Electric potential ($\Phi$)

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 ideal infinite sheet midway between parallel grounded plates has equal fields of magnitude $\sigma /(2{ϵ}_0)$ on its two sides, directed from positive sheet charge toward the plates. Integrating over the half-separation d/2 gives the displayed electric potential. Each plate carries induced inner-face surface charge density $-\sigma /2$. The potential is triangular between the plates, and the field jumps by $\sigma /{ϵ}_0$ across the sheet. Edge fields are excluded.

For an infinite circular cylinder of radius $R$, assume the electrostatic potential is independent of the axial coordinate. An angular Fourier mode $s_n\cos n\theta$, or $s_n\sin n\theta$, in the prescribed jump ${\partial }_r\Phi {|}_{R^{+}}-{\partial }_r\Phi {|}_{R^{-}}=-\sigma (\theta )$ produces the same angular mode in the potential, multiplied byHere $r_{<}=\min (r,R)$ and $r_{>}=\max (r,R)$ in the abbreviated header formula. Separation of variables selects $r^n$ inside by regularity and $r^{-n}$ outside by decay. Continuity equates their boundary values, and the derivative jump is minus $2n/R$ times that value. A nonzero angular mean of the surface charge density introduces a logarithmic term, so zero mean is necessary for a potential decaying at infinity in this two-dimensional reduction.

A voltmeter measures an electric potential difference between two contacts; its finite input electrical resistance can cause voltmeter loading.

Voltmeter loading is the change in an electric potential difference caused by the electric current drawn by the measuring voltmeter.

For a loaded voltage divider in steady direct current, a voltmeter connected across the lower resistor places its finite input electrical resistance $R_m$ in parallel with the lower arm's electrical resistance $R_2$ and any intended electrical load of electrical resistance $R_L$. With positive, fixed electrical resistances, an upper resistor of electrical resistance $R_1$, and an ideal voltage source supplying electric potential difference $V_{\mathrm{in}}$, Kirchhoff's current law and Ohm's law giveHere $V_m$ is the connected circuit's electric potential difference across the voltmeter; omit $1/R_L$ if there is no intended electrical load. The parallel connection and need for large input electrical resistance are standard voltmeter operating principles.^[ref]

Voltmeter loading is a bias relative to the target electrical circuit with the voltmeter absent and any intended electrical load still present. Voltage divider loading error instead compares the attached-load output with the wholly unloaded output. An accurate readout of $V_m$, supported by measurement calibration, can still differ from either target: calibration accuracy concerns the indication of the connected electric potential difference, whereas voltmeter loading changes that electric potential difference itself.

Conductive-paper electric potential mapping uses a voltmeter to locate equipotentials on a resistive sheet, then infers the in-plane electric field from the measured electric potential differences.

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].

A particle constrained to a surface can have a stable minimum of electrostatic potential restricted to that surface even when the unrestricted equilibrium is unstable. Away from source charges, Laplace's equation makes the full potential Hessian matrix trace zero, which rules out a positive-definite three-dimensional Hessian but does not rule out a positive-definite restriction to two dimensions. Equal positive charges at the corners of a square give a planar minimum at its centre and an unstable perpendicular direction.

For two electrostatic potentials satisfying the Poisson equation $\Delta {\varphi }_i=-{\rho }_i/{\epsilon }_0$, Green's second identity yields the displayed reciprocity formula. The product rule for divergence applied to ${\varphi }_1\nabla {\varphi }_2-{\varphi }_2\nabla {\varphi }_1$ cancels the cross-gradient terms. The normal points outward from the integration region; for an exterior region this means inward toward the excluded object on its inner boundary. Boundary terms may be removed only under suitable decay or boundary conditions.

In a grounded conducting enclosure, let the harmonic function $w_j$ equal one on boundary component $j$ and zero on the others. A point charge $q$ at ${\mathbf{r}}_0$ induces total electric charge $Q_j=-q{w}_j({\mathbf{r}}_0)$ on that component. The weighting harmonic function is its harmonic measure.

To prove the identity, let $\varphi$ be the zero-boundary electric potential, so ${\nabla }^2\varphi =-q{\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}_0)/{ϵ}_0$. Green's second identity gives $-q{w}_j({\mathbf{r}}_0)/{ϵ}_0={\int }_{\partial D}{w}_j{\partial }_n\varphi dS$, where $\mathbf{n}$ points out of the vacuum domain into the metal. The electrostatic boundary conditions at a conductor give $\sigma ={ϵ}_0{\partial }_n\varphi$, proving the result. In an unbounded domain require the additional boundary term at infinity to vanish. Between concentric spheres, $w_a(r)=a(b-r)/(r(b-a))$; between parallel planes $z=0,d$, $w_0=1-z/d$ and $w_d=z/d$.

An image charge is a fictitious point charge placed outside the physical solution region so that its potential, combined with the real sources, enforces the required conductor boundary values. The Uniqueness of the Dirichlet problem then identifies the constructed potential with the physical one. The image need not describe the actual induced surface-charge distribution.

For point electric charges at heights $h_j>0$ on one normal above an infinite grounded conducting plane, the method of images places opposite image charges at $-h_j$. At the point directly below them, tangential electric fields cancel and normal components add, giving the displayed one-sided field. The electric field inside the metal is zero instead. The boundary jump is supported by induced surface charge.

A normal electric dipole comprising $+q$ at height $h$ and $-q$ at height $h+d$, with $d\ll h$, induces the displayed leading surface charge density on a grounded plane. The image charge construction first gives ${\sigma }_q(a)=-qh/[2\pi (a^2+h^2{)}^{3/2}]$ for one point charge; subtraction at heights $h$ and $h+d$ gives $-d{\partial }_h{\sigma }_q$ to first order. The central negative region ends at $a=\sqrt{2}h$, outside which the induced surface charge density is positive. Integrating over a disk gives $Q(<a)=-qd{a}^2/(a^2+h^2{)}^{3/2}$ to this order, so the total induced electric charge is zero. At finite $d$, the exact sign-change radius obeys $a^2=h^{2/3}(h+d{)}^{2/3}[h^{2/3}+(h+d{)}^{2/3}]$.

For radii $a<b$ and a point charge $q$ at $r_0{\mathbf{e}}_z$, $a<r_0<b$, set $k=b/a$. The method of images gives the two families $(q{k}^n,r_0{k}^{2n}{\mathbf{e}}_z)$ and $(-qa{k}^n/r_0,a^2{k}^{2n}{\mathbf{e}}_z/r_0)$ for $n\in \mathbb{Z}$. The first family's $n=0$ term is the physical point charge. All other image charges lie outside the gap. Sum their Coulomb's law electric potentials with factor $1/(4\pi {ϵ}_0)$.

For a short proof, reflection of $(Q,z)$ in a grounded sphere of radius $R$ gives $(-RQ/z,R^2/z)$. On $r=a$ this pairs first-family index $n$ with second-family index $-n$; on $r=b$ it pairs $n$ with $1-n$. Each pair has zero boundary electric potential. The negative-index electric potential tails are $O(k^n)$ and the positive-index tails are $O(k^{-n})$, uniformly on the closed gap after excluding the physical singularity. These geometric series establish absolute convergence, boundary cancellation, and termwise differentiation away from the physical point charge. Uniqueness of Poisson equation identifies the solution. A spherical Gaussian surface just outside $a$ encloses the first family with $n\le -1$ and the second with $n\le 0$, so Gauss's law and the geometric series give $Q_a=-qa(b-r_0)/(r_0(b-a))$.

For conducting sheets $y=\pm x$ at common zero potential and a charge in $y>|x|$, successive reflections give two negative images and one positive double-reflection image. On the bisector, the negative images lie at $(\pm y,0)$ and the positive image at $(0,-y)$. Summing their fields gives the displayed downward force. Image forces exclude the real charge's self-field; replacing the induced charge by images does not make the images physical particles.

The electrostatic bisector force is laterally anti-restoring, with the positive stiffness coefficient on the right. A spring of tension $T$ and length $\ell$ adds transverse restoring stiffness $T/\ell$ when its fixed support is directly above the charge. The displayed inequality therefore distinguishes actual transverse stability from the saddle of a model with only a vertical spring force. Axial spring stiffness alone does not specify the transverse verdict. At fixed finite support geometry and positive gravitational load, sufficiently small electric charge preserves lateral stability: the destabilizing coefficient tends to zero as $q^2$, while the tension term approaches $mg/\ell >0$. The vertical equilibrium must separately satisfy its own positive-stiffness condition.

For an attractive force $-\beta /y^2$ and an otherwise linear vertical spring, differentiation about the charged equilibrium $y_e$ reduces the spring stiffness by $2\beta /y_e^3$. If the uncharged height is $y_0$ and the shift is small, the fractional frequency change is $-\beta /(k{y}_0^3)$. Using only the force value, without its derivative, would miss this frequency shift.

For a point electric charge $q$ in a homogeneous medium of permittivity ${ϵ}_1$ beside a plane interface with permittivity ${ϵ}_2$, its potential on the source side is reproduced by a mirrored image charge $q^{\prime }=q({ϵ}_1-{ϵ}_2)/({ϵ}_1+{ϵ}_2)$. The interface has no free surface charge; continuity of tangential electric field and normal electric displacement field fixes this coefficient.

For a real charge $Q$ in medium 1, the potential in medium 2 can be represented by the source $Q_t=2{ϵ}_{2}Q/({ϵ}_1+{ϵ}_2)$ using medium 2's Coulomb prefactor $1/(4\pi {ϵ}_2)$. The same potential uses $2{ϵ}_{1}Q/({ϵ}_1+{ϵ}_2)$ if medium 1's prefactor is retained instead. These different effective source labels do not change the physical field. A decomposition $Q_t=Q+Q^{\prime }$ must state which prefactor it uses.

The image of a charge $q$ at height $z$ is $-q$ at height $-z$. Their separation is $2z$, giving force $-q^2/(16\pi {\epsilon }_0{z}^2)$. Starting at rest at height $d$, energy conservation gives ${\dot{z}}^2=q^2(1/z-1/d)/(8\pi {\epsilon }_{0}m)$. Integrating the reciprocal speed to the plane yields the stated finite time in the electrostatic Newtonian model. The induced-conductor potential energy is $-q^2/(16\pi {\epsilon }_{0}z)$; using the full image-pair energy would double the force incorrectly.

A real charge of height $z$ has a reflected image charge at $-z$. Its acceleration is $\ddot{z}=-\kappa /z^2$, with $\kappa =q^2/(16\pi {\epsilon }_{0}m)$. Released from rest at height $d$, its ideal-model arrival time isThe image follows the instantaneous reflected position. The induced-charge energy carries one-half the full real-image pair interaction factor, consistent with this force.

For a charge $q$ at $\mathbf{p}$, $R=|\mathbf{p}|>a$, outside a grounded sphere of radius $a$, the displayed image gives zero potential on the sphere and decay at infinity. On $|\mathbf{x}|=a$, one has $|\mathbf{x}-{\mathbf{p}}^{\prime }|=(a/R)|\mathbf{x}-\mathbf{p}|$, proving boundary cancellation. The induced force is inward, of magnitude $q^{2}aR/[4\pi {\epsilon }_0(R^2-a^2{)}^2]$, from Coulomb's law applied to the image field. The self-field is excluded.

Place equal positive point charges at distances $d$ on opposite sides of a grounded conducting sphere of radius $a$, and set $x=d/a>1$. Each physical charge has an image $-qa/d$ at distance $a^2/d$ from the centre on its own side. The repulsion from the other physical charge balances attraction to both images precisely when the displayed equation holds. At large $x$, the two sides behave as $1/(4x^2)$ and $2/x^3$, explaining the approximate root $x\simeq 8$. A constant nonzero sphere potential adds a central image and changes the force balance.

A real charge $q$ at $d>a$ outside a spherical conductor of radius $a$ has the usual grounded-sphere image charge $q^{\prime }=-aq/d$ at $x^{\prime }=a^2/d$. Adding a central image $q_0=-q^{\prime }$ makes the total image charge zero, representing an isolated neutral conductor. The off-centre pair has zero potential on $r=a$; the central image therefore sets the constant conductor potential to $q/(4\pi {\epsilon }_{0}d)$. Its induced electric dipole moment is $p_x=q^{\prime }{x}^{\prime }=-q{a}^3/d^2$. Keeping the applied central field ${\mathbf{E}}_0=-q\hat{\mathbf{x}}/(4\pi {\epsilon }_0{d}^2)$ fixed gives $\mathbf{p}=4\pi {\epsilon }_0{a}^3{\mathbf{E}}_0$. As $d\to \infty$, the external field becomes uniform near the sphere; its divergent constant potential can be removed without affecting the electric field.

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.

The electric potential difference between the ground-contact points of the two feet. A ground current can make it nonzero.

An applied potential difference that sets a device operating condition or shifts reservoir electrochemical potentials. In a multi-terminal device the grounded lead, biased lead and gate reference must be stated; they determine the relation between measured slopes and capacitances.

A point charge idealizes electric charge concentrated at one point. In vacuum, a charge $Q$ at the origin has potential $Q/(4\pi {ϵ}_{0}r)$.

An equipotential is a level set of the electric potential. The electric field is normal to every smooth equipotential.