Electric Potential from the Field

You’ll be able to

Define potential and compute potential differences from V = −∫E·dl
Derive V = kq/r for a point charge and build V for rings and discs by integrating dV = k·dq/r
Exploit the scalar nature of potential in superposition problems

Potential is potential energy per unit charge

Divide the potential energy of a test charge by the charge itself and you get a property of space alone: the electric potential V = U/q₀, measured in volts (1 V = 1 J/C). Just as E is force per unit charge, V is energy per unit charge — and the two are linked by the same integral that linked force and energy: the potential difference between two points is the negative line integral of the field along any path between them.

Potential difference from the field

V_b − V_a = −∫ₐᵇ →E·d→l

Path-independent. Moving *with* the field, potential drops; moving against it, potential rises. For a uniform field along a straight displacement d: ΔV = −E·d.

Worked example

Derive the potential a distance r from a point charge q, taking V = 0 at infinity, then evaluate it 0.30 m from a +5.0 μC charge.

1.Integrate inward along a radial path from infinity to r. On this path →E·d→l = (kq/r′²) dr′, since both E and dl are radial.
2.V(r) − V(∞) = −∫∞^r (kq/r′²) dr′ = −kq·[−1/r′] from ∞ to r.
3.Evaluate: −kq·(−1/r − 0) = kq/r. So V(r) = kq/r with V(∞) = 0.
4.Note V carries the sign of q, and falls off as 1/r — slower than the 1/r² field.
5.Numerically: V = (8.99 × 10⁹)(5.0 × 10⁻⁶)/0.30 = (4.50 × 10⁴)/0.30 = 1.5 × 10⁵ V.

Answer: V(r) = kq/r; here V ≈ 1.5 × 10⁵ V

Potential of point charges and distributions

V = k·q/r, V = Σ k·qᵢ/rᵢ, V = ∫ k·dq/r

All plain scalar sums — signed numbers, never components. This is why V is often far easier to compute than E.

Superposition without vectors

Because potential is a scalar, superposition is just addition of signed numbers: V = Σ kqᵢ/rᵢ. No angles, no components, no cancellation diagrams. For a continuous body, slice it into elements dq and integrate dV = k·dq/r. Compare this with Unit 1’s field integrals, where every dE had to be resolved into components first — the potential integral skips that step entirely, which is exactly why the usual strategy is find V first, then differentiate to get E (next lesson).

Worked example

A ring of radius R = 0.40 m carries charge Q = +5.0 nC spread uniformly. Find the potential on its axis a distance x from the center, then evaluate at x = 0.30 m.

1.Every element dq of the ring sits at the same distance from the axial point: r = √(x² + R²).
2.So dV = k·dq/√(x² + R²) with a common denominator, and the integral is trivial: V = k/√(x² + R²) · ∫dq = kQ/√(x² + R²).
3.No symmetry argument about canceling components was needed — potential has no components to cancel.
4.Limits check: at x = 0, V = kQ/R (nonzero, even though E = 0 there); for x ≫ R, V ≈ kQ/x, a point charge.
5.Numerically: √(x² + R²) = √(0.09 + 0.16) = 0.50 m, so V = (8.99 × 10⁹)(5.0 × 10⁻⁹)/0.50 = 45.0/0.50 ≈ 90 V.

Answer: V(x) = kQ/√(x² + R²); at x = 0.30 m, V ≈ 90 V

From ring to disc: integrate over rings

A uniformly charged disc (surface density σ, radius R) is a stack of concentric rings. A ring of radius r′ and width dr′ carries dq = σ·(2πr′ dr′) and contributes dV = k·dq/√(x² + r′²). Using k·2πσ = σ/(2ε₀), the axial potential is V = (σ/2ε₀) ∫₀ᴿ r′ dr′/√(x² + r′²) = (σ/2ε₀)·(√(x² + R²) − x). The antiderivative of r′/√(x² + r′²) is just √(x² + r′²) — a substitution u = x² + r′² makes it a one-liner. Far away this reduces to kQ/x, and near the disc it approaches the infinite-plane result.

Watch out

Potential is a SCALAR. There is no such thing as “the x-component of V,” and potentials never “cancel by symmetry” the way field vectors do — two equal positive charges give double the potential at the midpoint, even though their fields cancel there. Adding potentials with cosines is the classic way to ruin an otherwise easy problem.

Checkpoint

A uniform field E = 400 N/C points in the +x direction. What is V(x = 0.50 m) − V(0)?

Checkpoint

An electron moves from a point at 50 V to a point at 150 V. Its electric potential energy: (e = 1.6 × 10⁻¹⁹ C)

On the exam

Sign conventions to lock in for the exam: ΔV = −∫→E·d→l (potential drops along the field), U = qV and ΔU = qΔV (signs of q included), and W_field = −ΔU. A positive charge released from rest falls toward lower V; a negative charge toward higher V. Nearly every potential FRQ point hinges on one of these signs.

Answer the 2 checkpoints as you read.