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A surface state is an electronic state localized near a material boundary. Its occupation can store electric charge and change the near-surface electric potential, producing band bending.

Walter Houser Brattain's surface studies combined changes of contact potential under light, electrical conductivity, and transient response to distinguish surface effects from bulk properties. His Nobel lecture distinguishes fast and slow charge responses and warns that slow behavior can involve the surrounding gas rather than a simple electronic state.^[ref]

A thermocouple uses two different electrical conductors and the Seebeck effect to infer a measuring-junction temperature from a voltage and a known reference temperature. The reference material and the reference temperature are distinct choices: B specifies the comparison Seebeck coefficient, while $T_c$ specifies the lower limit of the integral. The two-material experiment in Seebeck effect gives the geometry, sign convention, and calibration limitations.

The useful measurement is a pair: the raw terminal electric potential difference and the thermocouple reference junction temperature. For the A-positive terminal order and homogeneous-wire assumptions of Seebeck effect, write the open-circuit electromotive force asHere $T_h$ is the thermocouple measuring junction temperature and $T_c$ is the common temperature of the reference connections. The subscript $h$ names the measuring end; it need not be hotter. A voltmeter senses the complete connected thermoelectric path through the wires and its leads, in the electrochemical potential convention already described under Seebeck effect. It does not measure an absolute temperature, the individual Seebeck coefficient of A, or an isolated contact electric potential.

A type-I band alignment confines electrons and electron holes together: its inner conduction band edge is lower, its valence band edge higher.^[ref]

In the chosen band offset example, the electron barrier is $0.6$, while the electron hole barrier is $0.4$. A hole's excitation energy is measured downward from a valence band maximum: reversing that convention would incorrectly make the two barriers appear to repel opposite carriers. An electrostatic shift $-q\varphi$ moves both electron energy band edges together; changing the band gap also changes their separation. Thus material composition supplies a control beyond the electric potential $\varphi$ alone.

For real point charges $Q$ and $-2Q$ at $(0,b)$ and $(a,b)$ above the plane $y=0$, the image charges are $-Q$ and $+2Q$ at the reflected positions. The image charges enforce constant electric potential on the plane. The physical forces follow by summing ordinary Coulomb's law forces on each real charge; the electrostatic-energy factor of one half for induced charges does not multiply this force. The smaller positive charge can have an upward total force component, even though its own image attracts it downward.

In a homogeneous Ohmic conductor, a steady electric current density obeys $\nabla \cdot \mathbf{j}=0$ and $\nabla \times \mathbf{j}=0$. Uniform axial current in a straight cylinder follows if its side is insulated and its flat end faces are equipotentials. The electric potential then solves Laplace's equation with those boundary conditions, and uniqueness makes it linear in the axial coordinate. Steadiness alone does not impose these contact conditions: nonuniform end potentials can produce nonuniform current even with an insulating side. Bending also generally redistributes current; in an annular conductor the azimuthal current density is proportional to the inverse radius.

A sphere with constant electric polarization $\mathbf{P}=P\hat{\mathbf{z}}$ has zero volume bound charge and surface bound charge density ${\sigma }_b=P\cos \theta$. Integrating $\mathbf{r}{\sigma }_{b}dA$ gives the displayed electric dipole moment. The surface distribution is purely degree one in spherical harmonics, so its external static electric potential is exactly dipolar. Its interior field is finite and is not that of a point singularity.

A spherical cavity of radius $a$ filled with fixed uniform charge density $\rho$ and surrounded by a grounded conductor has electric field $\rho r\hat{\mathbf{r}}/(3{ϵ}_0)$ inside and zero electric field outside the conductor when no external sources are present. Gauss's law supplies the interior electric field; integration from $V(a)=0$ gives the displayed electric potential. Its stored electrostatic energy is $2\pi {\rho }^2{a}^5/(45{ϵ}_0)$, obtained by integrating ${ϵ}_0|\mathbf{E}{|}^2/2$ over the cavity. Grounding permits the conductor's net electric charge to change.

In a vacuum domain with specified charge density, fixed boundary electric potential and the appropriate condition at infinity, two solutions of the Poisson equation have difference $w$ satisfying ${\nabla }^{2}w=0$ with zero boundary data. Integration by parts gives $\int |\nabla w{|}^{2}dV=0$; hence their electric fields coincide, and their potentials coincide after fixing the constant. This is the Dirichlet case of the more general dielectric uniqueness result. It justifies the exterior field constructed by the method of images.

A circuit node held close to zero electric potential by active feedback, without being directly connected to the reference conductor. In an ideal linear operational amplifier with its non-inverting input grounded and stable negative feedback, $V_{-}=V_{+}=0$ while neither input draws electric current. External branch currents meeting the inverting node therefore obey Kirchhoff's current law without flowing into the amplifier input. Equality of the input potentials fails when the output saturates or the feedback ceases to enforce linear operation.

The volt is the SI unit of electric potential and electric potential difference: $1\mathrm{V}=1J/C$, one joule per coulomb. A positive electric charge of $2\mathrm{C}$ moved through a potential increase of $3\mathrm{V}$ gains $\Delta U=q\Delta \Phi =6\mathrm{J}$ of potential energy; the electric field does $-6\mathrm{J}$ of work. ^[ref].

First disconnect the detector. Each side is a voltage divider. Ohm's law and Kirchhoff's voltage law give the open-circuit intermediate electric potentials and their difference:Setting this difference to zero and cross-multiplying givesThe balance compares ratios, not necessarily equal arm values or equal branch electric currents. At balance the detector carries no electric current, so Kirchhoff's current law again makes the electric current constant along each two-arm path.^[ref]

For a worked null measurement, suppose $R_1=1.000\mathrm{k\Omega }$ and $R_2=2.000\mathrm{k\Omega }$, and the detector null occurs at adjustable $R_3=1.500\mathrm{k\Omega }$. ThenFor $V_s=3.000\mathrm{V}$, both intermediate electric potentials are $2.000\mathrm{V}$. The left branch carries $1.000\mathrm{mA}$ and the right branch $0.6667\mathrm{mA}$: the zero detector reading does not mean that the unknown carries no electric current. These are stipulated ideal values illustrating the calculation, not experimental observations.

A real voltmeter has finite input electrical resistance $R_d$; a galvanometer also has a finite winding electrical resistance. Away from balance, its electric current changes both intermediate electric potentials. This is voltmeter loading when the detector is a voltmeter. Model the detector as a passive resistor, with no offset or injected electric current. Thévenin's theorem supplies the full response without pretending the connected bridge remains two independent voltage dividers.

With the detector removed, the equivalent source is $V_{\mathrm{oc}}$. Replace the ideal supply by a wire to calculate the source electrical resistance seen between $B,C$: nodes $A,D$ merge, making $R_1,R_2$ parallel and $R_3,R_x$ parallel. ConsequentlySince the attenuation factor is strictly positive for $R_d>0$, the null remains $R_2{R}_3=R_1{R}_x$. A smaller detector electrical resistance reduces the measured electric potential difference off balance but does not shift the ideal null. Its electric current sensitivity instead depends on $1/(R_{\mathrm{th}}+R_d)$; detector gain and resolution also matter in deciding how closely a null can be located.

In the plotted example, $R_x=1.100\mathrm{k\Omega }$ gives $V_{\mathrm{oc}}=-119.048\mathrm{mV}$ and $R_{\mathrm{th}}=1.02381\mathrm{k\Omega }$. A detector with $R_d=1.000\mathrm{k\Omega }$ instead reads $V_d=-58.824\mathrm{mV}$. Using that loaded reading in the unloaded voltage divider formula would underestimate the deviation of $R_x$ from $1\mathrm{k\Omega }$. Balancing the bridge avoids this particular systematic measurement error.

The local derivative makes the sensitivity explicit. For $R_1=R_2=R_3=R$ and $R_x=R+\delta R$, the first-order Taylor expansion isAt the plotted null, the open-circuit slope is $-1.25mV/\Omega$, compared with $-0.625mV/\Omega$ for $R_d=1\mathrm{k\Omega }$. Increasing the supply improves this ideal sensitivity but also increases Joule heating. A finite source electrical resistance requires recomputing $R_{\mathrm{th}}$; the displayed parallel-pair expression assumes an ideal supply.