The secondary pyroelectric effect is the change of polarization density produced when thermal strain acts through piezoelectricity. It is additional to the primary change at fixed strain. A linear constitutive description along the polar axis iswhere $e_{3j}$ are piezoelectric stress coefficients and ${\alpha }_j$ the freely developing thermal strains per kelvin. Thermal-expansion mismatch with a support or a deliberately chosen compliant structure can change the secondary contribution. It enhances the total response only if its sign reinforces the primary response; mechanical boundary conditions and thermal cycling must be characterized rather than assuming every restraint is beneficial.

Preventing axial expansion removes the free axial-strain contribution, but does not automatically eliminate every secondary contribution. In a one-dimensional idealization with only ${\epsilon }_3$ coupling and perfect axial clamping, that strain-mediated term vanishes. A real rod may still expand transversely, and nonzero $e_{31}$ or $e_{32}$ can change $P_3$ through those strains. In a stress-based constitutive equation, the same restraint appears as thermal stress and a piezoelectric term; do not add it a second time to a coefficient already specified for the constrained condition. Full suppression of all relevant thermal strains leaves the primary coefficient at fixed strain. The answer therefore depends on which degrees of freedom are constrained, not simply on the word “nanorod”.

For electrode area $A$, the change in spontaneous polarization density supplies a charge scale $\Delta Q=pA\Delta T$, while the film's capacitance is $C=\epsilon A/d$. The voltage magnitude is therefore $|\Delta V|=|p|d|\Delta T|/\epsilon$. More explicitly, for an initially compensated film with no change in free electrode charge, $\Delta D=\epsilon \Delta E+p\Delta T=0$. Taking $z$ from the lower electrode to the upper one gives $\Delta V_{\mathrm{upper}-\mathrm{lower}}=-d\Delta E=pd\Delta T/\epsilon$.The measurable magnitude is $0.339\mathrm{V}$, with polarity reversed by reversing the poling direction or the temperature change. This is an ideal open-circuit estimate: leakage, the measuring instrument's input capacitance and finite charge-relaxation time can reduce or reshape the observed voltage.

A thermoelectric generator uses the Seebeck effect and needs a spatial temperature gradient: a couple develops $V\simeq (S_p-S_n)\Delta T$ between hot and cold junctions. It can deliver continuous power under a steady thermal gradient and has no moving parts, but requires sustained heat flow and adequate hot- and cold-side thermal contacts. Thermal conduction, internal electrical resistance and the material's thermoelectric figure of merit limit conversion.

A pyroelectric generator responds to a time-dependent spontaneous electric polarization. Under approximately uniform heating or cooling, $I=pAdT/dt$ when the electrical boundary condition allows charge flow. It can harvest temporal thermal fluctuations without a persistent spatial gradient. However, at a constant uniform temperature its ordinary pyroelectric current stops; practical extraction needs cycling, low-leakage electronics and a suitable load. A large open-circuit voltage is not evidence of a large available average power. Temperature-driven nanowire measurements provide a primary experimental example of this distinction.^[ref]

At unchanged intrinsic catalyst kinetics, increasing the electrochemically accessible true area by $100$ multiplies exchange current per geometric area by $100$. Equivalently, the local current density at a fixed total geometric current falls by $100$. Each Tafel equation activation loss decreases by $b{\log }_{10}100=2b$:Thus $V_{\mathrm{new}}=0.59948\mathrm{V}$ and $P_{\mathrm{new}}/A=5.99\mathrm{W}{\mathrm{m}}^{-2}$, an increase of $2.39\mathrm{W}{\mathrm{m}}^{-2}$, or about $66.3\%$. Power does not increase a hundredfold, because the voltage improvement is logarithmic and the specified geometric current is unchanged. This assumes both electrodes gain accessible active area, the original exchange currents use the original area basis, and the porous structure does not introduce additional transport or Ohmic losses.

The reversible voltage follows from the two-electron hydrogen reaction:At the specified geometric current density, the magnitude of each activation loss follows its Tafel equation, $\eta =b{\log }_{10}(j/j_0)$:Both losses reduce the delivered fuel cell voltage. Therefore $V=0.36048\mathrm{V}$ and $P/A=jV=3.60\mathrm{W}{\mathrm{m}}^{-2}$. The very small oxygen exchange current is responsible for most of the activation loss. The calculation uses current densities on a common area basis, unit water activity and the stated gas conditions, and neglects the losses the problem excludes.

In a galvanic cell, a spontaneous oxidation-reduction reaction supplies electrical work: ${\Delta }_{r}G<0$ for discharge, the anode is negative, and the cathode positive. A discharging electric battery and a hydrogen–oxygen fuel cell are two practical examples. The first stores its reactants internally; the second is supplied continuously with fuel and oxidant.

An electrolytic cell consumes externally supplied electrical work to drive electrode reactions or material transfer under an applied potential. Its anode is positive and cathode negative. water electrolysis for hydrogen production and metal electroplating are two practical examples. oxidation always occurs at the anode and reduction at the cathode in both cell types; the electrode signs change. The electrolyte conducts ions, while electrons flow through the external circuit. A rechargeable battery can operate in either regime depending on whether it is discharging or being charged.

The oxygen-pressure exponents can be explained by a change in the dominant charge state of metal vacancies, with holes carrying the p-type current. For a divalent metal oxide, a doubly negative metal vacancy is $V_{\mathrm{M}}^{\prime \prime }$ in Kröger–Vink notation. A conventional oxygen-incorporation reaction isTreating regular oxygen-site activity as constant, the law of mass action gives $K_2=[V_{\mathrm{M}}^{\prime \prime }]p_h^2/p_{{\mathrm{O}}_2}^{1/2}$. If these are the compensating defects, charge neutrality gives $p_h=2[V_{\mathrm{M}}^{\prime \prime }]$, hence $p_h^3\propto p_{{\mathrm{O}}_2}^{1/2}$ and $\sigma \propto p_{{\mathrm{O}}_2}^{1/6}$, assuming nearly constant hole mobility.

For singly negative vacancies, useNow $p_h=[V_{\mathrm{M}}^{\prime }]$ and $p_h^2\propto p_{{\mathrm{O}}_2}^{1/2}$, so $\sigma \propto p_{{\mathrm{O}}_2}^{1/4}$. The charge-state equilibrium $V_{\mathrm{M}}^{\prime }\rightleftharpoons V_{\mathrm{M}}^{\prime \prime }+h^{\bullet }$ gives $[V_{\mathrm{M}}^{\prime \prime }]/[V_{\mathrm{M}}^{\prime }]=K_{\mathrm{ion}}/p_h$: the singly charged fraction can therefore dominate as oxygen pressure and hole concentration increase. This is a consistent dilute-defect model, not a unique experimental identification of every defect. Changes in mobility, defect association or other compensating defects require independent tests.

For two equal spheres, the Derjaguin approximation uses $R_{\mathrm{eff}}=R/2$. A planar nonretarded Van der Waals force interaction has energy per area $W(h)=-A_H/(12\pi h^2)$, soThe force is attractive, with magnitude $13.0\mu \mathrm{N}$, taking separation increase as positive. Using the sphere–plane value of $R_{\mathrm{eff}}$ would overestimate it by a factor of two. The gap is much smaller than either radius, as required. Treat the supplied Hamaker constant as the effective one for the intervening medium; roughness, solvation and molecular-scale contact can modify a literal $1\mathrm{nm}$ experiment.

Balance upward surface-tension force $2\pi r\gamma \cos \theta$ against the weight $\rho g\pi r^{2}h$ of the liquid column. The capillary action isWith perfect wetting, $\cos \theta =1$, $D=10^{-3}\mathrm{m}$, $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $h=2.94\times 10^{-2}\mathrm{m}=29.4\mathrm{mm}$. The diameter, not the radius, is $1\mathrm{mm}$. This neglects the small meniscus-volume correction, gas-density buoyancy and contact-angle hysteresis.

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.

With $n=2$, the standard reaction Gibbs free energy and dimensionless equilibrium constant areFor the reaction as written, $Q=a_{\mathrm{F}{\mathrm{e}}^{2+}}/a_{\mathrm{S}{\mathrm{n}}^{2+}}$, and the Nernst equation gives $E=E^{\circ }-(RT/2F)\ln Q$. Consequently zero cell potential requires $[\mathrm{F}{\mathrm{e}}^{2+}]/[\mathrm{S}{\mathrm{n}}^{2+}]=K\simeq 1.93\times 10^{10}$ under the specified unit-activity-coefficient assumption. Reversing that ratio reverses the conclusion. This mathematical ratio does not by itself guarantee experimentally attainable concentrations without precipitation or other chemistry.