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Potential Gradients

Level 1 - Physics topic page in Electric Potential.

Principle

The electric field points in the direction where electric potential decreases fastest, with magnitude set by the potential gradient.

Notation

\(\vec E\)
electric field
\(\mathrm{N\,C^{-1}}\)
\(V\)
electric potential
\(\mathrm{V}\)
\(x,y,z\)
Cartesian coordinates
\(\mathrm{m}\)
\(\nabla V\)
gradient of electric potential
\(\mathrm{V\,m^{-1}}\)
\(\Delta s\)
small displacement along a chosen direction
\(\mathrm{m}\)
\(E_s\)
field component along that direction
\(\mathrm{N\,C^{-1}}\)

Method

Derivation 1: One-dimensional relation

Move a positive test charge a small distance \(dx\) along the \(x\)-axis. Electric work and potential change describe the same energy transfer.

Small electric work
\[dW=qE_x\,dx\]
Potential energy change
\[dU=q\,dV\]
Conservative relation
\[dU=-dW\]
Cancel charge
\[q\,dV=-qE_x\,dx\]
Field component
\[E_x=-\frac{dV}{dx}\]

Derivation 2: Three-dimensional gradient

In three dimensions, the same argument applies to each coordinate direction. Collecting the components gives the gradient relation.

Component relations
\[E_x=-\frac{\partial V}{\partial x},\quad E_y=-\frac{\partial V}{\partial y},\quad E_z=-\frac{\partial V}{\partial z}\]
Gradient
\[\nabla V=\frac{\partial V}{\partial x}\hat\imath+\frac{\partial V}{\partial y}\hat\jmath+\frac{\partial V}{\partial z}\hat k\]
Field from potential
\[\vec E=-\nabla V\]

Derivation 3: Uniform field as a limiting case

If the electric field is constant along one direction, potential changes linearly with distance.

Constant component
\[E_s=\mathrm{constant}\]
Integrate
\[\Delta V=-\int E_s\,ds\]
Uniform field
\[\Delta V=-E_s\Delta s\]

Rules

These are the compact gradient relations.

One dimension
\[E_x=-\frac{dV}{dx}\]
Gradient relation
\[\vec E=-\nabla V\]
Uniform field
\[\Delta V=-E_s\Delta s\]
Field unit
\[1\,\mathrm{V\,m^{-1}}=1\,\mathrm{N\,C^{-1}}\]

Examples

Question
Potential decreases by
\[12\,\mathrm{V}\]
over
\[0.30\,\mathrm{m}\]
in the
\[+x\]
direction. Find \(E_x\).
Answer
\[E_x=-\frac{\Delta V}{\Delta x}=-\frac{-12}{0.30}=40\,\mathrm{V\,m^{-1}}\]
The field points in the
\[+x\]
direction.

Checks

  • The minus sign means \(\vec E\) points downhill in potential.
  • A flat potential region has zero electric field.
  • A steep potential graph means a large field magnitude.
  • Use the field component along the displacement when applying \(\Delta V=-E_s\Delta s\).

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