathbf{x}}$. Each electron has electric charge $-e$, where the elementary charge $e=1.602176634\times 10^{-19}\mathrm{C}$ is exact in SI. These are defined model inputs, rather than measured parameters of a particular material. ^[ref] The current density and its magnitude areFor wire cross-sectional area $A=1\mathrm{m}{\mathrm{m}}^2$ and constant transport over time $10\mathrm{s}$, this givesThe two minus signs make conventional current point opposite to the electron drift velocity. A small drift velocity can carry an appreciable electric current because many charge carriers contribute; it does not specify the propagation speed of an electrical disturbance.

Define $\Delta V={\Phi }_{\mathrm{entry}}-{\Phi }_{\mathrm{exit}}$ as the electric potential drop along the particle's passage. If the electric field changes negligibly during that short crossing, the work givesThe sign matters: a positive ion gains kinetic energy while going to lower electric potential; a negative ion needs the opposite drop. A fixed-sign field would accelerate on one crossing and decelerate on the next. The alternating field must reverse before the particle comes back.

For a fixed electric charge sign and circulation direction, absorb this alternating crossing direction into the definition of the arrival phase ${\varphi }_n$. If $V_0$ is the peak electric potential difference across the whole gap, rather than the amplitude of either electrode relative to ground, a thin-gap model isArrival at ${\varphi }_n=0$ gives the largest accelerating kick; $|{\varphi }_n|<\pi /2$ gives a positive kick. Arrival at ${\varphi }_n=\pi$ gives deceleration. For an ideal classical particle initially on crest, the cyclotron resonance condition keeps successive arrival phases equal. It does not automatically bring an initially mistimed particle onto crest.

Take an ion of mass $m$ and electric charge $q\ne 0$, with velocity perpendicular to a uniform magnetic field of magnitude $B>0$. In the classical mechanics limit, the Lorentz force provides the centripetal acceleration:Here $r$ is the local radius of curvature and ${\omega }_c$ is the cyclotron frequency, expressed as an angular frequency. A half-turn takesIncreasing speed increases the path length in exactly the same proportion. Drive the gap at angular frequency ${\omega }_{\mathrm{RF}}={\omega }_c$, or ordinary frequency $f_{\mathrm{RF}}=|q|B/(2\pi m)$. In each half-turn the gap electric field reverses once, just as the particle reverses its direction of passage through the gap. The particle therefore receives two accelerating kicks per complete turn, using one oscillation per complete turn. This is the fundamental resonance; transit time through the gap and departures from a uniform magnetic field have been neglected.

A Daniell cell uses the zinc/${\mathrm{Zn}}^{2+}$ and ${\mathrm{Cu}}^{2+}$/copper pairs. In the separated-compartment teaching version, a zinc electrode contacts an aqueous chemical solution containing ${\mathrm{Zn}}^{2+}$, and a copper electrode contacts one containing ${\mathrm{Cu}}^{2+}$. A salt bridge supplies an ionic connection; a wire and load supply an electronic connection. During discharge,The zinc electrode loses material and copper is deposited at the cathode. Each half-reaction conserves chemical atom counts and electric charge: the first has net charge zero on each side, the second $+2e-2e=0$ on the left and zero on the right. The net reaction has $+2e$ on both sides, with no electrons left in its chemical equation. The Zn/Cu teaching cell and these half-reactions are described in Florida State University, Redox Reactions in Voltaic Cells: Construction of a Potential Series, lab background.

Creating one mole of ${\mathrm{Zn}}^{2+}$ adds $2F$ of positive electric charge to the anode compartment's ionic inventory; consuming one mole of ${\mathrm{Cu}}^{2+}$ removes $2F$ of positive electric charge from the cathode compartment. Here $F$ is the Faraday constant. In an ideal bookkeeping example using a potassium nitrate salt bridge, two moles of nitrate anions entering the anode compartment compensate the first change, and two moles of potassium cations entering the cathode compartment compensate the second. Other ions in the cell can share the actual transport. The two compensating inventory changes do not mean that $4F$ flowed through the load: the same series electric current transports $2F$ through each cross-section of the circuit per mole of net reaction. See salt bridge.

Under standard-state conditions near $298\mathrm{K}$, tabulated reduction potentials give a cell electromotive force of approximately $1.10\mathrm{V}$. This is a reversible reference value, not a promised loaded reading; actual composition matters through thermodynamic activity. The tabulated copper and zinc values and their subtraction are given in Purdue University, Electrochemical Reactions: standard-state reduction potentials. For the net chemical equation as written, the molar Gibbs free energy change obeysThus $\mathcal{E}=1.10\mathrm{V}$ corresponds to about $-212\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ of net reaction. At fixed temperature and pressure, $-{\Delta }_{r}G$ is the maximum reversible electrical work per mole for this cell reaction. It is not automatically all the heat released by a finite-current cell. See 17.4 Potential, Free Energy, and Equilibrium (§17.4: cell potential and free energy).

The motivating observation is that a chemical reaction can produce a signed electrical reading even when its reacting materials occupy separate vessels. UC San Diego's teaching demonstration places a copper strip in $0.1\mathrm{mol}{\mathrm{L}}^{-1}$ $\mathrm{CuS}{\mathrm{O}}_4$ and a zinc strip in $0.1\mathrm{mol}{\mathrm{L}}^{-1}$ $\mathrm{ZnS}{\mathrm{O}}_4$, using two $400\mathrm{mL}$ beakers with about $300\mathrm{mL}$ of chemical solution in each. The strips connect to a voltage meter. Initially there is no ionic connection; the demonstration specifies a zero reading. A filter-paper salt bridge containing aqueous $\mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4$ then connects the chemical solutions. With the meter's positive lead on copper and negative lead on zinc, its reported readings are about $+1.1\mathrm{V}$; exchanging the leads reverses the reading. These are observations reported by the university, not data generated for this article. See UC San Diego, demonstration 9B: apparatus, bridge control and meter polarity.

The controls answer different questions. Exchanging only the meter leads checks the sign convention; it does not reverse the chemical reaction or swap the anode and cathode. Interrupting only the salt bridge tests the need for an ionic return path. Its absence prevents sustained cell electric current, but a very high-input-resistance meter may still show a transient or unstable reading while interfaces store electric charge. The zero reading specified for the university's apparatus is therefore not a universal theorem for every meter. Reconnecting the same salt bridge distinguishes a reversible interruption from a disconnected wire or exhausted chemical solution. This interpretation follows from electric charge conservation in a model allowing interfacial electric charge storage; see IUPAC Gold Book: integral capacitance of an electrode for the capacitance associated with electrode charge.

To distinguish terminal electric potential difference from delivered power, compare the nearly open-circuit reading with readings across a known resistor load. In the nearly open-circuit condition, a meter with input electrical resistance much larger than the cell's effective internal electrical resistance draws little electric current. With a load, measure its terminal electric potential difference $V$ and infer $I=V/R$ using Ohm's law; track the readings over time and repeat the nearly open-circuit reading afterwards. Keep the temperature, chemical solution compositions, immersed electrode areas, bridge geometry and electrical contacts comparable. This is a model-based extension of the documented demonstration, not a claim that UC San Diego published a load sweep. A finite-current drop can reflect ionic electrical resistance, changes of interfacial reaction rates, or depletion near an electrode; a single voltage reading cannot identify which dominates. The interfacial and composition contributions are distinguished in IUPAC Gold Book: overpotential and ^[ref].

For a deliberately simplified calculation, model the cell by a constant electromotive force $\mathcal{E}$ in series with a constant internal electrical resistance $r$, and a passive load $R>0$. Take $I$ positive from the cathode through the load to the anode. ThenThe nearly open-circuit limit $R\to \infty$ gives $V\to \mathcal{E}$ but $I\to 0$ and $P_{\mathrm{load}}\to 0$. A straight $V$-versus-$I$ line is a prediction of constant $r$, not a general law for electrochemistry. See Ling, Sanny and Moebs, University Physics II, §10.1: internal resistance, terminal voltage and load.

For the chosen values, $I=5.50\mathrm{mA}$ and $V=0.550\mathrm{V}$. Holding these toy parameters fixed for $600\mathrm{s}$ gives transported electric charge $Q=It=3.30\mathrm{C}$ and an extent of reaction $\xi =Q/(2F)\simeq 17.1\mu \mathrm{mol}$, assuming all the electric current serves the two specified half-reactions. The same $\xi$ counts zinc consumed and copper deposited, rather than equal masses. The energy budget isThe load's electrical work also becomes Joule heating when the load is a resistor. Here $\mathcal{E}Q$ accounts for the reaction's available reversible electrical work; a full cell thermal budget would also require its thermodynamic entropy contribution. Real measurements need uncertainty estimates, a specified meter input electrical resistance and calibration, and allowance for changing composition and unwanted chemical reactions. The toy value $r=100\Omega$ is not a measured property of the university's cell.

The de Broglie wavelength is the distance over which a matter wave accumulates one full spatial quantum phase cycle in its propagation direction:Here $h$ is the Planck constant, $\mathbf{p}$ is the momentum, and $\mathbf{k}$ is the angular wavevector. The scalar $p$ in $\lambda =h/p$ means the positive momentum magnitude, not a signed component. Reversing a momentum reverses the wavevector without making the wavelength negative. At zero momentum there is no finite spatial phase period: the limiting wavelength diverges.

This relation makes microscopic diffraction accessible with particles: increasing the momentum shortens the wavelength until it is comparable with the spacings in a crystal scattering problem. The electron accelerating-voltage wavelength gives a practical way to tune this scale, while the Davisson–Germer experiment tests it by comparing electron diffraction directions with independently known crystal geometry. The same relation applies to neutral particles in a neutron interferometer; electric charge is not a requirement for a matter wave.

The wavelength belongs to the wavefunction, not to a sinusoidal path of a particle. A free-particle momentum convention makes the distinction especially clear: a plane wave has a rotating complex quantum phase but a constant probability density. A localized wave packet contains a range of momenta and therefore generally has a range of wavelengths, rather than one exact spatial period.

For a uniformly charged sphere of radius $a$, total electric charge $Q$, and an exterior dilute electrolyte of permittivity $\epsilon$, the Debye–Hückel approximation gives $(r^2{\varphi }^{\prime }{)}^{\prime }={\kappa }^2{r}^2\varphi$. Decay at infinity and Gauss's law, $-\epsilon {\varphi }^{\prime }(a)=Q/(4\pi a^2)$, select the displayed electric potential. Substituting $u=r\varphi$ reduces the proof to $u^{\prime \prime }={\kappa }^{2}u$. The screening parameter is ${\kappa }^2={\sum }_i{z}_i^2{e}^2{n}_{i,\infty }/(\epsilon k_{B}T)$, and the ion profiles are $n_i\simeq n_{i,\infty }(1-z_{i}e\varphi /(k_{B}T))$. The validity condition is $|z_{i}e\varphi (a)|\ll k_{B}T$ for every relevant species. Integrating the diffuse charge density using the radial equation gives total screening charge $-Q$.

Deferred charge in CCD readout reaches later output samples than its originating packet. George Elwood Smith describes surface states at the silicon–silicon dioxide interface capturing electrons and releasing them later, smearing the image. A buried channel keeps the stored packet away from that interface.^[ref]

An original comparison separates two interpretations of the same main-packet deficit. If a single transfer defers fraction $ϵ$ by exactly one sample and loses nothing, an isolated input packet $N$ produces samples ${\eta }_{\mathrm{tr}}N$ followed by $ϵN$: summing both recovers $N$. A truly lossy transfer yields only ${\eta }_{\mathrm{tr}}N$, with no later recovery. For $N=10000$ and $ϵ=0.01$, both give $9900$ in the intended sample, but only the deferred case gives a $100$-electron trail. Such a trail redistributes position information even when total electric charge is conserved. Repeated delays require a time-response model; $N_{\mathrm{in}}{\eta }_{\mathrm{tr}}^n$ alone cannot predict trail shape. Electron-hole recombination, escape to another contact, or release beyond the recorded window can instead remove signal from the measured sum.

Deprotonation removes a proton from a chemical species, changing its electric charge by $-1$. It can expose a donor lone pair or change donor charge without changing the metal oxidation state.

Desolvation removes some or all of the solvent environment of a species. Removing a hydration shell does not remove an ion's electric charge. Complete reversal of a specified solvation transfer reverses its Gibbs free energy and enthalpy changes when the same states and conventions are used. Partial desolvation, exchange of one solvent for another, and creation of a gaseous species have different endpoints; their costs need not match.

A dielectric is a material in which an electric field produces electric polarization while transport of electric charge is sufficiently small for the intended use. An insulating solid in a capacitor is an example; real dielectrics can still have dielectric loss and nonzero electrical conductivity.

The important distinction is between rearranging bound electric charge and transporting mobile charge carriers through the specimen. Bound electric charges can move a little within a chemical atom or molecule, or an existing electric dipole can turn. Their averaged electric dipole moments produce electric polarization even if there is negligible steady electric current between the electrodes. A changing electric polarization does give a bound current; the absence of steady conduction does not mean that the material is electrically unresponsive.^[ref]

For uniform electric polarization $\mathbf{P}=P\hat{\mathbf{x}}$ in a rectangular slab, polarization density gives zero bound volume charge density in the interior and opposite bound surface charge densities $-P,+P$ on the left and right faces. The electric charge has been redistributed, not created. More generally, the divergence theorem makes the cancellation explicit:Here $\mathbf{n}$ is the outward normal vector on the material boundary. This accounts for both bound volume charge density and bound surface charge density; omitting either can falsely suggest that an initially neutral dielectric acquired net electric charge.

An electrical insulator between capacitor plates supports a useful combination: appreciable electric polarization, little leakage electric current, and tolerance of the intended electric field. This lets a small parallel-plate capacitor store separated electric charge and recoverable capacitor energy. The comparison in dielectric capacitor response explains why specifying fixed electric charge or fixed electric potential difference changes the answer. High relative permittivity alone does not establish good insulation or a large usable energy density; dielectric loss and dielectric breakdown matter as well.

Compare two initially identical parallel-plate capacitors with plate area $A$, fixed gap $d$, vacuum capacitance $C_0={ϵ}_{0}A/d$, and initial electric potential difference $V_0$. Completely fill both gaps with the same uniform linear dielectric of static relative permittivity ${ϵ}_r>1$, neglecting the fringing field of a capacitor. Keep one connected to an ideal voltage source and isolate the other. The geometric capacitance becomes $C={ϵ}_r{C}_0$ in both cases; the electrical constraint determines what else changes.^[ref]

Write $Q_0=C_0{V}_0$, $E_0=V_0/d$ and $U_0=C_0{V}_0^2/2$. Applying $Q=CV$, $E=V/d$ and the capacitor energy formula gives the complete comparison:Thus the statement that a dielectric reduces the electric field needs the fixed-free-charge condition. With fixed electric potential difference, the voltage source supplies additional free plate electric charge, maintaining $E=V_0/d$ despite the new bound electric charges.

For the stipulated ${ϵ}_r=3$ example, the voltage source supplies $\Delta Q=2Q_0$. At constant $V_0$, its delivered electrical work is $V_0\Delta Q=4U_0$, whereas the increase in capacitor energy is $2U_0$. In an ideal reversible quasistatic insertion, the remaining $2U_0$ is mechanical work delivered by the assembly, as accounted for in fixed-voltage capacitor energy balance. For the isolated assembly, capacitor energy falls from $U_0$ to $U_0/3$, allowing $2U_0/3$ of mechanical work in the same ideal limit. Actual losses reduce recoverable work; energy increases in the connected case do not arise from the dielectric supplying free energy.

Modification of the interaction between electric charges by a medium's electrical response. In the simplest Wannier-Mott exciton model, the Coulomb potential is reduced by a constant relative permittivity; actual screening can depend on spatial and temporal scales.