Field from Potential & Equipotentials
Integrate E to get V; differentiate V to get E.
Lesson at a glanceShow the key points
- dV = −E⃗·dr⃗ (29.10); V₂ − V₁ = −∫E⃗·dr⃗ (29.11); V(r⃗) = −∫∞ E⃗·dr⃗ (29.12).
- Uniform field between plates: V = Ed.
- Eₓ = −∂V/∂x, E_y = −∂V/∂y, E_z = −∂V/∂z (29.13); −dV/dr = E cos θ (29.14).
- The field points where V falls fastest; V is constant perpendicular to E.
- Equipotential surfaces are perpendicular to field lines (spheres around a point charge).
Field and potential are two descriptions of one thing. Knowing either, you can find the other: integrate the field to get potential, differentiate the potential to get the field.
From field to potential
Displace a test charge q by dr⃗. The field does work qE⃗·dr⃗, so U changes by −qE⃗·dr⃗ and V by:
A uniform field of 5000 V/m points from plate B to plate A, 2.0 cm apart. Find V_B − V_A.
From potential to field
Writing dV = −Eₓdx − E_y dy − E_z dz and changing one coordinate at a time:
More generally, for a step dr at angle θ to the field, dV = −E dr cos θ, so
- −dV/dr gives the component of E in the direction you step.
- Stepping along the field (θ = 0), V falls fastest: the field points the way the potential decreases most steeply.
- Stepping at right angles to the field, V doesn’t change.
In a region V = (40 V/m²)x² − (30 V/m)y. Find the field at the point (1.0 m, 2.0 m).
Equipotential surfaces
A surface on which the potential is the same everywhere is an equipotential surface. Since V doesn’t change along it, the field has no component along it: the field is perpendicular to equipotentials everywhere. For a point charge they are concentric spheres.
- 1dV = −E⃗·dr⃗; V₂ − V₁ = −∫E⃗·dr⃗.
- 2Eₓ = −∂V/∂x etc.; −dV/dr is the field component along dr.
- 3The field points the way V falls fastest; uniform field between plates: V = Ed.
- 4Equipotentials are perpendicular to field lines.
Check yourself
Keep exploring
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