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Translational and reflection symmetry parallel to the slabs imply $\mathbf{E}=E_x(x)\hat{\mathbf{x}}$. The differential Gauss's law is $d{E}_x/dx=\rho /{ϵ}_0$. Starting at the left boundary with $E_x(-a)=0$ givesThere is no singular surface charge at $x=0$, so the normal electric field is continuous there: $E_x(0)=-C/{ϵ}_0$. Integrating through the positive slab givesThe second expression indeed returns to zero at $x=b$, because the two total charges per unit area are $-C$ and $+C$. For $C>0$, the electric field throughout the interior points from the positive slab towards the negative slab, along $-x$; reversing $C$ reverses the directions. The piecewise linear profile is a useful consequence of the uniform volume charge density.

The current density is the moving charge density times its velocity:For a cylindrical Gaussian surface of radius $r<R$ and length $\ell$, cylindrical symmetry makes the electric field radial. Gauss's law gives $E(2\pi r\ell )=\rho \pi r^2\ell /{ϵ}_0$. A circular Ampère's circuital law contour encloses electric current $I(r)=\rho v_0\pi r^2$, giving $B(2\pi r)={\mu }_{0}I(r)$. HenceThe right-hand rule fixes the positive azimuthal direction of the magnetic field for positive electric charge moving along $+z$. Both the electric field and the magnetic field vanish on the axis. These are the steady long-cylinder electric field and magnetic field; end effects and changes of the prescribed beam distribution are excluded.

The integral and differential forms of Gauss's law in electrostatics areHere $\partial \mathcal{V}$ is a closed surface bounding a volume $\mathcal{V}$, $d\mathbf{A}=\hat{\mathbf{n}}dA$ is its outward area vector, $\mathbf{E}$ is the electric field, $\rho$ is total volume charge density, and $Q_{\mathrm{enc}}={\int }_{\mathcal{V}}\rho d^{3}r$ is enclosed electric charge. The divergence theorem converts one form to the other. For ideal sheets or point charges, interpret the differential equation with their distributional charge densities.

Assemble the spherically symmetric charge distribution by adding thin concentric shells, with the final charge density kept uniform. When the assembled radius is $s$, the enclosed electric charge and the next increment areThe electric potential of the already assembled material at that shell is $k_{e}q(s)/s$. Consequently its assembly work is $dU=k_{e}q(s)dq/s$, and the electrostatic energy of a uniformly charged solid sphere isThere is no extra factor of one half in this assembly integral: each pair of charge elements has already been counted exactly once, when the later element was brought in. A solid uniformly charged insulator has this value; a conducting sphere has its electric charge on the surface and instead has electrostatic energy $k_e{Q}^2/(2r)$.

For this space-charge-filled parallel-plate capacitor, place the first plate at $x=0$ and the second at $x=d=x_o$, ignoring fringing field of a capacitor. With vacuum permittivity between the plates and fixed charge density $\rho$, Poisson's equation and the prescribed boundary electric potentials giveIntegrating twice and imposing both values yieldsThe electric field is minus the gradient of the electric potential, henceThus the electric potential is a parabola, concave downward for $\rho >0$ and upward for $\rho <0$. The signed electric field is a straight line. Its magnitude has a V-shaped corner if its zerolies inside the gap; otherwise its magnitude is a single straight segment. An interior zero requires $|V_1-V_2|<|\rho |d^2/(2{ϵ}_0)$. At that point the electric potential has its interior extremum. For $\rho =0$, these reduce to a linear electric potential and a uniform electric field.

In SI units, Maxwell's equations in vacuum, allowing charge density $\rho$ and electric current density $\mathbf{J}$, areThe source-free specialization sets $\rho =0$ and $\mathbf{J}=0$; it cannot describe the charged objects at their locations. Applying the divergence theorem to the first equation gives Gauss's law:For an isolated point charge $q_1$, spherical symmetry makes the electric field radial and constant in magnitude on a concentric sphere. Therefore $4\pi r^2{E}_r=q_1/{ϵ}_0$. Multiplying by a second point charge gives Coulomb's law,The force points away from the first charge for like signs and toward it for opposite signs.

For the spherical capacitor, a Gaussian sphere between the shells encloses $q$, so $E_r=q/(4\pi {ϵ}_0{r}^2)$ for $a<r<b$. Integrating the electric field inward from the outer shell yieldsAn arbitrary constant $V(b)$ remains unless a reference is specified. The capacitance is determined by the potential difference:For the subsequent charging operation, take the initial outer shell to be earthed, or interpret the stated $V$ as the inner-to-outer voltage with the outer reference zero. Isolating the inner shell fixes its electric charge at $q=CV$. Removing the outer shell leaves an isolated conducting sphere. Its electric field and electric potential, now with zero at infinity, areConsequentlyThese changes of electric potential are compatible with constant electric charge. Without the initial reference convention the absolute stated $V$ does not determine $q=CV$: for example, if the outer shell were initially isolated and neutral, the absolute inner electric potential would instead be $q/(4\pi {ϵ}_{0}a)$.

Finally take the connected spheres to be sufficiently far apart that the long wire's capacitance and their mutual influence can be neglected. Their common electric potential requiresso $q_a=q/3$ and $q_{2a}=2q/3$. The magnitudes of the surface electric fields areThis is the usual widely separated-sphere approximation. A wire enforces an equipotential, but alone does not make this ratio exact for arbitrary arrangements. As a concrete counterexample, if the second shell enclosed the first concentrically and they were connected, all net electric charge would reside on the exterior and the inner sphere's surface electric field would be zero.

In vacuum, Maxwell's equations in differential form areHere $\rho$ is charge density, $\mathbf{J}$ is electric current density, $\mathbf{E}$ is electric field, and $\mathbf{B}$ is magnetic flux density. In a source-free region, taking the curl of the third equation and using $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$ gives the wave equationThe same argument gives the wave equation for $\mathbf{B}$. A concrete family satisfying all four Maxwell's equations isThe function $F$ may be any twice differentiable profile. These are transverse travelling waves; reverse the magnetic field direction for propagation in the opposite direction. Because the source-free Maxwell's equations are homogeneous linear differential equations, any finite linear combination of such paired electric fields and magnetic fields is again a solution. Infinite superpositions also work whenever the required derivatives converge. A solution of the component wave equations alone must still satisfy the divergence and mutual-curl constraints.

In SI units, the Maxwell equations in vacuum with prescribed charge density $\rho$ and current density $\mathbf{J}$ areWhen $\rho =0$ and $\mathbf{J}=0$, take the curl of the third equation and substitute the fourth. Using the curl of the curl identity and $\nabla \cdot \mathbf{E}=0$ gives the electromagnetic wave equation:For example, $\mathbf{E}=E_0\hat{\mathbf{x}}\cos (kz-\omega t)$, $\mathbf{B}=(E_0/c)\hat{\mathbf{y}}\cos (kz-\omega t)$ is a plane electromagnetic wave provided $\omega =ck$. The electric field, magnetic field and propagation direction are mutually perpendicular.

The Maxwell equations are linear, so the superposition principle applies to fields. However, the Poynting vector is quadratic:One may add the average wave intensities only when the cross terms vanish under the actual detector average. For co-propagating plane electromagnetic waves of parallel linear polarization, averaging over optical cycles but potentially resolving the difference frequency givesIf the detector integrates for $T$, the last term acquires $\mathrm{sinc}[({\omega }_1-{\omega }_2)T/2]$, with sinc function convention $\mathrm{sinc}x=\sin x/x$, evaluated at the midpoint phase. Thus $T\gg 1/|f_1-f_2|$ suppresses beats, and the infinite-time average adds the wave intensities exactly for distinct fixed frequencies. Random mutually uncorrelated wave phases also remove the cross terms in an ensemble average; orthogonal linear polarizations remove them for co-propagating waves in polarization-insensitive detection. Merely having different frequencies does not remove a resolved beat signal: two nearly equal frequencies with fixed relative wave phase are a concrete counterexample. For different propagation directions, the relevant addition is of average Poynting vectors, or their components through the chosen detector surface.

Let the original cylinder axis be $\mathbf{r}=0$, and let the hole axis be at transverse vector $\mathbf{a}$, with $|\mathbf{a}|=a$. The charge density is reproduced by a full cylinder of radius $R_1$ and charge density $+\rho$, superposed with a cylinder of radius $R_2$ centred at $\mathbf{a}$ and charge density $-\rho$. This uses the superposition principle for fixed insulating electric charges; the hole is not a hole in an electrical conductor.

For a complete uniformly charged infinite cylinder of radius $R$, a coaxial Gaussian surface of length $\ell$ gives, by Gauss's law,Consequently defineThe complete electric field, including the exterior if desired, isIn the insulating material, $|\mathbf{r}|<R_1$ and $s=|\mathbf{r}-\mathbf{a}|>R_2$, soInside the hole both contributions use their interior formulas, and all dependence on $\mathbf{r}$ cancels:The electric field in the hole is uniform, points from the original axis towards the hole axis for $\rho >0$, and vanishes for a concentric hole. The two expressions agree at the hole wall, as expected with no imposed surface charge density there. Its uniformity is the useful uniform electric field in an off-axis cylindrical cavity result.

Take an abrupt equilibrium p-n junction with the p side at $x<0$ and the n side at $x>0$. In the quasi-neutral bulk, electron holes are the majority carriers on the p side and electrons are the majority carriers on the n side. With ionized dopant densities $N_A,N_D$ and nondegenerate statistics,where $n_i$ is the intrinsic carrier concentration. The semiconductor mass-action law $np=n_i^2$ applies at equilibrium. Both mobile populations are small in the depletion region compared with the dopant densities.

The charge and carrier density of an abrupt p-n junction must distinguish mobile populations from net charge. Diffusion leaves negatively charged ionized acceptor impurities on the p side and positively charged ionized donor impurities on the n side. Under the depletion approximation,The neutral bulks have large carrier densities but nearly zero net charge density. The electric field points from the positive donor region toward the negative acceptor region, balancing diffusion at equilibrium.

The figure uses equal dopant densities only to show a simple symmetric sketch. Generally the more lightly doped side has the wider depleted interval. The carrier profiles are smooth, whereas the step-shaped net charge density is the ideal depletion approximation.

Semiconductor doping introduces controlled impurities to alter equilibrium carrier populations and electrical conductivity. In an intrinsic semiconductor, thermally excited electrons in the conduction band and electron holes in the valence band have equal densities $n=p=n_i$. A shallow donor impurity, such as phosphorus in silicon, supplies an electron from an impurity level just below the conduction-band edge. Once ionized it is positively charged, and the material is n type. An acceptor impurity, such as boron in silicon, accepts a valence electron into a level just above the valence-band edge, leaving a mobile hole and a negatively charged ionized impurity; the material is p type.

With dopant densities $N_D,N_A$, charge neutrality requiresFor fully ionized, nondegenerate dopants at equilibrium, the semiconductor mass-action law gives $np=n_i^2$. In a substantially n-type sample, $N_D-N_A\gg n_i$, soThe corresponding p-type result is $p\simeq N_A-N_D$ and $n\simeq n_i^2/(N_A-N_D)$. The majority population grows while the minority population decreases; adding donors does not simply add equal numbers of free electrons and holes.

The Fermi energy shifts toward the conduction band for n-type material and toward the valence band for p-type material. In the nondegenerate approximation,where $N_C,N_V$ are effective band-edge state densities. At very low temperature, dopants may not be ionized and carrier freeze-out occurs. At intermediate temperatures, ionized dopants dominate the carrier density; at sufficiently high temperature, intrinsic excitation can dominate. Heavy doping can produce degenerate carrier statistics, invalidating these Boltzmann expressions.

The electrical conductivity is $\boxed{\sigma =e(n{\mu }_e+p{\mu }_h)}$. Doping usually raises the majority carrier density, but ionized impurity scattering can lower mobility, so conductivity need not scale indefinitely with dopant concentration. Joining p-type and n-type regions creates a p-n junction, with a depletion region and built-in electric field established by carrier diffusion and compensating drift. The fixed ionized dopants, not an absence of all charge, account for the depleted region's net charge density.

Gauss's law connects electric flux through a closed surface to its enclosed charge:The differential form follows from the divergence theorem. The law is exact; extracting a field magnitude by simple division additionally needs symmetry. For spherical symmetry, $E_r(r)=Q(r)/(4\pi {ϵ}_0{r}^2)$; a uniformly charged ball therefore has a field linear in $r$ inside and an inverse-square field outside. An infinite line of linear charge density $\Lambda$ gives $E_r=\Lambda /(2\pi {ϵ}_{0}r)$ by a cylindrical surface. A uniformly charged infinite sheet gives $E=\sigma /(2{ϵ}_0)$ on either side by a pillbox.

An electrostatic conductor has zero interior field and constant potential; excess charge occupies its surface. A pillbox gives $E_{n,\mathrm{out}}=\sigma /{ϵ}_0$ when the inside is conducting. More generally, $\hat{\mathbf{n}}\cdot ({\mathbf{E}}_2-{\mathbf{E}}_1)=\sigma /{ϵ}_0$. For dielectric matter, $\nabla \cdot \mathbf{D}={\rho }_{\mathrm{free}}$ separates bound polarization charge from free charge. Gauss's law alone does not establish a uniform field for an arbitrary finite plate or recover a detailed nonsymmetric charge distribution from one flux measurement.

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

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

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

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

In a linear isotropic dielectric of relative permittivity $ϵ$, the electric displacement field and electric field satisfyHere ${\rho }_f$ denotes the free charge density called $\rho$ in the question. For spatially constant $ϵ$, these also give $\nabla \cdot \mathbf{E}={\rho }_f/({ϵ}_0ϵ)$. Equivalently, $\mathbf{E}=-\nabla \phi$ and ${\nabla }^2\phi =-{\rho }_f/({ϵ}_0ϵ)$. The electric displacement field has units $\mathrm{C}{\mathrm{m}}^{-2}$ and the electric field has units $\mathrm{V}{\mathrm{m}}^{-1}$.

The local equations determine the fields once their geometry and boundary conditions are specified; a density value alone does not determine a unique vector field without that information.

In SI units, with microscopic charge density ${\rho }_e$ and current density $\mathbf{J}$, Maxwell's equations can be writtenThe first relates electric flux to charge. The second states that the net magnetic flux through a closed surface is zero. Faraday's law relates electric circulation to changing magnetic flux. The Ampère-Maxwell equation includes both conduction current and displacement current; the latter is essential to time-dependent fields and local charge conservation.

For fixed contours and spanning surfaces, the integral curl equations are $\oint \mathbf{E}\cdot d\mathbf{l}=-d{\Phi }_B/dt$ and $\oint \mathbf{B}\cdot d\mathbf{l}={\mu }_0{I}_{\mathrm{enc}}+{\mu }_0{\epsilon }_{0}d{\Phi }_E/dt$. For a moving material contour the first must include the motional force term; its total emf is $\oint (\mathbf{E}+{\mathbf{v}}_{\mathrm{loop}}\times \mathbf{B})\cdot d\mathbf{l}$. This prevents confusing a changing circuit area with a partial time derivative of a static magnetic field.

Taking the divergence of the Ampère-Maxwell equation and using electric Gauss's law gives ${\partial }_t{\rho }_e+\nabla \cdot \mathbf{J}=0$. In a source-free vacuum, combining the two curl equations gives ${\nabla }^2\mathbf{E}-{\mu }_0{\epsilon }_0{\partial }_t^2\mathbf{E}=0$, and similarly for $\mathbf{B}$, predicting electromagnetic waves of speed $c=1/\sqrt{{\mu }_0{\epsilon }_0}$. The Lorentz force law then specifies how these fields act on a charge. Macroscopic matter descriptions may instead use $\mathbf{D}$ and $\mathbf{H}$ with free sources and constitutive relations; those should not be silently mixed with the microscopic-source equations above.

Multiplying each ion number density by its signed charge gives the local charge densityThe fixed surface charge is supplied separately as a boundary condition; it is not part of this mobile charge density in the liquid.

A charge density wave is a spatial modulation of the electron charge density, often accompanied by a periodic lattice distortion. Write a weak modulation as $\rho (x)={\rho }_0+{\rho }_1\cos (Qx+\phi )$. In a one-dimensional metal, the Peierls distortion couples states near opposite Fermi points most strongly for $Q=2k_F$. Opening a gap near the Fermi energy lowers occupied electronic energies, while the lattice deformation has a positive elastic energy cost. Their competition can make a periodic state favorable below a transition temperature.

The modulation can be commensurate or incommensurate with the underlying Bravais lattice. Diffraction detects additional satellite peaks, while transport can show reduced low-energy carrier density. Impurities can pin its phase; above a depinning field a collectively sliding modulation can contribute nonlinear current. These effects distinguish collective density modulation from simply assigning independent static charges to isolated lattice sites.

In a conductor, mobile electrons redistribute in response to an electric potential, producing an induced charge density that opposes its source. Finite electronic compressibility means a finite density change requires a change of local kinetic Fermi energy. Screening is therefore not perfect at every microscopic distance. On scales comparable to atomic structure or the Fermi wavelength, a continuum local-density model itself fails.

Thomas-Fermi screening treats electrons as a locally uniform degenerate free electron gas, replaces the ionic background by a smooth neutralizing charge, and neglects nonlocal response, exchange and correlation corrections. The electrochemical chemical potential is uniform in equilibrium. Because an electron has charge $-e$, its electrostatic energy is $-eV$, givingAt zero potential $\mu =E_{F0}$. Linearizing at fixed $\mu$, with $|eV|\ll E_{F0}$, givesInsert this in Poisson's equation to obtain $({\nabla }^2-{\xi }^{-2})V=-{\rho }_{\mathrm{free}}/{ϵ}_0$, where the Thomas-Fermi screening length isAway from a point source, direct radial differentiation shows ${\nabla }^2(e^{-r/\xi }/r)={\xi }^{-2}{e}^{-r/\xi }/r$. The decaying solution is therefore $V=C{e}^{-r/\xi }/r$. Integrating Poisson's equation over a small sphere containing charge $Q$ fixes $C=Q/(4\pi {ϵ}_0)$. For a sufficiently large source, the linearized formula is only a far-field approximation, not a justified description of its singular core.

For monovalent sodium, the electron number density is its atomic number density,This gives $E_{F0}=3.15\mathrm{eV}$ andRoom temperature is far below the corresponding Fermi temperature, so degeneracy is well justified, and sodium's simple conduction band makes a free-electron estimate useful. However $k_F\xi =0.615$ and the mean atomic spacing is $n_0^{-1/3}=0.340\mathrm{nm}$. The potential varies on a scale shorter than $k_F^{-1}$ and the atomic spacing, so the slowly varying local-density and continuum assumptions are not quantitatively controlled at the screening length itself. A nonlocal treatment includes Friedel oscillations, absent from the pure exponential.

Nearly unchanged interlayer force constants in two-layer and twenty-layer films are consistent with very local screening and bonding forces: reducing the distant bulk need not strongly change the force between adjacent layers. They also indicate that metallic electronic response can survive in the thin film. The observation does not by itself validate the numerical Thomas-Fermi approximation, establish an identical full phonon spectrum, or rule out surface-induced electronic changes; it directly constrains the measured local force constant.

Equilibrium junction. Put the p-type side on the left and n-type side on the right. When they are joined, electrons diffuse from n to p and electron holes diffuse from p to n. This leaves negative ionized acceptors on the p-side and positive ionized donors on the n-side of the depletion region. The resulting built-in electric field points from n to p, opposing further diffusion. At equilibrium there is one constant Fermi level, and drift and diffusion currents cancel for each carrier species.

In the nondegenerate, fully ionized limit, the neutral-region carrier densities areThe conduction band and valence band lie higher on the p-side and bend down towards the n-side. The gap between them stays fixed for a homojunction. If $\psi$ increases towards the n-side, both edges shift by $-e\psi$, whileThus $n$ rises and $p$ falls continuously across the junction; neither mobile population is an exact step function. Their depletion exposes fixed dopant charge density, rather than an absence of every kind of charge. The built-in voltage is

Light emission. Forward bias lowers the barrier and injects minority carriers across the p-n junction. In a light-emitting diode, electron-hole recombination converts some of their energy into photons. A direct band gap permits efficient radiative transitions with nearly conserved crystal momentum; nonradiative channels compete. Near-edge emission has $\hslash \omega \simeq E_g$ and $\lambda \simeq hc/E_g$. Material composition and confinement select the emission colour. Reverse-bias saturation is not the ordinary LED operating mode.

Reverse current. Take $V=V_R>0$ and $I_R>0$ to mean reverse voltage and reverse current magnitudes. The Shockley diode equation in this convention isThe bias-independent $I_S$ term is the supply of thermally excited minority carriers from the neutral regions, swept across the junction by its electric field. The opposing injection current is reduced by the extra barrier energy $e{V}_R$ and is $I_S{e}^{-e{V}_R/(k_{B}T)}$. Their equality gives zero net current at zero bias; for $e{V}_R\gg k_{B}T$, the reverse current approaches $I_S$ until leakage or breakdown invalidates this ideal law. The collecting field need not create the minority carriers; their thermal supply limits the diffusion-dominated current.

For an ideal long junction of area $\mathcal{A}$, steady diffusion with recombination gives the diode minority-carrier diffusion currentIt depends on doping, geometry, carrier diffusion coefficients and lifetimes, and especially the intrinsic carrier concentration. Depletion-region generation current and surface leakage can add nonideal contributions to the measured reverse current.

Band-gap estimate from the data. Read a nearly flat region well above the small-bias knee and below the high-bias upturn, rather than estimating $I_S$ in the breakdown region. For a diffusion-dominated junction with nearly temperature-independent gap and effective masses,so $I_S=C(T)T^3{e}^{-E_g/(k_{B}T)}$, where $C(T)$ includes transport and doping factors. Extracting a semiconductor band gap from diode saturation current amounts to measuring the slope of $\ln [I_S/(C(T)T^3)]$ against $1/T$: it is $-E_g/k_B$. If the prefactor is taken approximately constant over a short interval,Approximate readings from the original PDF plateaus areUsing the two warmer curves and $k_B=8.6173\times 10^{-5}eV/K$ givesKeeping the $T^3$ factor but taking the residual transport factor constant changes this estimate to approximately $0.57\mathrm{eV}$. The justified graphical estimate is thereforeThe colder curve is a genuine limitation of the supplied data: the $0$ to $-45{}^{\circ }\mathrm{C}$ pair gives only about $0.32\mathrm{eV}$ with a constant prefactor, and the endpoint pair gives about $0.42\mathrm{eV}$. Thus all three cannot support one precise diffusion-only activation slope. A depletion-region generation current, approximately proportional to $n_i$ rather than $n_i^2$, can have a roughly half-gap activation energy; other leakage and temperature-dependent transport can also alter the slope. The PDF alone does not identify their relative contributions. A reliable precision gap measurement would use more temperatures and verify the transport regime instead of forcing one fit through these three readings.

Write ${\epsilon }_s={ϵ}_0ϵ$ and take $e>0$, $\alpha >0$, so the p side is $x<0$. In the depletion approximation, mobile electrons and electron holes are negligible compared with the fixed ionized dopants. Their charge density isApplying Poisson equation givesOutside the depletion region, mobile carriers approximately neutralize the dopants, so $\rho \simeq 0$ and $V^{\prime \prime }\simeq 0$. Full dopant ionization alone does not remove the mobile-carrier charge: the result also requires the depletion approximation.