Point a multimeter's two probes at a battery and the screen reads something like 1.5\ \text{V}. That single number — the voltage — quietly runs your whole world: it decides which way current flows, how much energy each electron carries out of a socket, whether a spark will jump a gap. Yet nowhere in that reading is there an arrow. The electric field \vec{E} is a vector at every point in space, a thicket of directions you would have to add up head-to-tail to get anywhere. Voltage is just a number. This page is about that astonishing simplification: how the messy vector field collapses into a single scalar you can read off a dial, and why that is not cheating but a deep fact about electricity.

The whole idea rests on one sentence worth reading twice: the electric force is conservative, so instead of tracking forces we can just do energy bookkeeping. Bookkeeping with numbers is easy; steering vectors around is hard. The number we keep in the ledger is the electric potential V — the potential energy per unit charge — and it will turn out to carry all the information of the field, in a form you can add up with ordinary arithmetic.

Why bookkeeping beats vectors: the field is conservative

Lift a book and gravity does negative work on the way up; drop it and gravity hands the work back. The total work gravity does around any closed loop is zero, and the work between two points depends only on where you start and finish — never on the route. Any force with that property is called conservative, and it lets us define a potential energy: a stored number that the force spends and refunds.

The electric force from static charges behaves in exactly the same way. Carry a small test charge from point a to point b along a straight dash, or along a wild scenic detour, and the electric force does the same total work either way. This is not obvious — it is a theorem about the field, that \vec{E} is a conservative vector field (its curl vanishes, so every closed-loop integral is zero). Because of it we are allowed to define an electric potential energy U that depends only on position, and to speak of the work as a simple difference U(a) - U(b) rather than an integral we have to redo for every path.

The masterstroke is to strip out the test charge. If you double the test charge you double both the force and the energy, so the energy per unit charge is a property of the location alone. That per-charge quantity is the electric potential:

Potential is the field, integrated along a line

Force and energy are two views of the same thing, tied together by work. Moving a charge q a tiny step d\vec{l} against the field costs energy -q\,\vec{E}\cdot d\vec{l}. Divide by q and add up the steps from a to b, and you get the difference in potential:

In words: walk against the field and the potential climbs; walk with it and the potential falls. Because the field is conservative, this integral does not care which path you take between the two points, so it is a genuine function of position.

The relationship runs both ways. If potential is the field summed up along a line, then the field is the potential's rate of change in space — its gradient, with a minus sign:

Why the minus sign? Think of V as the height of a landscape and a positive charge as a ball. A ball rolls downhill, from high ground to low. The gradient \nabla V points in the direction of steepest increase of V — uphill — so the field, which pushes a positive charge downhill toward lower potential, must point the opposite way. The field is the downhill arrow of the potential hill; the minus sign is what turns "steepest ascent" into "steepest descent". Where the potential is steep, the field is strong; where V is flat, the field vanishes.

The star example: a single point charge

Do the line integral for the field of a lone point charge Q, taking the zero of potential infinitely far away (the natural reference — infinitely far from all charges, nothing is going on), and a beautifully simple result drops out:

Stare at the exponent. The field of a point charge falls off as 1/r^2, but the potential falls off only as 1/r — one power gentler, exactly as the line integral of 1/r^2 should. And because V is a scalar, its sign simply follows the sign of Q: a positive charge builds a "hill" of positive potential, a negative charge digs a "well" of negative potential.

Here is where scalars pay off spectacularly. For many charges you do not add vectors — you add numbers. The total potential at a point is just the algebraic sum of each charge's contribution, superposition:

where r_i is the distance from the i-th charge to your point. No angles, no components, no head-to-tail arrows — just add the terms up with their signs. Finding the field the same way would mean resolving every contribution into x and y pieces first. This is the single biggest practical reason physicists reach for potential.

The graph below shows both falls for one positive charge (taking kQ = 1). Slide Q and watch the two curves scale together: the potential V = kQ/r (the gentler 1/r) and the field magnitude E = kQ/r^2 (the steeper 1/r^2).

Equipotential surfaces: contour lines of the electric landscape

On a hiking map, contour lines join points of equal height. Do the same for potential and you get equipotential surfaces — the sets of points that all share one value of V. They have two lovely properties that fall straight out of everything above:

• No work to move along them. Since the potential does not change along an equipotential, \Delta V = 0, and the work W = q\,\Delta V = 0. You can slide a charge around an equipotential for free — like walking a level contour, never climbing or descending.
• Perpendicular to the field everywhere. If the field had any component along the surface it would do work as you moved, contradicting the first point. So \vec{E} must cross every equipotential at a right angle — field lines and equipotentials meet at 90° all over space.

For a single point charge the equipotential V = kQ/r is constant wherever r is constant — i.e. on spheres centred on the charge (in the plane, circles). The field lines shoot radially outward, and they pierce every one of those spheres dead perpendicular. Reveal the figure to see the radial field lines and the ring of equipotentials cross at right angles.

Worked examples

Example 1 — potential from two charges (scalar sum). A charge Q_1 = +3\ \text{nC} sits 2\ \text{m} from a point P, and a charge Q_2 = -5\ \text{nC} sits 1\ \text{m} from the same point. What is the potential at P? Just add the two numbers (with k \approx 9\times 10^{9} and charges in nanocoulombs, each term is 9\,Q/r volts):

No angles, no components — the negative charge is closer and wins, so the net potential is negative. A vector field sum would have taken far more work for the same information.

Example 2 — work to move a charge, W = q\,\Delta V. How much work does the field do dragging a charge q = +2\ \text{C} from a place where V = 8\ \text{V} to a place where V = 3\ \text{V}? The field does work as the charge falls through the potential drop:

The charge moved "downhill" (to lower potential), so the field did positive work — energy the charge can now spend, exactly the energy a battery hands to the electrons it pushes round a circuit. Bringing a charge in from infinity is the same sum with V_{\text{start}} = 0.

Example 3 — recover the field from the potential. Start from the point-charge potential V(r) = kQ/r and take E = -\,dV/dr to get the radial field:

The two minus signs cancel and out pops Coulomb's 1/r^2 field, pointing radially outward for positive Q — precisely down the steepest descent of the potential hill. Differentiating the easy scalar recovers the hard vector: this is the everyday route from a measured voltage map to the field.