Electric Field from Potential

Potential came from the field by integration: V_B − V_A = −∫E⋅ds, taken along any path joining the two points. Differentiation undoes it. Move a short distance ds in a chosen direction ŝ; the change in potential is dV = −E⋅ds = −Eₛ ds, where Eₛ is the component of E along ŝ. Divide by ds and Eₛ = −dV/ds. No new physics has entered — it is the same statement read in the other direction, and it is exact rather than approximate, because ds is a differential. The units agree: V m⁻¹ = J C⁻¹ m⁻¹ = N C⁻¹, the unit the field has carried since Coulomb's law. A field of 1 V m⁻¹ means the potential drops by one volt for every metre you advance along the field.

On a graph of V against x the field component is minus the gradient at that point, so what you read is a slope, never a height. Suppose V falls steadily from 60 V at x = 0.10 m to 20 V at x = 0.30 m. The slope is (20 − 60) V ÷ (0.30 − 0.10) m = −200 V m⁻¹, so Eₓ = +200 V m⁻¹: the field points toward +x, which is downhill in potential. Three habits follow. A flat stretch means zero field however high V sits there. A steeper stretch means a stronger field. And the sign takes care of itself — a descending graph gives a positive Eₓ, an ascending one a negative Eₓ — so the field always runs from high potential to low, which is exactly the way a positive charge is pushed.

In two or three dimensions the rule runs once per axis: Eₓ = −∂V/∂x, E_y = −∂V/∂y and E_z = −∂V/∂z, each derivative taken with the other coordinates held fixed. Written compactly, E = −∇V. Take the saddle-shaped map V(x, y) = A x y with A = 250 V m⁻². Then Eₓ = −A y and E_y = −A x, and at the point (0.020 m, 0.040 m) these give Eₓ = −250 × 0.040 = −10.0 V m⁻¹ and E_y = −250 × 0.020 = −5.0 V m⁻¹, a field of magnitude √(10.0² + 5.0²) = 11.2 V m⁻¹ aimed back into the third quadrant. Notice what the arithmetic exposes: Eₓ depends on y. A field component need not vary only with its own coordinate.

Step along an equipotential and V does not change, so dV = 0 and Eₛ = 0 in that direction: the field has no component along an equipotential and must cross it at right angles. Step perpendicular to it instead and you take the fastest change V can offer, which is why E points along the steepest descent with magnitude |dV/dn|, the derivative along that normal. That turns a contour map into a field map by eye. If equipotentials are drawn every 5.0 V and two neighbours lie 2.5 mm apart, the field between them is about 5.0 V ÷ 2.5 × 10⁻³ m = 2.0 × 10³ V m⁻¹, directed from the higher-voltage contour to the lower. Crowded contours mean a strong field, widely spaced ones a weak field, and parallel evenly spaced contours a uniform field.

This is why the detour through potential pays. Adding fields means adding vectors, component by component, for every charge present; adding potentials means adding numbers. So the efficient route to a complicated field is to sum V over all the charges and then differentiate the single answer. Far out along the axis of a dipole of moment p the two potentials almost cancel, leaving V = kp/z². One differentiation gives E_z = −dV/dz = 2kp/z³, the inverse-cube fall-off, with no Coulomb vectors resolved at all. For a water molecule, p = 6.2 × 10⁻³⁰ C m, so at z = 1.0 nm this is V = 56 mV and E_z = 1.1 × 10⁸ V m⁻¹. The same move checks familiar results: V = kq/r differentiates to Eᵣ = kq/r², and anywhere V is constant every derivative vanishes, so E = 0.

A charge q sitting in the potential has energy U(x) = qV(x) and feels Fₓ = qEₓ = −q dV/dx. Equilibrium needs Fₓ = 0, so dV/dx = 0 — a flat point on the V–x graph, whatever value V happens to take there. Stability is settled by the next derivative: the point is stable when q d²V/dx² > 0. A positive charge is therefore trapped by a minimum of V and thrown off a maximum, while a negative charge is trapped by a maximum. Put +4.0 nC at x = −5.0 cm and another at x = +5.0 cm. Midway V = 2kQ/a = 1.44 kV, and along the line the curvature is 4kQ/a³ = 1.15 × 10⁶ V m⁻², positive, so a positive test charge nudged along x is pushed back. Nudge it sideways and it runs away: the midpoint is a valley along the axis and a ridge across it.