02 · Intermediate → advanced

Electric potential

Start from the mechanical work of a fixed electrostatic field, derive potential differences with their sign, recover the point-charge potential and E=−∇V, use superposition and energy bookkeeping, and state when choosing V(∞)=0 is invalid (notably the infinite uniformly charged line).

230–290 minIntermediate → advanced

In this lesson

Derive ΔU=−Wfield for a quasistatically displaced charge, separating the test charge from the source.
Define ΔV=ΔU/q and obtain V(B)−V(A)=−∫A^B E·dℓ with a consistent sign.
Prove electrostatic path independence for a stationary point-source field away from its location.
Integrate a point-charge field radially to obtain V(r)=kq/r with reference at infinity under the stated assumptions.
Use the electrostatic condition to derive E=−∇V and identify the units of a potential gradient.
Superpose scalar potentials for discrete and continuous localized distributions.
Explain the additive reference freedom, why equipotentials are perpendicular to nonzero E, and why absolute potential is not an observable.
Compute potential differences for an infinite uniform line without assigning V(∞)=0.
Derive a finite on-axis dipole potential and its leading far-field term with the correct 1/r² decay.
Solve quantitative problems with sign, units, energy and limiting-case checks.

The sign question

Electric potential is not a second force law. It is the scalar energy-per-charge description of a conservative electrostatic field. We begin with the work done by the field, fix the sign once, and then derive both V and E. The sources are held static; all potential differences refer to the same chosen field configuration.

02.1Work by a static field and potential-energy change

For a charge qₜ moving by an infinitesimal displacement dℓ in a prescribed electric field E, the electrostatic force is qₜE. Its infinitesimal work on the test charge is dWfield=qₜE·dℓ. For a conservative interaction the associated potential energy decreases by the work performed by the field:

dW_field=F·dℓ=qₜE·dℓ

dU=−dW_field; U(B)−U(A)=−qₜ∫_A^BE·dℓ

Positive work by the field lowers U. If an external agent moves the charge very slowly so that its kinetic energy does not change, the work required by that agent is +ΔU, provided it counters the field force and other forces are negligible. These sign statements must not be interchanged.

02.2Define potential difference and retain the sign of the path integral

Divide the source-related energy difference by the nonzero test charge. In the ideal small-probe limit, the ratio does not depend on the sign or magnitude of the chosen probe. Define the scalar potential difference between B and A by

V(B)−V(A)=[U(B)−U(A)]/qₜ

V(B)−V(A)=−∫_A^BE·dℓ

[V]=J/C; ΔU=qₜΔV

Moving in the direction of E gives E·dℓ positive and therefore V decreases. A positive probe loses potential energy when moving that way; a negative probe gains it because its charge multiplies ΔV with the opposite sign. V is a property of the prescribed source field, not a property of the selected probe.

02.3Why the integral is independent of path in electrostatics

For a stationary point charge q at the origin, E=kq r̂/r² away from r=0. Along any differentiable path avoiding the origin, the differential of the radius is dr=r̂·dℓ. Hence

E·dℓ=kq dr/r²=−d(kq/r)

∫_A^BE·dℓ=kq(1/r_A−1/r_B)

∮E·dℓ=0

The endpoint-only result survives linear superposition of stationary charges. Locally, ∇×E=0 in electrostatics away from the idealized singular source points. In Maxwell–Faraday theory, ∮E·dℓ=−dΦB/dt: a time-varying magnetic flux can make the electric field non-conservative. We are not applying electrostatic path independence to that situation.

02.4Potential of an isolated point charge: reference at infinity

For a point source in vacuum, choose the additive reference V(∞)=0. This is permissible because kq/r tends to zero at infinity. Using the path integral radially from infinity to finite r gives

V(r)−V(∞)=−∫_∞^r(kq/r′²)dr′

V(r)=kq/r, r>0

V(B)−V(A)=kq(1/r_B−1/r_A)

V has the sign of the source charge q. For q<0, V is negative relative to the selected reference, while E points toward the negative source. At r=0, the ideal point-source potential diverges; a finite physical charge distribution requires its own model.

02.5From a scalar V back to a vector E

For an infinitesimal displacement dℓ=(dx,dy,dz), the potential difference is dV=−E·dℓ, while multivariable calculus gives dV=(∂V/∂x)dx+(∂V/∂y)dy+(∂V/∂z)dz. Since the equality holds for independently chosen infinitesimal displacements,

E=−∇V

E_x=−∂V/∂x; E_y=−∂V/∂y; E_z=−∂V/∂z

V=kq/r ⇒ E=kq r̂/r²

Check the last step in radial coordinates: −d(kq/r)/dr=+kq/r². The minus sign in E=−∇V is not arbitrary; it already followed from the work definition. A potential slope is measured in V/m, equal to N/C.

02.6Superpose scalar potentials and choose one reference

Because V is scalar, the potential of finitely many localized point charges with common zero at infinity is the algebraic sum of their individual potentials. For continuous localized distributions, integrate the source element rather than its vector direction:

V(r)=kΣ_i=1^Nqᵢ/|r−rᵢ|+C

V(r)=k∫dq(r′)/|r−r′|+C

C is one global arbitrary additive constant for the chosen field region, not a separately adjustable constant for every source after the physical reference has been fixed. E is unchanged when V→V+C because ∇C=0. Unlike fields, potentials add as signed scalars; zero potential at a point does not imply zero E there.

V₁ > 0

0 < V₂ < V₁

0 < V₃ < V₂

E = −∇V

Equipotentials: electric field crosses constant-potential surfaces

For an isolated q>0 in vacuum with V(∞)=0, V(r)=q/(4πε₀r), so Vinner>Vouter>0. The electric field E=−∇V points OUTWARD, normal to the spherical equipotentials, in the direction of decreasing V. These are two-dimensional sections of 3D equipotential spheres; electric field lines are not the closed rings.

02.7Equipotentials and energy bookkeeping

Along a displacement tangent to V=constant, dV=0. Therefore E·dℓ=0: at regular points where E is nonzero, E is normal to an equipotential surface. Neither the drawing of a field line nor a color map of V determines a particle trajectory by itself.

V=constant ⇒ dV=0 ⇒ E·dℓ_tangent=0

ΔU=qₜ[V(B)−V(A)]

W_field,A→B=−ΔU

A negative probe reverses the sign of ΔU at fixed ΔV; it does not change the equipotential surfaces created by the sources. A positive probe released from rest can accelerate down a potential gradient locally, but inertia and geometry must still be included when determining its full path.

02.8Infinite line: its potential difference exists, V(∞)=0 does not

Chapter 01 derived the ideal infinite-line radial field E(s)=2kλ/s ŝ for s>0. Integrate between two finite positive radial distances sA and sB:

V(s_B)−V(s_A)=−2kλ∫_sA^sBds/s

=−2kλ ln(s_B/s_A)

V(s)=−2kλ ln(s/s₀) if V(s₀)=0

As s→∞, this logarithm does not approach a finite limit when λ is nonzero. Thus the convenient point-charge convention V(∞)=0 is mathematically invalid for an infinite line. The potential difference between two finite radii remains well-defined. Real finite wires require their finite geometry for absolute-reference calculations.

02.9Dipole potential: exact axial expression and its far field

Put +q at z=+a and −q at z=−a with q>0. On the axis z>a choose V(∞)=0 for this localized finite pair. Superpose the two signed point-source potentials:

V(z)=kq[1/(z−a)−1/(z+a)]

=2kqa/(z²−a²)=kp/(z²−a²); p=2qa

z≫a ⇒ V(z)≈kp/z²

E_z=−dV/dz=2kpz/(z²−a²)²

The 1/z² potential and 1/z³ field are consistent: differentiating a reciprocal power adds one more inverse power. The far-field approximation cannot be used near z=a. The result is an on-axis formula only; the general angular dipole potential needs the angle between p and the observation direction.

02.10Mastery exercises with full sign checks

A +2.00 nC point source is at the origin. Find V(B)−V(A) for rA=0.100 m and rB=0.200 m. Specify which point has higher potential.
For that same move, a +3.00 nC probe travels quasistatically from A to B. Find ΔU and the work done by the field. Repeat the sign of ΔU for a negative probe.
For λ=+1.00 nC/m, compute V(0.040 m)−V(0.020 m) for the infinite-line idealization; explain why the answer does not use V(∞).
Prove from V=kq/r that the radial electric field is kq/r², and state the domain excluded by the ideal model.
For a pair +q at +a and −q at −a with q=1.00 nC, a=0.010 m and z=0.100 m, compute the exact axial V and compare with kp/z².
Explain why V=0 at a point need not imply E=0 there, using a symmetric pair of equal and opposite charges.

Worked solutions

1. ΔV=k(2.00×10⁻⁹)(1/0.200−1/0.100)=−8.99×10¹ V. The nearer point A is at higher potential for a positive source.

2. ΔU=qₜΔV=(3.00×10⁻⁹)(−89.9)≈−2.70×10⁻⁷ J; Wfield≈+2.70×10⁻⁷ J. For a negative probe ΔU reverses sign.

3. ΔV=−2k(1.00×10⁻⁹)ln(2)≈−12.5 V. The reference at infinity diverges logarithmically for nonzero λ.

4. Er=−d(kq/r)/dr=kq/r². Exclude r=0, where the point-source idealization diverges.

5. p=2qa=2.00×10⁻¹¹ C·m. Exact V=kp/(z²−a²)≈18.2 V; far field kp/z²≈18.0 V. Their small difference is consistent with a/z=0.1.

6. At the midpoint of a finite equal-and-opposite pair, scalar point potentials cancel by sign, while both vector fields point from +q toward −q and add. Thus V=0 need not imply ∇V=0.

Chapter 03 introduces Gauss's law from electrostatic flux and checks exactly when symmetry lets a flux integral determine E.