UY1: Potential Gradient

Relate electric field to potential gradient with clear sign meaning, component formulas, and derivative-based problem solving.

University Physics Year 1

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Learning objectives

Construct electric-field and potential models for discrete and continuous charge distributions.

This page gives the UY1 working model/result for Potential Gradient. You reuse it when you build fields/potentials by symmetry or superposition, and when you connect fields to forces, energy, and circuits.

Module path: Electromagnetism (UY1)
Practice: UY1 Electromagnetism Quiz
Full routing: UY1 Assessment Map
Math toolkit: Mathematics for Undergraduate Physics

1) At a glance

Prerequisites: potential definition (Electric Potential) and partial derivatives (Partial Derivatives)
Outcomes: compute vector E from a given V(x,y,z), interpret the minus sign physically, and check units/signs quickly
Key result: vector E = -∇ V
Common trap: missing the minus sign or mixing dV/dx (1D) with ∂ V/∂ x (multi-D)
Electric field and potential are linked by:

vector E = -∇ V

Component form:

Eₓ = -(∂ V)/(∂ x), E_y = -(∂ V)/(∂ y), E_z = -(∂ V)/(∂ z)

Minus sign means field points toward decreasing potential.

Motivation / intuition

The gradient points in the direction of steepest increase of V. The electric field points “downhill” (toward decreasing potential), so it is the negative gradient.

2) Setup

Assume electrostatics (time-independent fields from static charges).
Choose coordinate axes and state variable units (V in volts, coordinates in meters).
In 1D along x, relation simplifies to:

Eₓ = -dV/dx

with units V m⁻¹ = N C⁻¹.

3) Core derivation/explanation

Start with potential difference definition:

V_b-Vₐ = -∫ₐ^b vector E · d vector l

In differential form:

dV = - vector E · d vector l

Also,

dV = ((∂ V)/(∂ x))dx + ((∂ V)/(∂ y))dy + ((∂ V)/(∂ z))dz

Comparing terms gives:

vector E = -(i hat (∂ V)/(∂ x) + j hat (∂ V)/(∂ y) + k hat (∂ V)/(∂ z)) = -∇ V

Sign meaning:

If V decreases as x increases, then dV/dx < 0 and Eₓ > 0.
If V increases as x increases, then Eₓ < 0.

Units: ∇ V has units V/m, and V/m = N/C, so it matches electric-field units.
Limits/signs: if V is constant everywhere, then ∇ V = vector 0 so vector E = vector 0; if V decreases with increasing x, then dV/dx < 0 so Eₓ > 0.

4) Worked example(s)

Given potential

V(x,y) = 120-40x + 5y²

(with x, y in m and V in volts), find field at (x,y) = (1,2).

Compute partial derivatives:

(∂ V)/(∂ x) = -40, (∂ V)/(∂ y) = 10y

At y = 2:

(∂ V)/(∂ y) = 20

Therefore:

vector E = -∇ V = 40i hat -20j hat N C⁻¹

Magnitude:

| vector E| = square root of (40² + (-20)²) = 44.7 N C⁻¹

5) Practice set (with hints + answers)

If V(x) = 50-8x, find Eₓ. Hint: Eₓ = -dV/dx. Answer: Eₓ = +8 N C⁻¹.
If V(x) = 3x², find Eₓ at x = 2 m. Hint: differentiate first. Answer: Eₓ = -6x = -12 N C⁻¹.
In a region where V is constant everywhere, what is vector E? Hint: gradient of constant is zero. Answer: vector E = vector 0.

6) Summary + next steps

Potential gradient is the most direct bridge from scalar potential to vector field.
The minus sign is physical: fields point “downhill” in potential.
This relation explains why equipotential surfaces are perpendicular to field lines.

Next: Electric Potential Of A Ring Of Charge Previous: Electric Potential Back To Electromagnetism (UY1)