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An electrical reference potential fixes the arbitrary additive constant of an electric potential. Assigning zero to ground makes other potentials differences from that baseline; it does not assert that the electric field vanishes there. If a source calculation gives $\Phi (a)=89.9\mathrm{V}$ and $\Phi (b)=179.8\mathrm{V}$ relative to infinity, choosing $a$ as the zero instead gives ${\Phi }^{\prime }(a)=0$ and ${\Phi }^{\prime }(b)=89.9\mathrm{V}$, with the same electric potential difference. Grounding an electrical conductor additionally connects it to a reservoir that can exchange electric charge to maintain the prescribed electric potential; that physical boundary condition differs from merely relabeling an unchanged solution.^[ref]; ^[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.

Electrostatics studies time-independent electric charge distributions and their electric fields. In vacuum, its local equations in SI units areHere $\rho$ is charge density, ${ϵ}_0$ is vacuum permittivity, and $\Phi$ is electric potential. Gauss's law supplies the divergence equation; the static form of Faraday's law of induction supplies the zero curl. The Poisson equation becomes Laplace's equation where there is no electric charge. Prescribed electric charges give electric fields through Coulomb's law and the superposition principle; mobile electric charges on an electrical conductor instead require solving for their distribution together with the electric potential. ^[ref].

The useful distinction is between prescribed sources and prescribed boundary conditions. On a grounded electrical conductor, an external connection fixes its electric potential relative to an electrical reference potential and allows electric charge to enter or leave. On an isolated electrical conductor, total electric charge is fixed and its constant electric potential is an unknown. A neutral isolated electrical conductor can therefore have a nonzero local surface charge density through electrostatic induction. Equal electric potentials also do not imply equal electric charges: charge sharing between distant connected spheres gives a simple geometric example. ^[ref].

Electrostatics is useful because a spatial boundary value problem replaces the need to follow every moving charge carrier during charge relaxation. It describes the settled state of a capacitor and explains electrostatic shielding of sensitive equipment. Its assumptions must match the apparatus: stationary boundaries and sources, negligible time variation of the magnetic field, and enough time for charge relaxation. A metal carrying a maintained electric current through finite electrical resistance generally needs a nonzero internal electric field by Ohm's law; it is not a conductor in electrostatic equilibrium. Static macroscopic electric charge also does not mean that every electron is motionless. ^[ref].

A voltage divider uses two resistors connected in series to obtain a fraction of an applied electric potential difference. Let the upper electrical resistance be $R_1>0$, the lower one $R_2>0$, and the input be $V_{\mathrm{in}}=V_s$ relative to a common electrical reference potential. The output is the junction's electric potential difference relative to that same reference, across $R_2$. The familiar ratio assumes zero output electric current: no external electrical load draws from the junction. In a steady direct current model with fixed resistors obeying Ohm's law and an ideal voltage source,Here $V_0$ is the open-circuit voltage; $R_1,R_2$ correspond to $R_a,R_b$ in the header formula. With no junction branch, Kirchhoff's current law makes the same electric current pass through both resistors, and Ohm's law gives the two voltage drops. For $V_s\ne 0$, division gives $V_0/V_s=R_2/(R_1+R_2)$; if $V_s=0$, the predicted output is zero but this ratio is undefined.

An original stipulated example has $R_1=R_2=10\mathrm{k\Omega }$ and $V_{\mathrm{in}}=10\mathrm{V}$. With zero output electric current, $I=0.5\mathrm{mA}$ and $V_0=5\mathrm{V}$. An attached $10\mathrm{k\Omega }$ electrical load instead gives $V_{\mathrm{out}}=10/3\mathrm{V}$; a $1\mathrm{M\Omega }$ load gives $V_{\mathrm{out}}=1000/201\mathrm{V}=4.975124378\dots \mathrm{V}$. These are exact model predictions, not measured readings or specifications of a particular multimeter or oscilloscope.

The divider is useful for scaling a voltage or converting a change in a sensor's electrical resistance into an electric potential difference. Its beauty is also its trap: multiplying both electrical resistances by the same positive factor preserves the unloaded ratio, but changes the effect of any fixed electrical load. A divider therefore supplies a predictable ratio under specified loading conditions; it is not a source that maintains that voltage for arbitrary output electric currents. A Wheatstone bridge compares two divider junctions, while Thévenin's theorem describes the divider's response to many different loads.

Southern Illinois University's ET 304A guide, Lab 4: Voltage Measurements and Meter Loading Effects, documents an instructive test: build equal-arm voltage dividers on a $20\mathrm{V}$ supply, then measure across the lower arm with a digital multimeter, analog multimeter, and oscilloscope. Starting with two $10\mathrm{k\Omega }$ resistors, repeat with equal pairs from $220\mathrm{k\Omega }$ to $4.7\mathrm{M\Omega }$. It requires separate recorded readings, unloaded predictions, and predictions including each instrument's input electrical resistance. The guide obtains meter resistances from specification sheets and scope resistance from the input-panel marking. Equal ratios control the unloaded target while changing susceptibility to voltmeter loading. The provided tables are blank: the guide specifies a procedure, not measured outcomes.^[ref]

Preparation and accuracy checks are made concrete by Berkeley's Lab 1: Introductory Experiments and Linear Circuits I. Its different divider uses nominal $470\mathrm{k\Omega }$ and $10\mathrm{k\Omega }$ resistors and a $24\mathrm{V}$ supply on a breadboard. Measure isolated resistors first, record actual supply voltage, recalculate with these measured values, then record the junction voltage with the digital multimeter. It explicitly compares actual values with resistor tolerance and the final voltage with a range-dependent accuracy bound from the instrument manual. This is an uncertainty check, not an independent measurement calibration.^[ref]

For a controlled comparison, additionally record the selected instrument range, probe factor, and input coupling; connect each instrument separately to the same electrical reference potential, and monitor the input rather than treating the supply label as an exact voltage. These are proposed controls for interpreting the documented procedures. A voltage indication and a resistance indication from the same digital multimeter do not independently validate its measurement calibration; a traceable reference comparison would be needed for that. Resistor tolerance describes component variation, while measurement uncertainty describes what is known about measured values. Using nominal arms alone can confuse mismatch with loading: equal nominal arms each within one-percent tolerance can give unloaded ratios from $0.495$ to $0.505$.

The inference to test is specific: once actual arm values, input voltage, and meter input electrical resistance are accounted for, the readout should approach the loaded voltage divider prediction within the remaining measurement uncertainty. A discrepancy can instead reflect contacts, changing temperature, source droop, or an input that is not a fixed passive resistor. A source with positive series electrical resistance $R_s$ can be modeled by replacing $R_1$ with $R_1+R_s$ only when $V_s$ denotes its ideal internal voltage; when the actual voltage across the divider is measured, that terminal value is already $V_{\mathrm{in}}$. During settling, or with appreciable input capacitance and changing signals, the steady direct current calculation is insufficient. Neither institutional procedure supplies measured values for the original $10\mathrm{V}$ example above.

The Wheatstone bridge has four resistor arms and two diagonals. Here the supply holds node $A$ at constant electric potential difference $V_s>0$ relative to node $D$, our electrical reference potential. The left path is $A\to R_1\to B\to R_2\to D$; the right path is $A\to R_3\to C\to R_x\to D$. The unknown is $R_x$, and $R_3$ is adjustable. A voltmeter or galvanometer connects $B$ to $C$. We define its positive reading as $V_B-V_C$; exchanging its leads reverses the sign. All four electrical resistances are positive, the supply is ideal, and the calculation describes steady electric currents at fixed temperature.

Charles Wheatstone described the instrument in his 1843 Bakerian lecture while explicitly crediting Samuel Hunter Christie. The account from King's College London's collection identifies that attribution; the familiar name should not be read as an assertion that Wheatstone alone originated the bridge.^[ref]