Electric potential
Electric potential is a way to describe an electric field using energy per unit charge. Because potential is a scalar, many multi-charge problems become easier: add potentials algebraically first, then use the result to reason about energy or recover the field.
Voltage is energy per unit charge
ΔV = ΔU/q and therefore ΔU = qΔV
1 volt = 1 joule per coulomb.
Potential difference belongs to two locations in the field; electric potential energy belongs to a particular charge placed there. The same ΔV can increase U for a positive charge and decrease U for a negative one.
Potential of point charges
V = kQ/r
For several point charges, potential obeys superposition: V = Σ(kQi/ri). Unlike electric-field vectors, these terms are added as signed scalars.
Point-charge potential
For an isolated point charge, V = kQ/r with zero potential chosen at infinity.
From potential back to the electric field
The electric field points toward decreasing electric potential. In one dimension, Ex = −dV/dx. A steep change in V with position corresponds to a strong field; a flat potential corresponds locally to a small field.
Moving along an equipotential gives ΔV = 0, so electrostatic work associated with that displacement is zero. Field lines therefore meet equipotential surfaces at right angles.
Choosing a zero of potential
Only potential differences affect measurable energy changes, so the zero level is a reference choice. For isolated charges, V = 0 at infinity is convenient. In circuits and laboratory systems, another point may be defined as ground or zero volts.
Changing the reference shifts all potentials by the same constant but leaves ΔV, electric fields and physical predictions unchanged.
Energy reasoning and common sign mistakes
Electric potential itself is not electric potential energy. A negative potential is not automatically a low energy state for every particle: U = qV reverses the ordering for negative q. Likewise, a point can have V = 0 while E is nonzero if positive and negative source contributions cancel in potential but not in field.
When comparing two points, work done by the electric field is Wfield = −ΔU = −qΔV. A positive charge released from rest tends to accelerate toward lower V; a negative charge tends toward higher V. This energy rule is often quicker than trying to infer force from a memorized sign pattern.
Worked examples
1. Potential of a point charge
Find V at 0.30 m from a +4.0 nC point charge, taking V = 0 at infinity.
Solution
V = kQ/r = (8.99 × 109)(4.0 × 10−9)/0.30 ≈ 120 V.
2. Energy change from voltage
A +2.0 μC charge moves through ΔV = −150 V. Find ΔU.
Solution
ΔU = qΔV = (2.0 × 10−6)(−150) = −3.0 × 10−4 J. Its electric potential energy decreases.
3. Uniform field from potential change
Potential falls by 600 V over 0.20 m in a region where the field is uniform and parallel to the displacement. Find the field magnitude.
Solution
For a uniform one-dimensional field, |E| = |ΔV|/Δx = 600/0.20 = 3.0 × 103 V m−1, directed toward lower potential.