Physics · Ch 2 — Electrostatic Potential and Capacitance
Relation Between Electric Field Intensity and Potential
Relation Between Electric Field Intensity and Potential
Because electric field and electric potential describe exactly the same electrostatic field, one as a vector and the other as a scalar, an exact mathematical relation must connect the two -- and it takes the form of a derivative in one direction and an integral in the other.
From potential to field (differentiation). Consider a small displacement along the direction in which the field points, moving a test charge through it. The work done by the field itself on the charge over this small step is ; the work an EXTERNAL agent must supply to move the charge quasi-statically (balancing the field's force at every instant) is therefore . Dividing by , this external work per unit charge is, by definition, the change in potential over that step, :
So the component of the electric field along any direction equals minus the rate at which potential falls off in that direction. Two consequences follow immediately: first, the electric field points in the direction of the STEEPEST DECREASE of potential (never toward increasing potential, for a positive test charge to be pushed the way it actually is); second, wherever the field is strong, potential must be changing rapidly with position, and wherever the field is weak (or zero), potential must be changing slowly (or be constant) -- exactly the picture equipotential surfaces make visual in Section 2.7.
From field to potential (integration). Running the same relation the other way, the potential difference between two points and is recovered by integrating the field along any path joining them:
Special case: a uniform field. Between the plates of a parallel plate capacitor (Section 2.13), the field is uniform and directed straight from the positive plate to the negative plate. Integrating the relation above along a straight path of length between the plates, in the direction of , gives simply …