Electric potential and potential difference

Put a small test charge q₀ at a point in a fixed arrangement of source charges and it has electric potential energy U there. Double q₀ and U doubles, because the force on it doubled and so did the work of bringing it in; triple q₀ and U triples. The ratio U/q₀ therefore does not depend on the probe at all, and that ratio is the electric potential V at the point. One joule per coulomb is one volt. A 2.0 nC probe carrying 3.0 × 10⁻⁶ J sits where V = 1500 V, and swapping it for a 4.0 nC probe gives 6.0 × 10⁻⁶ J and the same 1500 V. Two details keep the definition honest. The probe must be small enough not to shove the sources around, so strictly V is the q₀ → 0 limit; and V belongs to every point of the region, occupied or not, so the map exists before any charge is placed on it.

Going the other way, from E to V, is a line integral: V_B − V_A = −∫_A^B E⋅dl, where dl is a step along whatever path you choose from A to B. Operationally, chop the path into steps, take E⋅dl = E dl cos θ on each one, add them, and flip the sign. That minus sign is bookkeeping for work: if the field pushes a positive charge along the path it does positive work, the charge loses potential energy, and V must fall. Two features make the sum easy. A step perpendicular to E contributes nothing, so only displacement along the field counts, and for a uniform field the sum collapses to V_B − V_A = −E d. Across plates with E = 2.0 × 10⁴ V m⁻¹ and a 5.0 mm gap that is −100 V. Walk instead along a slanted 0.080 m path at 60° to a 1500 V m⁻¹ field and you get −1500 × 0.080 × cos 60° = −60 V, exactly what the 0.040 m of along-field displacement gives on its own.

Nothing in the definition fixes where V = 0. Replace V by V + C everywhere, with C any constant, and every difference V_B − V_A, every work q(V_A − V_B) and every field is untouched, because the constant cancels in each. So the zero is chosen for convenience. For a charge distribution of finite size the usual choice is V → 0 infinitely far away, which is what lets the potential near an isolated charge be quoted as a single number. In a circuit the chosen point is the ground terminal; in a car it is the chassis. Change the choice and every potential shifts together: a 9.0 V battery can sit at 0 V and 9.0 V, or at −4.5 V and +4.5 V, or at 1000 V and 1009 V if the whole apparatus is raised on an insulated platform, and nothing it does changes. It is also why a bird on an 11 kV line is safe — both feet are on one conductor, a few millivolts apart — and why a voltmeter has two leads. One lead would have nothing to measure.

Potential energy comes back by multiplying: U = qV, and the work the field does in carrying a charge from A to B is W = q(V_A − V_B), which is the same as −q ΔV. Track the sign of q. A +3.0 μC charge falling through a 200 V drop has the field do W = 3.0 × 10⁻⁶ × 200 = 6.0 × 10⁻⁴ J on it, and its potential energy falls by that amount. Send a −3.0 μC charge through the same drop and W = −6.0 × 10⁻⁴ J: something external has to push it. Left alone, a positive charge accelerates from high potential to low and a negative charge from low to high, and both lose potential energy, because U = qV falls in both cases. That is the whole sign rule. Downhill is a statement about U, not about V, and the two coincide only when q is positive.

Release a charge from rest in vacuum and let the field do all the work. Energy conservation gives ½mv² = |q||ΔV|, so v = √(2|q||ΔV|/m). An electron crossing 500 V picks up 500 × 1.602 × 10⁻¹⁹ = 8.01 × 10⁻¹⁷ J, and with m = 9.11 × 10⁻³¹ kg that is v = 1.33 × 10⁷ m s⁻¹, about 4.4% of the speed of light, fast but still safely non-relativistic. Because that first step is always charge times voltage, particle physics keeps the voltage as the unit of energy: one electronvolt is what one elementary charge gains across 1 V, and 1 eV = 1.602 × 10⁻¹⁹ J. A proton and an electron crossing the same 500 V each gain 500 eV. What differs is the direction they move and the speed they reach, since the proton's mass is 1836 times larger and its speed smaller by √1836 ≈ 43.

The construction rests on one property: for charges at rest, the work done in carrying a test charge around any closed loop is zero, so ∮E⋅dl = 0. That is what makes −∫E⋅dl depend on the endpoints alone, and it is what lets one number V be attached to each point. Were the loop integral non-zero you could go round and return to a different potential at the same place, and V would not be a function of position at all. Electrostatic fields pass the test because Coulomb's law is a central inverse-square force. A field driven by a changing magnetic flux does not: its loop integral equals minus the rate of change of that flux, and no single-valued potential exists for it. Keep the units apart while you are here, too. V is joules per coulomb, E is volts per metre, U is joules, and no algebra turns one into another without supplying a charge or a length.