= Electric potential
{title2=$\Phi$}
{wiki}
= Electrostatic potential
{synonym}
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>,
$$
\Delta\Phi=-\int_A^B\mathbf E\cdot d\mathbf l,\qquad
\Delta U=q\Delta\Phi,\qquad
W_{\mathrm{field}}=-q\Delta\Phi,\qquad
\Delta K=-\Delta U.
$$
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>,
$$
1\,\mathrm V=1\,\mathrm{J/C}.
$$
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.<University Physics Volume 2 by Samuel J. Ling and others/7.2 Electric Potential and Potential Difference>[§7.2, equations 7.4–7.7: energy per charge, work and potential difference]{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>:
$$
\mathbf E=-\nabla\Phi,\qquad E_x=-\frac{\partial\Phi}{\partial x},\quad E_y=-\frac{\partial\Phi}{\partial y},\quad E_z=-\frac{\partial\Phi}{\partial z}.
$$
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> $U=q\Phi$.<University Physics Volume 2 by Samuel J. Ling and others/7.4 Determining Field from Potential>[§7.4, equations 7.13–7.15: components and negative gradient]{ref}
Only differences fix physical <electrostatic potentials>: adding the same constant $C_0$ everywhere, $\Phi'=\Phi+C_0$, 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>.<University Physics Volume 2 by Samuel J. Ling and others/7.3 Calculations of Electric Potential>[§7.3, equation 7.8 and Figure 7.18: infinity and ground references]{ref}
For a fixed <point charge> $Q$ at the origin in <vacuum>, choose zero at infinity. <Coulomb's law> gives
$$
\Phi(r)=\frac{kQ}{r},\qquad
\mathbf E(r)=\frac{kQ}{r^2}\widehat{\mathbf r},\qquad
k=\frac{1}{4\pi\varepsilon_0},\qquad r>0,
$$
where $\varepsilon_0$ is the <vacuum permittivity>. Use the stipulated approximation $k=8.99\times10^9\,\mathrm{N\,m^2/C^2}$ and $Q=+2\,\mathrm{nC}$. At $r=0.20\,\mathrm m$,
$$
\Phi=89.9\,\mathrm V,\qquad E_r=449.5\,\mathrm{N/C}.
$$
A test <electric charge> $q=-3\,\mathrm{nC}$ has $U=-2.697\times10^{-7}\,\mathrm J$ and radial <electric force> $F_r=-1.3485\times10^{-6}\,\mathrm N$, 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>:
$$
\Delta\Phi=+89.9\,\mathrm V,\qquad
\Delta U=-2.697\times10^{-7}\,\mathrm J,\qquad
W_{\mathrm{field}}=+2.697\times10^{-7}\,\mathrm J.
$$
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.<University Physics Volume 2 by Samuel J. Ling and others/7.3 Calculations of Electric Potential>[§7.3: point-charge potential and scalar superposition]{ref}
For a uniform <electric field> $\mathbf E=300\,\mathrm{V/m}\,\widehat{\mathbf x}$, choose $\Phi(0)=12\,\mathrm V$. Then
$$
\Phi(x)=12\,\mathrm V-(300\,\mathrm{V/m})x,\qquad
\Delta\Phi=-\mathbf E\cdot(\mathbf r_B-\mathbf r_A).
$$
Moving 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.<University Physics Volume 2 by Samuel J. Ling and others/7.4 Determining Field from Potential>[§7.4: voltage gradient in a uniform field]{ref}
\Image[/electric-potential-work-and-equipotentials.png]
{title=Analytic <electric potential> models: <point charge> <equipotentials>, a uniform <electric field>, and the sign of the change in <potential energy>}
{id=electric-potential-work-and-equipotentials}
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.<University Physics Volume 2 by Samuel J. Ling and others/7.5 Equipotential Surfaces and Conductors>[§7.5, Figures 7.30–7.32: equipotentials, perpendicular field and zero work]{ref}
The <superposition principle> adds <electric potentials> as signed <scalars>. For finitely many stationary <point charges> $q_j$ at $\mathbf r_j$ in <vacuum>,
$$
\Phi(\mathbf r)=k\sum_j\frac{q_j}{|\mathbf r-\mathbf r_j|},\qquad
\mathbf E(\mathbf r)=k\sum_jq_j\frac{\mathbf r-\mathbf r_j}{|\mathbf r-\mathbf r_j|^3}.
$$
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 $\mathbf E=-(2kq/a^2)\widehat{\mathbf x}$. With $q=2\,\mathrm{nC}$ and $a=0.10\,\mathrm m$, this is $-3596\,\mathrm{N/C}\,\widehat{\mathbf x}$. At $(2a,0,0)$, the same model gives $\Phi=2kq/(3a)\approx119.867\,\mathrm V$ and $E_x=8kq/(9a^2)\approx1598.222\,\mathrm{N/C}$. Cancellation of <electric potentials> is quite different from cancellation of <vector> <electric fields>.
\Image[/electrostatic-dipole-potential.png]
{title=Computed dipole equipotentials and electric-field directions: zero potential on the vertical midline still permits a nonzero field}
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.<Physics 2212 laboratory manual by Eric Murray/Physics 2212, Lab 5: Electric Potential and Field>{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 $\Phi(x)=(50\,\mathrm{V/m})x$. 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 $E_x=-50\,\mathrm{V/m}$. 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.<Physics 2212 laboratory manual by Eric Murray/Physics 2212, Lab 5: Electric Potential and Field>{ref}
\Image[/electric-potential-voltmeter-mapping.png]
{title=<Voltmeter> setup for <conductive-paper electric potential mapping> and a synthetic profile whose negative <derivative> gives the <electric field>}
{id=electric-potential-voltmeter-mapping}
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>.<Electricity and Magnetism laboratory manual/EC-2: Electric Fields>{ref}<University Physics Volume 2 by Samuel J. Ling and others/10.4 Electrical Measuring Instruments>[§10.4, Measuring Voltage with a Voltmeter: parallel connection and large input resistance]{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.<Electricity and Magnetism laboratory manual/EC-2: Electric Fields>{ref}<Physics 2212 laboratory manual by Eric Murray/Physics 2212, Lab 5: Electric Potential and Field>{ref}
The scalar description has a crucial limit. <Faraday's law of induction> for a stationary closed path $\Gamma$ says
$$
\oint_\Gamma\mathbf E\cdot d\mathbf l=-\frac{d\Phi_B}{dt},\qquad
\nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t},
$$
where $\Phi_B$ 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 give
$$
\mathbf E=-\nabla\Phi-\frac{\partial\mathbf A}{\partial t},\qquad
\mathbf B=\nabla\times\mathbf A.
$$
Consequently, 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>.<The Feynman Lectures on Physics, Volume II by Richard Feynman and others/Chapter 21: Solutions of Maxwell’s Equations with Currents and Charges>[§21–3]{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.<The Feynman Lectures on Physics, Volume II by Richard Feynman and others/Chapter 4: Electrostatics>{ref}
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